IB DP Maths Topic 5.6 Poisson distribution, its mean and variance HL Paper 2

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Question

The lifts in the office buildings of a small city have occasional breakdowns. The breakdowns at any given time are independent of one another and can be modelled using a Poisson Distribution with mean 0.2 per day.

(a) Determine the probability that there will be exactly four breakdowns during the month of June (June has 30 days).

(b) Determine the probability that there are more than 3 breakdowns during the month of June.

(d) Find the probability that the first breakdown in June occurs on June ${3^{{\text{rd}}}}$.

(e) It costs 1850 Euros to service the lifts when they have breakdowns. Find the expected cost of servicing lifts for the month of June.

(f) Determine the probability that there will be no breakdowns in exactly 4 out of the first 5 days in June.

Answer/Explanation

Markscheme

(a) mean for 30 days: $30 \times 0.2 = 6$ . (A1)

${\text{P}}(X = 4) = \frac{{{6^4}}}{{4!}}{{\text{e}}^{ – 6}} = 0.134$ (M1)A1 N3

[3 marks]

(b) ${\text{P}}(X > 3) = 1 – {\text{P}}(X \leqslant 3) = 1 – {{\text{e}}^{ – 6}}(1 + 6 + 18 + 36) = 0.849$ (M1)A1 N2

[2 marks]

mean for five days: $5 \times 0.2 = 1$ (A1)

${\text{P}}(X = 0) = {{\text{e}}^{ – 1}}\,\,\,\,\,( = 0.368)$ A1 N2

mean for one day: 0.2 (A1)

${\text{P}}(X = 0) = {({{\text{e}}^{ – 0.2}})^5} = {{\text{e}}^{ – 1}}\,\,\,\,\,( = 0.368)$ A1 N2

[2 marks]

(d) Required probability $ = {{\text{e}}^{ – 0.2}} \times {{\text{e}}^{ – 0.2}} \times (1 – {{\text{e}}^{ – 0.2}})$ M1A1

= 0.122 A1 N3

[3 marks]

(e) Expected cost is $1850 \times 6 = {\text{11}}\,{\text{100 Euros}}$ A1

[1 mark]

(f) On any one day ${\text{P}}(X = 0) = {{\text{e}}^{ – 0.2}}$

Therefore, $\left( {\begin{array}{*{20}{c}}
5 \\
1
\end{array}} \right){({{\text{e}}^{ – 0.2}})^4}(1 – {{\text{e}}^{ – 0.2}}) = 0.407$ M1A1 N2

[2 marks]

Total [13 marks]

Examiners report

Many candidates showed familiarity with the Poisson Distribution. Parts (a), (b), and (c) were straightforward, as long as candidates multiplied 0.2 by 30 to get the mean. Part (e) was answered successfully by most candidates. Parts (d) and (f) were done very poorly. In part (d), most candidates calculated ${\text{P}}(X = 1)$ rather than ${\text{P}}(X \leqslant 1)$. Although some candidates realized the need for the Binomial in part (e), some incorrectly used 0.8 and 0.2.

Question

The distance travelled by students to attend Gauss College is modelled by a normal distribution with mean 6 km and standard deviation 1.5 km.

(i) Find the probability that the distance travelled to Gauss College by a randomly selected student is between 4.8 km and 7.5 km.

(ii) 15 % of students travel less than d km to attend Gauss College. Find the value of d.

[7]

At Euler College, the distance travelled by students to attend their school is modelled by a normal distribution with mean $\mu $ km and standard deviation $\sigma $ km.

If 10 % of students travel more than 8 km and 5 % of students travel less than 2 km, find the value of $\mu $ and of $\sigma $ .

[6]

The number of telephone calls, T, received by Euler College each minute can be modelled by a Poisson distribution with a mean of 3.5.

(i) Find the probability that at least three telephone calls are received by Euler College in each of two successive one-minute intervals.

(ii) Find the probability that Euler College receives 15 telephone calls during a randomly selected five-minute interval.

[8]

Answer/Explanation

Markscheme

(i) ${\text{P}}(4.8 < X < 7.5) = {\text{P}}( – 0.8 < Z < 1)$ (M1)

= 0.629 A1 N2

Note: Accept ${\text{P}}(4.8 \leqslant X \leqslant 7.5) = {\text{P}}( – 0.8 \leqslant Z \leqslant 1)$ .

(ii) Stating ${\text{P}}(X < d) = 0.15$ or sketching an appropriately labelled diagram. A1

$\frac{{d – 6}}{{1.5}} = – 1.0364…$ (M1)(A1)

d = (−1.0364…)(1.5) + 6 (M1)

= 4.45 (km) A1 N4

[7 marks]

Stating both ${\text{P}}(X > 8) = 0.1$ and ${\text{P}}(X < 2) = 0.05$ or sketching an appropriately labelled diagram. R1

Setting up two equations in $\mu $ and $\sigma $ (M1)

8 = $\mu $ + (1.281…)$\sigma $ and 2 = $\mu $ − (1.644…)$\sigma $ A1

Attempting to solve for $\mu $ and $\sigma $ (including by graphical means) (M1)

$\sigma $ = 2.05 (km) and $\mu $ = 5.37 (km) A1A1 N4

Note: Accept $\mu $ = 5.36, 5.38 .

[6 marks]

(i) Use of the Poisson distribution in an inequality. M1

${\text{P}}(T \geqslant 3) = 1 – {\text{P}}(T \leqslant 2)$ (A1)

= 0.679… A1

Required probability is ${(0.679…)^2} = 0.461$ M1A1 N3

Note: Allow FT for their value of ${\text{P}}(T \geqslant 3)$ .

(ii) $\tau \sim {\text{Po(17.5)}}$ A1

${\text{P}}(\tau = 15) = \frac{{{{\text{e}}^{ – 17.5}}{{(17.5)}^{15}}}}{{15!}}$ (M1)

= 0.0849 A1 N2

[8 marks]

Examiners report

This question was generally well done despite a large proportion of candidates being awarded an accuracy penalty. Candidates found part (a) (i) to be quite straightforward and was generally done very well. In part (a) (ii), a number of candidates used $\frac{{d – 6}}{{1.5}} = 1.0364…$ instead of $\frac{{d – 6}}{{1.5}} = – 1.0364…$ . In part (b), a pleasingly high number of candidates were able to set up and solve a pair of simultaneous linear equations to correctly find the values of $\mu $ and $\sigma $. Some candidates prematurely rounded intermediate results. In part (c), a number of candidates were unable to express a correct Poisson inequality. Common errors included stating ${\text{P}}(T \geqslant 3) = 1 – {\text{P}}(T \leqslant 3)$ and using $\mu = 7$.

Question

(a) Ahmed is typing Section A of a mathematics examination paper. The number of mistakes that he makes, X , can be modelled by a Poisson distribution with mean 3.2 . Find the probability that Ahmed makes exactly four mistakes.

(b) His colleague, Levi, is typing Section B of this paper. The number of mistakes that he makes, Y , can be modelled by a Poisson distribution with mean m.

(i) If ${\text{E}}({Y^2}) = 5.5$ , find the value of m.

(ii) Find the probability that Levi makes exactly three mistakes.

(c) Given that X and Y are independent, find the probability that Ahmed makes exactly four mistakes and Levi makes exactly three mistakes.

Answer/Explanation

Markscheme

(a) $X \sim {\text{Po(3.2)}}$

${\text{P}}(X = 4) = \frac{{{{\text{e}}^{ – 3.2}}{{3.2}^4}}}{{4!}}$

= 0.178 A1

(b) (i) ${\text{Var}}(Y) = {\text{E}}({Y^2}) – {{\text{E}}^2}(Y)$ (M1)

$m = 5.5 – {m^2}$ A1

m = 1.90 (or m = –2.90 which is valid) A1

(ii) $Y \sim {\text{Po(1.90)}}$

${\text{P}}(Y = 3) = \frac{{{{\text{e}}^{ – 1.90}}{{1.90}^4}}}{{3!}}$ (M1)

= 0.171 A1

[8 marks]

Examiners report

Part (a) was correctly solved by most candidates, either using the formula or directly from their GDC. Solutions to (b), however, were extremely disappointing with the majority of candidates giving $\sqrt 5 $, incorrectly, as their value of m . It was possible to apply follow through in (b) (ii) and (c) which were well done in general.

Question

Mr Lee is planning to go fishing this weekend. Assuming that the number of fish caught per hour follows a Poisson distribution with mean $0.6$, find

(a) the probability that he catches at least one fish in the first hour;

(b) the probability that he catches exactly three fish if he fishes for four hours;

(c) the number of complete hours that Mr Lee needs to fish so that the probability of catching more than two fish exceeds 80 %.

Answer/Explanation

Markscheme

(a) $X{\text{ ~ Po(0}}{\text{.6)}}$

${\text{P}}(X \geqslant 1) = 1 – {\text{P}}(X = 0)$ M1

$ = 0.451$ A1 N1

(b) $Y{\text{ \~ Po(2}}{\text{.4)}}$ (M1)

${\text{P}}(Y = 3) = 0.209$ A1

${\text{P}}(Z \geqslant 3) = 1 – {\text{P}}(Z \leqslant 2) > 0.8$ (M1)

Note: Only one of these M1 marks may be implied.

$n \geqslant 7.132…$ (hours)

so, Mr Lee needs to fish for at least $8$ complete hours A1 N2

Note: Accept a shown trial and error method that leads to a correct solution.

[7 marks]

Examiners report

It was clear that many students had not been taught the topic and were consequently unable to make an attempt at the question. Of those students who were able to start, common errors were in a misunderstanding of the language. Many had difficulties in part (c) and “at least” in part (a) was sometimes misinterpreted.

Question

Testing has shown that the volume of drink in a bottle of mineral water filled by Machine A at a bottling plant is normally distributed with a mean of $998$ ml and a standard deviation of $2.5$ ml.

(a) Show that the probability that a randomly selected bottle filled by Machine A contains more than $1000$ ml of mineral water is $0.212$.

(b) A random sample of $5$ bottles is taken from Machine A. Find the probability that exactly $3$ of them each contain more than $1000$ ml of mineral water.

(c) Find the minimum number of bottles that would need to be sampled to ensure that the probability of getting at least one bottle filled by Machine A containing more than $1000$ ml of mineral water, is greater than $0.99$.

(d) It has been found that for Machine B the probability of a bottle containing less than $996$ ml of mineral water is $0.1151$. The probability of a bottle containing more than $1000$ ml is $0.3446$. Find the mean and standard deviation for the volume of mineral water contained in bottles filled by Machine B.

(e) The company that makes the mineral water receives, on average, m phone calls every $10$ minutes. The number of phone calls, $X$ , follows a Poisson distribution such that ${\text{P}}(X = 2) = {\text{P}}(X = 3) + {\text{P}}(X = 4)$ .

(i) Find the value of $m$ .

(ii) Find the probability that the company receives more than two telephone calls in a randomly selected $10$ minute period.

Answer/Explanation

Markscheme

(a) $X \sim {\text{N}}$($998$, ${2.5^2}$ ) M1

${\text{P}}(X > 1000) = 0.212$ AG

[1 mark]

(b) $X \sim {\text{B}}$($5$, $0.2119…$)

evidence of binomial (M1)

${\text{P}}(X = 3) = \left( {\begin{array}{*{20}{c}}
5 \\
3
\end{array}} \right){\left( {0.2119…} \right)^3}{\left( {0.7881…} \right)^2} = 0.0591$ (accept $0.0592$) (M1)A1

[3 marks]

$1 – {\left( {0.7881…} \right)^n} > 0.99$

${\left( {0.7881…} \right)^n} < 0.01$ A1

Note: Award A1 for line 2 or line 3 or equivalent.

$n > 19.3$ (A1)
minimum number of bottles required is $20$ A1N2

[4 marks]

(d) $\frac{{996 – \mu }}{\sigma } = – 1.1998$ (accept $1.2$) M1A1

$\frac{{1000 – \mu }}{\sigma } = 0.3999$ (accept $0.4$) M1A1

$\mu = 999$(ml), $\sigma = 2.50$(ml) A1A1

[6 marks]

(e) (i) $\frac{{{{\text{e}}^{ – m}}{m^2}}}{{2!}} = \frac{{{{\text{e}}^{ – m}}{m^3}}}{{3!}} + \frac{{{{\text{e}}^{ – m}}{m^4}}}{{4!}}$ M1A1

$\frac{{{m^2}}}{2} = \frac{{{m^3}}}{6} + \frac{{{m^4}}}{{24}}$

$12{m^2} – 4{m^3} – {m^4} = 0$ (A1)

$m = – 6$, $0$, $2$

$ \Rightarrow m = 2$ A1N2

(ii) ${\text{P}}(X > 2) = 1 – {\text{P}}(X \leqslant 2)$ (M1)

$ = 1 – {\text{P}}(X = 0) – {\text{P}}(X = 1) – {\text{P}}(X = 2)$

$ = 1 – {{\text{e}}^{ – 2}} – 2{{\text{e}}^{ – 2}} – \frac{{{2^2}{{\text{e}}^{ – 2}}}}{{2!}}$

$ = 0.323$ A1

[6 marks]

Total [20 marks]

Examiners report

This was the best done of the section B questions, with the majority of candidates making the correct choice of probability distribution for each part. The main sources of errors: (b) missing out the binomial coefficient in the calculation; (c) failure to rearrange ‘at least one bottle’ in terms of the probability of obtaining no bottles; (d) using $1.2$ rather than $ – 1.2$ in the inverse Normal or not performing an inverse Normal at all; (e)(ii) misinterpreting ‘more than two’.

Question

Casualties arrive at an accident unit with a mean rate of one every 10 minutes. Assume that the number of arrivals can be modelled by a Poisson distribution.

(a) Find the probability that there are no arrivals in a given half hour period.

(b) A nurse works for a two hour period. Find the probability that there are fewer than ten casualties during this period.

(c) Six nurses work consecutive two hour periods between 8am and 8pm. Find the probability that no more than three nurses have to attend to less than ten casualties during their working period.

(d) Calculate the time interval during which there is a 95 % chance of there being at least two casualties.

Answer/Explanation

Markscheme

Note: Accept exact answers in parts (a) to (c).

(a) number of patients in 30 minute period = X (A1)

$X \sim {\text{Po(3)}}$ (M1)A1

[3 marks]

(b) number of patients in working period = Y (A1)

$Y \sim {\text{Po(12)}}$ (M1)A1

[3 marks]

$W \sim {\text{B}}(6,{\text{ }}0.2424 \ldots )$ (M1)A1

[4 marks]

(d) number of patients in t minute interval = X

$X \sim {\text{Po}}(T)$

${\text{P}}(X \geqslant 2) = 0.95$

${\text{P}}(X = 0) + {\text{P}}(X = 1) = 0.05$ (M1)(A1)

${{\text{e}}^{ – T}}(1 + T) = 0.05$ (M1)

$T = 4.74$ (A1)

t = 47.4 minutes A1

[5 marks]

Total [15 marks]

Examiners report

Parts (a) and (b) were well answered, but many students were unable to recognise the Binomial distribution in part (c) and were unable to form the correct equation in part (d). There were many accuracy errors in this question.

Question

The random variable X follows a Poisson distribution with mean $\lambda $.

(a) Find $\lambda $ if ${\text{P}}(X = 0) + {\text{P}}(X = 1) = 0.123$.

(b) With this value of $\lambda $, find ${\text{P}}(0 < X < 9)$.

Answer/Explanation

Markscheme

(a) required to solve ${{\text{e}}^{ – \lambda }} + \lambda {{\text{e}}^{ – \lambda }} = 0.123$ M1A1

solving to obtain $\lambda = 3.63$ A2 N2

Note: Award A2 if an additional negative solution is seen but A0 if only a negative solution is seen.

(b) ${\text{P}}(0 < X < 9)$

$ = {\text{P}}(X \leqslant 8) – {\text{P}}(X = 0)$ (or equivalent) (M1)

$ = 0.961$ A1

[6 marks]

Examiners report

Part (a) – Well done by most, although there were some answers that ignored the requirement of mathematical notation.

Part (b) – Not successfully answered by many. The main problem was not correctly interpreting the inequalities in the probability.

Question

The random variable X follows a Poisson distribution with mean m and satisfies

\[{\text{P}}(X = 1) + {\text{P}}(X = 3) = {\text{P}}(X = 0) + {\text{P}}(X = 2).\]

(a) Find the value of m correct to four decimal places.

(b) For this value of m, calculate ${\text{P}}(1 \leqslant X \leqslant 2)$.

Answer/Explanation

Markscheme

(a) ${\text{P}}(X = 1) + {\text{P}}(X = 3) = {\text{P}}(X = 0) + {\text{P}}(X = 2)$

$m{{\text{e}}^{ – m}} + \frac{{{m^3}{{\text{e}}^{ – m}}}}{6} = {{\text{e}}^{ – m}} + \frac{{{m^2}{{\text{e}}^{ – m}}}}{2}$ (M1)(A1)

${m^3} – 3{m^2} + 6m – 6 = 0$ (M1)

$m = 1.5961$ (4 decimal places) A1

(b) $m = 1.5961 \Rightarrow {\text{P}}(1 \leqslant X \leqslant 2) = m{{\text{e}}^{ – m}} + \frac{{{m^2}{{\text{e}}^{ – m}}}}{2} = 0.582$ (M1)A1

[6 marks]

Examiners report

Most candidates correctly stated the required equation for m. However, many algebraic errors in the simplification of this equation led to incorrect answers. Also, many candidates failed to find the value of m to the required accuracy, with many candidates giving answers correct to 4 sf instead of 4 dp. In part (b) many candidates did not realize that they needed to calculate ${\text{P}}(X = 1) + {\text{P}}(X = 2)$ and many attempts to calculate other combinations of probabilities were seen.

Question

A student arrives at a school $X$ minutes after 08:00, where X may be assumed to be normally distributed. On a particular day it is observed that 40% of the students arrive before 08:30 and 90% arrive before 08:55.

Consider the function $f(x) = \frac{{\ln x}}{x}$ , $0 < x < {{\text{e}}^2}$ .

Find the mean and standard deviation of $X$.

[5]

The school has 1200 students and classes start at 09:00. Estimate the number of students who will be late on that day.

[3]

Maelis had not arrived by 08:30. Find the probability that she arrived late.

[2]

At 15:00 it is the end of the school day and it is assumed that the departure of the students from school can be modelled by a Poisson distribution. On average 24 students leave the school every minute.

Find the probability that at least 700 students leave school before 15:30.

[3]

There are 200 days in a school year. Given that $Y$ denotes the number of days in the year that at least 700 students leave before 15:30, find

(i) ${\text{E}}(Y)$ ;

(ii) $P(Y > 150)$ .

[4]

Answer/Explanation

Markscheme

${\text{P}}(X < 30) = 0.4$

${\text{P}}(X < 55) = 0.9$

or relevant sketch (M1)

given $Z = \frac{{X – \mu }}{\sigma }$

${\text{P}}(Z < z) = 0.4 \Rightarrow \frac{{30 – \mu }}{\sigma } = – 0.253…$ (A1)

${\text{P}}(Z < z) = 0.9 \Rightarrow \frac{{55 – \mu }}{\sigma } = 1.28…$ (A1)

$\mu = 30 + \left( {0.253…} \right) \times \sigma = 55 – \left( {1.28…} \right) \times \sigma $ M1

$\sigma = 16.3$ , $\mu = 34.1$ A1

Note: Accept 16 and 34.

Note: Working with 830 and 855 will only gain the two M marks.

[5 marks]

$X{\text{ ~ N}}$($34.12…$, $16.28…{^2}$ )

late to school $ \Rightarrow X > 60$

${\text{P}}(X > 60) = 0.056…$ (A1)

number of students late ${\text{ = 0}}{\text{.0560}}… \times {\text{1200}}$ (M1)

$= 67$ (to nearest integer) A1

Note: Accept $62$ for use of $34$ and $16$.

[3 marks]

${\text{P}}(X > 60|X > 30) = \frac{{{\text{P}}(X > 60)}}{{{\text{P}}(X > 30)}}$ M1

$ = 0.0935$ (accept anything between $0.093$ and $0.094$) A1

Note: If $34$ and $16$ are used $0.0870$ is obtained. This should be accepted.

[2 marks]

let $L$ be the random variable of the number of students who leave school in a 30 minute interval

since $24 \times 30 = 720$ A1

${\text{L ~ Po(720)}}$

${\text{P(}}L \geqslant 700) = 1 – {\text{P}}(L \leqslant 699)$ (M1)

$ = 0.777$ A1

Note: Award M1A0 for $P(L > 700) = 1 – P(L \leqslant 700)$ (this leads to $0.765$).

[3 marks]

(i) $Y{\text{ ~ B}}$($200$, $0.7767…$) (M1)

${\text{E}}(Y) = 200 \times 0.7767… = 155$ A1

Note: On ft, use of $0.765$ will lead to $153$.

(ii) ${\text{P}}(Y > 150) = 1 – {\text{P}}(Y \leqslant 150)$ (M1)

$ = 0.797$ A1

Note: Accept $0.799$ from using rounded answer.

Note: On ft, use of $0.765$ will lead to $0.666$.

[4 marks]

Examiners report

Candidates who had been prepared to solve questions from this part of the syllabus did well on the question. As a general point, candidates did not always write down clearly which distribution was being used. There were many candidates who seemed unfamiliar with the concept of Normal distributions as well as the Poisson and Binomial distributions and did not attempt the question. Parts (a) – (c) of the question were a variation on similar problems seen on previous examinations and there were a disappointing number of candidates who seemed unable to start the question. The use of 830 and 850 rather than minutes after 8am was seen and this caused students to lose marks despite knowing the method required. In general technology was used well and this was seen in (d) when solving a problem that involved a Poisson distribution. A number of candidates were unable to identify the Binomial distribution in (e).

Question

The number of accidents that occur at a large factory can be modelled by a Poisson distribution with a mean of 0.5 accidents per month.

Find the probability that no accidents occur in a given month.

[1]

Find the probability that no accidents occur in a given 6 month period.

[2]

Find the length of time, in complete months, for which the probability that at least 1 accident occurs is greater than 0.99.

[6]

To encourage safety the factory pays a bonus of $1000 into a fund for workers if no accidents occur in any given month, a bonus of $500 if 1 or 2 accidents occur and no bonus if more than 2 accidents occur in the month.

(i) Calculate the expected amount that the company will pay in bonuses each month.

(ii) Find the probability that in a given 3 month period the company pays a total of exactly $2000 in bonuses.

[9]

Answer/Explanation

Markscheme

P(x = 0) = 0.607 A1

[1 mark]

EITHER

Using $X \sim {\text{Po}}(3)$ (M1)

Using ${(0.6065…)^6}$ (M1)

THEN

P(X = 0) = 0.0498 A1

[2 marks]

$X \sim {\text{Po}}(0.5t)$ (M1)

${\text{P}}(x \geqslant 1) = 1 – {\text{P}}(x = 0)$ (M1)

${\text{P}}(x = 0) < 0.01$ A1

${{\text{e}}^{ – 0.5t}} < 0.01$ A1

$ – 0.5t < \ln (0.01)$ (M1)

$t > 9.21{\text{ months}}$

therefore 10 months A1N4

Note: Full marks can be awarded for answers obtained directly from GDC if a systematic method is used and clearly shown.

[6 marks]

(i) P(1 or 2 accidents) = 0.37908… A1

${\text{E}}(B) = 1000 \times 0.60653… + 500 \times 0.37908…$ M1A1

$ = \$ 796\,\,\,\,\,$(accept $797 or $796.07) A1

(ii) P(2000) = P(1000, 1000, 0) + P(1000, 0, 1000) + P(0, 1000, 1000) +

P(1000, 500, 500) + P(500, 1000, 500) + P(500, 500, 1000) (M1)(A1)

Note: Award M1 for noting that 2000 can be written both as $2 \times 1000 + 1 \times 0$ and $2 \times 500 + 1 \times 1000$ .

$ = 3{(0.6065…)^2}(0.01437…) + 3{(0.3790…)^2}(0.6065…)$ M1A1

$ = 0.277\,\,\,\,\,$(accept 0.278) A1

[9 marks]

Examiners report

Most candidates successfully answered (a) and (b). Although many found the correct answer to (c), communication of their reasoning was weak. This was also true for (d)(i). Answers to (d)(ii) were mostly scrappy and rarely worthy of credit.

Question

The number of accidents that occur at a large factory can be modelled by a Poisson distribution with a mean of 0.5 accidents per month.

Find the probability that no accidents occur in a given month.

[1]

Find the probability that no accidents occur in a given 6 month period.

[2]

Find the length of time, in complete months, for which the probability that at least 1 accident occurs is greater than 0.99.

[6]

(i) Calculate the expected amount that the company will pay in bonuses each month.

(ii) Find the probability that in a given 3 month period the company pays a total of exactly $2000 in bonuses.

[9]

Answer/Explanation

Markscheme

P(x = 0) = 0.607 A1

[1 mark]

EITHER

Using $X \sim {\text{Po}}(3)$ (M1)

Using ${(0.6065…)^6}$ (M1)

THEN

P(X = 0) = 0.0498 A1

[2 marks]

$X \sim {\text{Po}}(0.5t)$ (M1)

${\text{P}}(x \geqslant 1) = 1 – {\text{P}}(x = 0)$ (M1)

${\text{P}}(x = 0) < 0.01$ A1

${{\text{e}}^{ – 0.5t}} < 0.01$ A1

$ – 0.5t < \ln (0.01)$ (M1)

$t > 9.21{\text{ months}}$

therefore 10 months A1N4

Note: Full marks can be awarded for answers obtained directly from GDC if a systematic method is used and clearly shown.

[6 marks]

(i) P(1 or 2 accidents) = 0.37908… A1

${\text{E}}(B) = 1000 \times 0.60653… + 500 \times 0.37908…$ M1A1

$ = \$ 796\,\,\,\,\,$(accept $797 or $796.07) A1

(ii) P(2000) = P(1000, 1000, 0) + P(1000, 0, 1000) + P(0, 1000, 1000) +

P(1000, 500, 500) + P(500, 1000, 500) + P(500, 500, 1000) (M1)(A1)

Note: Award M1 for noting that 2000 can be written both as $2 \times 1000 + 1 \times 0$ and $2 \times 500 + 1 \times 1000$ .

$ = 3{(0.6065…)^2}(0.01437…) + 3{(0.3790…)^2}(0.6065…)$ M1A1

$ = 0.277\,\,\,\,\,$(accept 0.278) A1

[9 marks]

Examiners report

Question

The number of accidents that occur at a large factory can be modelled by a Poisson distribution with a mean of 0.5 accidents per month.

Find the probability that no accidents occur in a given month.

[1]

Find the probability that no accidents occur in a given 6 month period.

[2]

Find the length of time, in complete months, for which the probability that at least 1 accident occurs is greater than 0.99.

[6]

(i) Calculate the expected amount that the company will pay in bonuses each month.

(ii) Find the probability that in a given 3 month period the company pays a total of exactly $2000 in bonuses.

[9]

Answer/Explanation

Markscheme

P(x = 0) = 0.607 A1

[1 mark]

EITHER

Using $X \sim {\text{Po}}(3)$ (M1)

Using ${(0.6065…)^6}$ (M1)

THEN

P(X = 0) = 0.0498 A1

[2 marks]

$X \sim {\text{Po}}(0.5t)$ (M1)

${\text{P}}(x \geqslant 1) = 1 – {\text{P}}(x = 0)$ (M1)

${\text{P}}(x = 0) < 0.01$ A1

${{\text{e}}^{ – 0.5t}} < 0.01$ A1

$ – 0.5t < \ln (0.01)$ (M1)

$t > 9.21{\text{ months}}$

therefore 10 months A1N4

Note: Full marks can be awarded for answers obtained directly from GDC if a systematic method is used and clearly shown.

[6 marks]

(i) P(1 or 2 accidents) = 0.37908… A1

${\text{E}}(B) = 1000 \times 0.60653… + 500 \times 0.37908…$ M1A1

$ = \$ 796\,\,\,\,\,$(accept $797 or $796.07) A1

(ii) P(2000) = P(1000, 1000, 0) + P(1000, 0, 1000) + P(0, 1000, 1000) +

P(1000, 500, 500) + P(500, 1000, 500) + P(500, 500, 1000) (M1)(A1)

Note: Award M1 for noting that 2000 can be written both as $2 \times 1000 + 1 \times 0$ and $2 \times 500 + 1 \times 1000$ .

$ = 3{(0.6065…)^2}(0.01437…) + 3{(0.3790…)^2}(0.6065…)$ M1A1

$ = 0.277\,\,\,\,\,$(accept 0.278) A1

[9 marks]

Examiners report

Question

The number of vehicles passing a particular junction can be modelled using the Poisson distribution. Vehicles pass the junction at an average rate of 300 per hour.

Find the probability that no vehicles pass in a given minute.

[2]

Find the expected number of vehicles which pass in a given two minute period.

[1]

Find the probability that more than this expected number actually pass in a given two minute period.

[2]

Answer/Explanation

Markscheme

$m = \frac{{300}}{{60}} = 5$ (A1)

${\text{P}}(X = 0) = 0.00674$ A1

or ${{\text{e}}^{ – 5}}$

[2 marks]

${\text{E}}(X) = 5 \times 2 = 10$ A1

[1 mark]

${\text{P}}(X > 10) = 1 – {\text{P}}(X \leqslant 10)$ (M1)

= 0.417 A1

[2 marks]

Examiners report

Parts (a) and (b) were answered successfully by many candidates. Some candidates had difficulty obtaining the correct inequality in (c).

Question

The number of vehicles passing a particular junction can be modelled using the Poisson distribution. Vehicles pass the junction at an average rate of 300 per hour.

Find the probability that no vehicles pass in a given minute.

[2]

Find the expected number of vehicles which pass in a given two minute period.

[1]

Find the probability that more than this expected number actually pass in a given two minute period.

[2]

Answer/Explanation

Markscheme

$m = \frac{{300}}{{60}} = 5$ (A1)

${\text{P}}(X = 0) = 0.00674$ A1

or ${{\text{e}}^{ – 5}}$

[2 marks]

${\text{E}}(X) = 5 \times 2 = 10$ A1

[1 mark]

${\text{P}}(X > 10) = 1 – {\text{P}}(X \leqslant 10)$ (M1)

= 0.417 A1

[2 marks]

Examiners report

Parts (a) and (b) were answered successfully by many candidates. Some candidates had difficulty obtaining the correct inequality in (c).

Question

The number of vehicles passing a particular junction can be modelled using the Poisson distribution. Vehicles pass the junction at an average rate of 300 per hour.

Find the probability that no vehicles pass in a given minute.

[2]

Find the expected number of vehicles which pass in a given two minute period.

[1]

Find the probability that more than this expected number actually pass in a given two minute period.

[2]

Answer/Explanation

Markscheme

$m = \frac{{300}}{{60}} = 5$ (A1)

${\text{P}}(X = 0) = 0.00674$ A1

or ${{\text{e}}^{ – 5}}$

[2 marks]

${\text{E}}(X) = 5 \times 2 = 10$ A1

[1 mark]

${\text{P}}(X > 10) = 1 – {\text{P}}(X \leqslant 10)$ (M1)

= 0.417 A1

[2 marks]

Examiners report

Parts (a) and (b) were answered successfully by many candidates. Some candidates had difficulty obtaining the correct inequality in (c).

Question

A fisherman notices that in any hour of fishing, he is equally likely to catch exactly two fish, as he is to catch less than two fish. Assuming the number of fish caught can be modelled by a Poisson distribution, calculate the expected value of the number of fish caught when he spends four hours fishing.

Answer/Explanation

Markscheme

$X \sim {\text{Po}}(m)$

${\text{P}}(X = 2) = {\text{P}}(X < 2)$ (M1)

$\frac{1}{2}{m^2}{{\text{e}}^{ – m}} = {{\text{e}}^{ – m}}(1 + m)$ (A1)(A1)

$m = 2.73 {\text{ }}\left( {1 + \sqrt 3 } \right)$ A1

in four hours the expected value is 10.9$\,\,\,\,\left( {4 + 4\sqrt 3 } \right)$ A1

Note: Value of m does not need to be rounded.

[5 marks]

Examiners report

Many candidates did not attempt this question and many others did not go beyond setting the equation up. Among the ones who attempted to solve the equation, once again, very few candidates took real advantage of GDC use to obtain the correct answer.

Question

A ski resort finds that the mean number of accidents on any given weekday (Monday to Friday) is 2.2 . The number of accidents can be modelled by a Poisson distribution.

Find the probability that in a certain week (Monday to Friday only)

(i) there are fewer than 12 accidents;

(ii) there are more than 8 accidents, given that there are fewer than 12 accidents.

[6]

a(i)(ii).

Due to the increased usage, it is found that the probability of more than 3 accidents in a day at the weekend (Saturday and Sunday) is 0.24.

Assuming a Poisson model,

(i) calculate the mean number of accidents per day at the weekend (Saturday and Sunday);

(ii) calculate the probability that, in the four weekends in February, there will be more than 5 accidents during at least two of the weekends.

[10]

b(i)(ii).

It is found that 20 % of skiers having accidents are at least 25 years of age and 40 % are under 18 years of age.

Assuming that the ages of skiers having accidents are normally distributed, find the mean age of skiers having accidents.

[6]

Answer/Explanation

Markscheme

(i) $X \sim {\text{Po}}(11)$ (M1)

${\text{P}}(X \leqslant 11) = 0.579$ (M1)A1

(ii) ${\text{P}}(X > 8\left| {x < 12) = } \right.$ (M1)

$ = \frac{{{\text{P}}(8 < X < 12)}}{{{\text{P}}(X < 12)}}{\text{ }}\left( {{\text{or }}\frac{{{\text{P}}(X \leqslant 11) – {\text{P}}(X \leqslant 8)}}{{{\text{P}}(X \leqslant 11)}}{\text{ or }}\frac{{0.3472…}}{{0.5792…}}} \right)$ A1

$ = 0.600$ A1 N2

[6 marks]

a(i)(ii).

(i) $Y \sim {\text{Po}}(m)$

${\text{P}}(Y > 3) = 0.24$ (M1)

${\text{P}}(Y \leqslant 3) = 0.76$ (M1)

${{\text{e}}^{ – m}}\left( {1 + m + \frac{1}{2}{m^2} + \frac{1}{6}{m^3}} \right) = 0.76$ (A1)

Note: At most two of the above lines can be implied.

Attempt to solve equation with GDC (M1)

$m = 2.49$ A1

(ii) $A \sim {\text{Po}}(4.98)$

${\text{P}}(A > 5) = 1 – {\text{P}}(A \leqslant 5) = 0.380…$ M1A1

$W \sim {\text{B}}(4,\,0.380…)$ (M1)

${\text{P}}(W \geqslant 2) = 1 – {\text{P}}(W \leqslant 1) = 0.490$ M1A1

[10 marks]

b(i)(ii).

${\text{P}}(A < 25) = 0.8,{\text{ P}}(A < 18) = 0.4$

$\frac{{25 – \mu }}{\sigma } = 0.8416…$ (M1)(A1)

$\frac{{18 – \mu }}{\sigma } = -0.2533…{\text{ (or}} -0.2534{\text{ from tables)}}$ (M1)(A1)

solving these equations (M1)

$\mu = 19.6$ A1

Note: Accept just 19.6, 19 or 20; award A0 to any other final answer.

[6 marks]

Examiners report

Generally, candidates had difficulties with this question, mainly in applying conditional probability and interpreting the expressions ‘more than’, ‘at least’ and ‘under’ to obtain correct expressions. Although many candidates identified the binomial distribution in part (b) (ii), very few succeeded in answering this question due to incorrect interpretation of the question or due to accuracy errors.

a(i)(ii).

b(i)(ii).

Question

A ski resort finds that the mean number of accidents on any given weekday (Monday to Friday) is 2.2 . The number of accidents can be modelled by a Poisson distribution.

Find the probability that in a certain week (Monday to Friday only)

(i) there are fewer than 12 accidents;

(ii) there are more than 8 accidents, given that there are fewer than 12 accidents.

[6]

a(i)(ii).

Due to the increased usage, it is found that the probability of more than 3 accidents in a day at the weekend (Saturday and Sunday) is 0.24.

Assuming a Poisson model,

(i) calculate the mean number of accidents per day at the weekend (Saturday and Sunday);

(ii) calculate the probability that, in the four weekends in February, there will be more than 5 accidents during at least two of the weekends.

[10]

b(i)(ii).

It is found that 20 % of skiers having accidents are at least 25 years of age and 40 % are under 18 years of age.

Assuming that the ages of skiers having accidents are normally distributed, find the mean age of skiers having accidents.

[6]

Answer/Explanation

Markscheme

(i) $X \sim {\text{Po}}(11)$ (M1)

${\text{P}}(X \leqslant 11) = 0.579$ (M1)A1

(ii) ${\text{P}}(X > 8\left| {x < 12) = } \right.$ (M1)

$ = 0.600$ A1 N2

[6 marks]

a(i)(ii).

(i) $Y \sim {\text{Po}}(m)$

${\text{P}}(Y > 3) = 0.24$ (M1)

${\text{P}}(Y \leqslant 3) = 0.76$ (M1)

${{\text{e}}^{ – m}}\left( {1 + m + \frac{1}{2}{m^2} + \frac{1}{6}{m^3}} \right) = 0.76$ (A1)

Note: At most two of the above lines can be implied.

Attempt to solve equation with GDC (M1)

$m = 2.49$ A1

(ii) $A \sim {\text{Po}}(4.98)$

${\text{P}}(A > 5) = 1 – {\text{P}}(A \leqslant 5) = 0.380…$ M1A1

$W \sim {\text{B}}(4,\,0.380…)$ (M1)

${\text{P}}(W \geqslant 2) = 1 – {\text{P}}(W \leqslant 1) = 0.490$ M1A1

[10 marks]

b(i)(ii).

${\text{P}}(A < 25) = 0.8,{\text{ P}}(A < 18) = 0.4$

$\frac{{25 – \mu }}{\sigma } = 0.8416…$ (M1)(A1)

$\frac{{18 – \mu }}{\sigma } = -0.2533…{\text{ (or}} -0.2534{\text{ from tables)}}$ (M1)(A1)

solving these equations (M1)

$\mu = 19.6$ A1

Note: Accept just 19.6, 19 or 20; award A0 to any other final answer.

[6 marks]

Examiners report

a(i)(ii).

b(i)(ii).

Question

The random variable X has the distribution ${\text{Po}}(m)$ .

Given that ${\text{P}}(X = 5) = {\text{P}}(X = 3) + {\text{P}}(X = 4)$, find

the value of m ;

[3]

P (X > 2) .

[2]

Answer/Explanation

Markscheme

${\text{P}}(X = 5) = {\text{P}}(X = 3) + {\text{P}}(X = 4)$

$\frac{{{{\text{e}}^{ – m}}{m^5}}}{{5!}} = \frac{{{{\text{e}}^{ – m}}{m^3}}}{{3!}} + \frac{{{{\text{e}}^{ – m}}{m^4}}}{{4!}}$ M1(A1)

${m^2} – 5m – 20 = 0$

$ \Rightarrow m = \frac{{5 + \sqrt {105} }}{2} = (7.62)$ A1

[3 marks]

${\text{P}}(X > 2) = 1 – {\text{P}}(X \leqslant 2)$ (M1)

$ = 1 – 0.018…$

$ = 0.982$ A1

[2 marks]

Examiners report

Again this proved to be a successful question for many candidates with a good proportion of wholly correct answers seen. It was good to see students making good use of the calculator.

Question

The random variable X has the distribution ${\text{Po}}(m)$ .

Given that ${\text{P}}(X = 5) = {\text{P}}(X = 3) + {\text{P}}(X = 4)$, find

the value of m ;

[3]

P (X > 2) .

[2]

Answer/Explanation

Markscheme

${\text{P}}(X = 5) = {\text{P}}(X = 3) + {\text{P}}(X = 4)$

$\frac{{{{\text{e}}^{ – m}}{m^5}}}{{5!}} = \frac{{{{\text{e}}^{ – m}}{m^3}}}{{3!}} + \frac{{{{\text{e}}^{ – m}}{m^4}}}{{4!}}$ M1(A1)

${m^2} – 5m – 20 = 0$

$ \Rightarrow m = \frac{{5 + \sqrt {105} }}{2} = (7.62)$ A1

[3 marks]

${\text{P}}(X > 2) = 1 – {\text{P}}(X \leqslant 2)$ (M1)

$ = 1 – 0.018…$

$ = 0.982$ A1

[2 marks]

Examiners report

Again this proved to be a successful question for many candidates with a good proportion of wholly correct answers seen. It was good to see students making good use of the calculator.

Question

The number of visitors that arrive at a museum every minute can be modelled by a Poisson distribution with mean 2.2.

If the museum is open 6 hours daily, find the expected number of visitors in 1 day.

[2]

Find the probability that the number of visitors arriving during an hour exceeds 100.

[3]

Find the probability that the number of visitors in each of the 6 hours the museum is open exceeds 100.

[2]

The ages of the visitors to the museum can be modelled by a normal distribution with mean $\mu $ and variance ${\sigma ^2}$ . The records show that 29 % of the visitors are under 35 years of age and 23 % are at least 55 years of age.

Find the values of $\mu $ and $\sigma $ .

[6]

One day, 100 visitors under 35 years of age come to the museum. Estimate the number of visitors under 50 years of age that were at the museum on that day.

[5]

Answer/Explanation

Markscheme

$2.2 \times 6 \times 60 = 792$ (M1)A1

[2 marks]

$V \sim {\text{Po}}(2.2 \times 60)$ (M1)

${\text{P}}(V > 100) = 0.998$ (M1)A1

[3 marks]

${(0.997801…)^6} = 0.987$ (M1)A1

[2 marks]

$A \sim {\text{N}}(\mu ,{\text{ }}{\sigma ^2})$

${\text{P}}(A < 35) = 0.29{\text{ and P}}(A > 55) = 0.23 \Rightarrow {\text{P}}(A < 55) = 0.77$

${\text{P}}\left( {Z < \frac{{35 – \mu }}{\sigma }} \right) = 0.29{\text{ and P}}\left( {Z < \frac{{55 – \mu }}{\sigma }} \right) = 0.77$ (M1)

use of inverse normal (M1)

$\frac{{35 – \mu }}{\sigma } = – 0.55338…{\text{ and }}\frac{{55 – \mu }}{\sigma } = 0.738846…$ (A1)

solving simultaneously (M1)

$\mu = 43.564…{\text{ and }}\sigma = 15.477…$ A1A1

$\mu = 43.6{\text{ and }}\sigma = 15.5{\text{ (3sf)}}$

[6 marks]

$0.29n = 100 \Rightarrow n = 344.82…$ (M1)(A1)

${\text{P}}(A < 50) = 0.66121…$ (A1)

expected number of visitors under 50 = 228 (M1)A1

[5 marks]

Examiners report

This question was generally well done by most candidates. It was evident that candidates had been well prepared in Poisson and normal distribution. In parts (a)-(d) candidates were usually successful and appropriate methods were shown although many candidates used labored algebraic approaches to solving simultaneous equations and wasted time answering part (d). Part (e) was very well answered by a smaller number of candidates but it was obviously more demanding in its level of abstraction.

Question

The number of visitors that arrive at a museum every minute can be modelled by a Poisson distribution with mean 2.2.

If the museum is open 6 hours daily, find the expected number of visitors in 1 day.

[2]

Find the probability that the number of visitors arriving during an hour exceeds 100.

[3]

Find the probability that the number of visitors in each of the 6 hours the museum is open exceeds 100.

[2]

Find the values of $\mu $ and $\sigma $ .

[6]

One day, 100 visitors under 35 years of age come to the museum. Estimate the number of visitors under 50 years of age that were at the museum on that day.

[5]

Answer/Explanation

Markscheme

$2.2 \times 6 \times 60 = 792$ (M1)A1

[2 marks]

$V \sim {\text{Po}}(2.2 \times 60)$ (M1)

${\text{P}}(V > 100) = 0.998$ (M1)A1

[3 marks]

${(0.997801…)^6} = 0.987$ (M1)A1

[2 marks]

$A \sim {\text{N}}(\mu ,{\text{ }}{\sigma ^2})$

${\text{P}}(A < 35) = 0.29{\text{ and P}}(A > 55) = 0.23 \Rightarrow {\text{P}}(A < 55) = 0.77$

${\text{P}}\left( {Z < \frac{{35 – \mu }}{\sigma }} \right) = 0.29{\text{ and P}}\left( {Z < \frac{{55 – \mu }}{\sigma }} \right) = 0.77$ (M1)

use of inverse normal (M1)

$\frac{{35 – \mu }}{\sigma } = – 0.55338…{\text{ and }}\frac{{55 – \mu }}{\sigma } = 0.738846…$ (A1)

solving simultaneously (M1)

$\mu = 43.564…{\text{ and }}\sigma = 15.477…$ A1A1

$\mu = 43.6{\text{ and }}\sigma = 15.5{\text{ (3sf)}}$

[6 marks]

$0.29n = 100 \Rightarrow n = 344.82…$ (M1)(A1)

${\text{P}}(A < 50) = 0.66121…$ (A1)

expected number of visitors under 50 = 228 (M1)A1

[5 marks]

Examiners report

Question

The number of visitors that arrive at a museum every minute can be modelled by a Poisson distribution with mean 2.2.

If the museum is open 6 hours daily, find the expected number of visitors in 1 day.

[2]

Find the probability that the number of visitors arriving during an hour exceeds 100.

[3]

Find the probability that the number of visitors in each of the 6 hours the museum is open exceeds 100.

[2]

Find the values of $\mu $ and $\sigma $ .

[6]

One day, 100 visitors under 35 years of age come to the museum. Estimate the number of visitors under 50 years of age that were at the museum on that day.

[5]

Answer/Explanation

Markscheme

$2.2 \times 6 \times 60 = 792$ (M1)A1

[2 marks]

$V \sim {\text{Po}}(2.2 \times 60)$ (M1)

${\text{P}}(V > 100) = 0.998$ (M1)A1

[3 marks]

${(0.997801…)^6} = 0.987$ (M1)A1

[2 marks]

$A \sim {\text{N}}(\mu ,{\text{ }}{\sigma ^2})$

${\text{P}}(A < 35) = 0.29{\text{ and P}}(A > 55) = 0.23 \Rightarrow {\text{P}}(A < 55) = 0.77$

${\text{P}}\left( {Z < \frac{{35 – \mu }}{\sigma }} \right) = 0.29{\text{ and P}}\left( {Z < \frac{{55 – \mu }}{\sigma }} \right) = 0.77$ (M1)

use of inverse normal (M1)

$\frac{{35 – \mu }}{\sigma } = – 0.55338…{\text{ and }}\frac{{55 – \mu }}{\sigma } = 0.738846…$ (A1)

solving simultaneously (M1)

$\mu = 43.564…{\text{ and }}\sigma = 15.477…$ A1A1

$\mu = 43.6{\text{ and }}\sigma = 15.5{\text{ (3sf)}}$

[6 marks]

$0.29n = 100 \Rightarrow n = 344.82…$ (M1)(A1)

${\text{P}}(A < 50) = 0.66121…$ (A1)

expected number of visitors under 50 = 228 (M1)A1

[5 marks]

Examiners report

Question

A ferry carries cars across a river. There is a fixed time of T minutes between crossings. The arrival of cars at the crossing can be assumed to follow a Poisson distribution with a mean of one car every four minutes. Let X denote the number of cars that arrive in T minutes.

Find T, to the nearest minute, if ${\text{P}}(X \leqslant 3) = 0.6$.

[3]

It is now decided that the time between crossings, T, will be 10 minutes. The ferry can carry a maximum of three cars on each trip.

One day all the cars waiting at 13:00 get on the ferry. Find the probability that all the cars that arrive in the next 20 minutes will get on either the 13:10 or the 13:20 ferry.

[4]

Answer/Explanation

Markscheme

$X \sim {\text{Po(0.25T)}}$ (A1)

Attempt to solve ${\text{P}}(X \leqslant 3) = 0.6$ (M1)

$T = 12.8453 \ldots = 13{\text{ (minutes)}}$ A1

Note: Award A1M1A0 if T found correctly but not stated to the nearest minute.

[3 marks]

let ${X_1}$ be the number of cars that arrive during the first interval and ${X_2}$ be the number arriving during the second.

${X_1}$ and ${X_2}$ are Po(2.5) (A1)

P (all get on) $ = {\text{P}}({X_1} \leqslant 3) \times {\text{P}}({X_2} \leqslant 3) + {\text{P}}({X_1} = 4) \times {\text{P}}({X_2} \leqslant 2)$

$ + {\text{P}}({X_1} = 5) \times {\text{P}}({X_2} \leqslant 1) + {\text{P}}({X_1} = 6) \times {\text{P}}({X_2} = 0)$ (M1)

$ = 0.573922 \ldots + 0.072654 \ldots + 0.019192 \ldots + 0.002285 \ldots $ (M1)

$ = 0.668{\text{ }}(053 \ldots )$ A1

[4 marks]

Examiners report

There were some good answers to part (a), although poor calculator use frequently let down the candidates.

Very few candidates were able to access part (b).

Question

Find T, to the nearest minute, if ${\text{P}}(X \leqslant 3) = 0.6$.

[3]

It is now decided that the time between crossings, T, will be 10 minutes. The ferry can carry a maximum of three cars on each trip.

One day all the cars waiting at 13:00 get on the ferry. Find the probability that all the cars that arrive in the next 20 minutes will get on either the 13:10 or the 13:20 ferry.

[4]

Answer/Explanation

Markscheme

$X \sim {\text{Po(0.25T)}}$ (A1)

Attempt to solve ${\text{P}}(X \leqslant 3) = 0.6$ (M1)

$T = 12.8453 \ldots = 13{\text{ (minutes)}}$ A1

Note: Award A1M1A0 if T found correctly but not stated to the nearest minute.

[3 marks]

let ${X_1}$ be the number of cars that arrive during the first interval and ${X_2}$ be the number arriving during the second.

${X_1}$ and ${X_2}$ are Po(2.5) (A1)

P (all get on) $ = {\text{P}}({X_1} \leqslant 3) \times {\text{P}}({X_2} \leqslant 3) + {\text{P}}({X_1} = 4) \times {\text{P}}({X_2} \leqslant 2)$

$ + {\text{P}}({X_1} = 5) \times {\text{P}}({X_2} \leqslant 1) + {\text{P}}({X_1} = 6) \times {\text{P}}({X_2} = 0)$ (M1)

$ = 0.573922 \ldots + 0.072654 \ldots + 0.019192 \ldots + 0.002285 \ldots $ (M1)

$ = 0.668{\text{ }}(053 \ldots )$ A1

[4 marks]

Examiners report

There were some good answers to part (a), although poor calculator use frequently let down the candidates.

Very few candidates were able to access part (b).

Question

A small car hire company has two cars. Each car can be hired for one whole day at a time. The rental charge is US$60 per car per day. The number of requests to hire a car for one whole day may be modelled by a Poisson distribution with mean 1.2.

Find the probability that on a particular weekend, three requests are received on Saturday and none are received on Sunday.

[2]

Over a weekend of two days, it is given that a total of three requests are received.

Find the expected total rental income for the weekend.

[5]

Answer/Explanation

Markscheme

$X \sim {\text{Po}}(1.2)$

${\text{P}}(X = 3) \times {\text{P}}(X = 0)$ (M1)

$ = 0.0867 \ldots \times 0.3011 \ldots $

$ = 0.0261$ A1

[2 marks]

Three requests over two days can occur as (3, 0), (0, 3), (2, 1) or (1, 2). R1

using conditional probability, for example

$\frac{{{\text{P}}(3,{\text{ }}0)}}{{{\text{P}}(3{\text{ requests, }}m = 2.4)}} = 0.125{\text{ or }}\frac{{{\text{P}}(2,{\text{ }}1)}}{{{\text{P}}(3{\text{ requests, }}m = 2.4)}} = 0.375$ M1A1

expected income is

$2 \times 0.125 \times {\text{US}}\$ 120 + 2 \times 0.375 \times {\text{US}}\$ 180$ M1

Note: Award M1 for attempting to find the expected income including both (3, 0) and (2, 1) cases.

$ = {\text{US}}\$ 30 + {\text{US}}\$ 135$

$ = {\text{US}}\$ 165$ A1

[5 marks]

Examiners report

Part (a) was generally well done although a number of candidates added the two probabilities rather than multiplying the two probabilities. A number of candidates specified the required probability correct to two significant figures only.

Part (b) challenged most candidates with only a few candidates able to correctly employ a conditional probability argument.

Question

The number of cats visiting Helena’s garden each week follows a Poisson distribution with mean $\lambda = 0.6$.

Find the probability that

(i) in a particular week no cats will visit Helena’s garden;

(ii) in a particular week at least three cats will visit Helena’s garden;

(iii) over a four-week period no more than five cats in total will visit Helena’s garden;

(iv) over a twelve-week period there will be exactly four weeks in which at least one cat will visit Helena’s garden.

[9]

A continuous random variable $X$ has probability distribution function $f$ given by

$f(x) = k\ln x$ $1 \leqslant x \leqslant 3$

$f(x) = 0$ otherwise

(i) Find the value of $k$ to six decimal places.

(ii) Find the value of ${\text{E}}(X)$.

(iii) State the mode of $X$.

(iv) Find the median of $X$.

[9]

Answer/Explanation

Markscheme

(i) $X \sim {\text{Po(0.6)}}$

${\text{P}}(X = 0) = 0.549{\text{ }}\left( { = {{\text{e}}^{ – 0.6}}} \right)$ A1

(ii) ${\text{P}}(X \geqslant 3) = 1 – {\text{P}}(X \leqslant 2)$ (M1)(A1)

$ = 1 – \left( {{{\text{e}}^{ – 0.6}} + {{\text{e}}^{ – 0.6}} \times 0.6 + {{\text{e}}^{ – 0.6}} \times \frac{{{{0.6}^2}}}{2}} \right)$

$ = 0.0231$ A1

(iii) $Y \sim {\text{Po(2.4)}}$ (M1)

${\text{P}}(Y \leqslant 5) = 0.964$ A1

(iv) $Z \sim {\text{B(12, 0.451}} \ldots )$ (M1)(A1)

Note: Award M1 for recognising binomial and A1 for using correct parameters.

${\text{P}}(Z = 4) = 0.169$ A1

[9 marks]

(i) $k\int_1^3 {\ln x{\text{d}}x = 1} $ (M1)

$(k \times 1.2958 \ldots = 1)$

$k = 0.771702$ A1

(ii) ${\text{E}}(X) = \int_1^3 {kx\ln x{\text{d}}x} $ (A1)

attempting to evaluate their integral (M1)

$ = 2.27$ A1

(iii) $x = 3$ A1

(iv) $\int_1^m {k\ln x{\text{d}}x = 0.5} $ (M1)

$k[x\ln x – x]_1^m = 0.5$

attempting to solve for m (M1)

$m = 2.34$ A1

[9 marks]

Examiners report

Parts (a) and (b) were generally well done by a large proportion of candidates. In part (a) (ii), some candidates used an incorrect inequality (e.g. ${\text{P}}(X \geqslant 3) = 1 – {\text{P}}(X \leqslant 3)$) while in (a) (iii) some candidates did not use $\mu = 2.4$. In part (a) (iv), a number of candidates either did not realise that they needed to consider a binomial random variable or did so using incorrect parameters.

Parts (a) and (b) were generally well done by a large proportion of candidates.

In (b) (i), some candidates gave their value of k correct to three significant figures rather than correct to six decimal places. In parts (b) (i), (ii) and (iv), a large number of candidates unnecessarily used integration by parts. In part (b) (iii), a number of candidates thought the mode of X was $f(3)$ rather than $x = 3$. In part (b) (iv), a number of candidates did not consider the domain of f when attempting to find the median or checking their solution.

Question

The number of birds seen on a power line on any day can be modelled by a Poisson distribution with mean 5.84.

Find the probability that during a certain seven-day week, more than 40 birds have been seen on the power line.

[2]

On Monday there were more than 10 birds seen on the power line. Show that the probability of there being more than 40 birds seen on the power line from that Monday to the following Sunday, inclusive, can be expressed as:

$\frac{{{\text{P}}(X > 40) + \sum\limits_{r = 11}^{40} {{\text{P}}(X = r){\text{P}}(Y > 40 – r)} }}{{{\text{P}}(X > 10)}}$ where $X \sim {\text{Po}}(5.84)$ and $Y \sim {\text{Po}}(35.04)$.

[5]

Answer/Explanation

Markscheme

mean for week is 40.88 (A1)

${\text{P}}(S > 40) = 1 – {\text{P}}(S \leqslant 40) = 0.513$ A1

[2 marks]

$\frac{{{\text{probability there were more than 10 on Monday AND more than 40 over the week}}}}{{{\text{probability there were more than 10 on Monday}}}}$ M1

possibilities for the numerator are:

there were more than 40 birds on the power line on Monday R1

11 on Monday and more than 29 over the course of the next 6 days R1

12 on Monday and more than 28 over the course of the next 6 days … until

40 on Monday and more than 0 over the course of the next 6 days R1

hence if X is the number on the power line on Monday and Y, the number on the power line Tuesday – Sunday then the numerator is M1

${\text{P}}(X > 40) + {\text{P}}(X = 11) \times {\text{P}}(Y > 29) + {\text{P}}(X = 12) \times {\text{P}}(Y > 28) + \ldots $

$ + {\text{P}}(X = 40) \times {\text{P}}(Y > 0)$

$ = {\text{P}}(X > 40) + \sum\limits_{r = 11}^{40} {{\text{P}}(X = r){\text{P}}(Y > 40 – r)} $

hence solution is $\frac{{{\text{P}}(X > 40) + \sum\limits_{r = 11}^{40} {{\text{P}}(X = r){\text{P}}(Y > 40 – r)} }}{{{\text{P}}(X > 10)}}$ AG

[5 marks]

Examiners report

[N/A]

Question

The random variable $X$ has a Poisson distribution with mean $\mu $.

Given that ${\text{P}}(X = 2) + {\text{P}}(X = 3) = {\text{P}}(X = 5)$,

(a) find the value of $\mu $;

(b) find the probability that X lies within one standard deviation of the mean.

Answer/Explanation

Markscheme

(a) $\frac{{{\mu ^2}{{\text{e}}^{ – \mu }}}}{{2!}} + \frac{{{\mu ^3}{{\text{e}}^{ – \mu }}}}{{3!}} = \frac{{{\mu ^5}{{\text{e}}^{ – \mu }}}}{{5!}}$ (M1)

$\frac{{{\mu ^2}}}{2} + \frac{{{\mu ^3}}}{6} – \frac{{{\mu ^5}}}{{120}} = 0$

$\mu = 5.55$ A1

[2 marks]

(b) $\sigma = \sqrt {5.55 \ldots } = 2.35598 \ldots $ (M1)

${\text{P}}(3.19 \leqslant X \leqslant 7.9)$

${\text{P}}(4 \leqslant X \leqslant 7)$

$ = 0.607$ A1

[2 marks]

Total [4 marks]

Examiners report

[N/A]

Question

The number of complaints per day received by customer service at a department store follows a Poisson distribution with a mean of $0.6$.

On a randomly chosen day, find the probability that

(i) there are no complaints;

(ii) there are at least three complaints.

[3]

In a randomly chosen five-day week, find the probability that there are no complaints.

[2]

On a randomly chosen day, find the most likely number of complaints received.

Justify your answer.

[3]

The department store introduces a new policy to improve customer service. The number of complaints received per day now follows a Poisson distribution with mean $\lambda $.

On a randomly chosen day, the probability that there are no complaints is now $0.8$.

Find the value of $\lambda $.

[2]

Answer/Explanation

Markscheme

(i) $P(X = 0) = 0.549{\text{ }}( = {{\text{e}}^{ – 0.6}})$ A1

(ii) $P(X \ge 3) = 1 – P(X \le 2)$ (M1)

$P(X \ge 3) = 0.0231$ A1

[3 marks]

EITHER

using $Y \sim {\text{Po(3)}}$ (M1)

using ${(0.549)^5}$ (M1)

THEN

${\text{P}}(Y = 0) = 0.0498{\text{ }}\left( { = {{\text{e}}^{ – 3}}} \right)$ A1

[2 marks]

${\text{P}}(X = 0)$ (most likely number of complaints received is zero) A1

EITHER

calculating ${\text{P}}(X = 0) = 0.549$ and ${\text{P}}(X = 1) = 0.329$ M1A1

sketching an appropriate (discrete) graph of ${\text{P}}(X = x)$ against $x$ M1A1

finding ${\text{P}}(X = 0) = {e^{ – 0.6}}$ and stating that ${\text{P}}(X = 0) > 0.5$ M1A1

using ${\text{P}}(X = x) = {\text{P}}(X = x – 1) \times \frac{\mu }{x}$ where $\mu < 1$ M1A1

[3 marks]

${\text{P}}(X = 0) = 0.8{\text{ (}} \Rightarrow {e^{ – \lambda }} = 0.8)$ (A1)

$\lambda = 0.223\left( { = \ln \frac{5}{4}, = – \ln \frac{4}{5}} \right)$ A1

[2 marks]

Total [10 marks]

Examiners report

Parts (a), (b) and (d) were generally well done. In (a) (ii), some candidates calculated $1 – {\text{P}}(X \le 3)$.

Parts (a), (b) and (d) were generally well done.

A number of candidates offered clear and well-reasoned solutions to part (c). The two most common successful approaches used to justify that the most likely number of complaints received is zero were either to calculate ${\text{P}}(X = x)$ for $x = 0,{\text{ }}1,{\text{ }} \ldots $ or find that ${\text{P}}(X = 0) = 0.549{\text{ }}( > 0.5)$. A number of candidates stated that the most number of complaints received was the mean of the distribution $(\lambda = 0.6)$.

Parts (a), (b) and (d) were generally well done.

Question

The number of complaints per day received by customer service at a department store follows a Poisson distribution with a mean of $0.6$.

On a randomly chosen day, find the probability that

(i) there are no complaints;

(ii) there are at least three complaints.

[3]

In a randomly chosen five-day week, find the probability that there are no complaints.

[2]

On a randomly chosen day, find the most likely number of complaints received.

Justify your answer.

[3]

The department store introduces a new policy to improve customer service. The number of complaints received per day now follows a Poisson distribution with mean $\lambda $.

On a randomly chosen day, the probability that there are no complaints is now $0.8$.

Find the value of $\lambda $.

[2]

Answer/Explanation

Markscheme

(i) $P(X = 0) = 0.549{\text{ }}( = {{\text{e}}^{ – 0.6}})$ A1

(ii) $P(X \ge 3) = 1 – P(X \le 2)$ (M1)

$P(X \ge 3) = 0.0231$ A1

[3 marks]

EITHER

using $Y \sim {\text{Po(3)}}$ (M1)

using ${(0.549)^5}$ (M1)

THEN

${\text{P}}(Y = 0) = 0.0498{\text{ }}\left( { = {{\text{e}}^{ – 3}}} \right)$ A1

[2 marks]

${\text{P}}(X = 0)$ (most likely number of complaints received is zero) A1

EITHER

calculating ${\text{P}}(X = 0) = 0.549$ and ${\text{P}}(X = 1) = 0.329$ M1A1

sketching an appropriate (discrete) graph of ${\text{P}}(X = x)$ against $x$ M1A1

finding ${\text{P}}(X = 0) = {e^{ – 0.6}}$ and stating that ${\text{P}}(X = 0) > 0.5$ M1A1

using ${\text{P}}(X = x) = {\text{P}}(X = x – 1) \times \frac{\mu }{x}$ where $\mu < 1$ M1A1

[3 marks]

${\text{P}}(X = 0) = 0.8{\text{ (}} \Rightarrow {e^{ – \lambda }} = 0.8)$ (A1)

$\lambda = 0.223\left( { = \ln \frac{5}{4}, = – \ln \frac{4}{5}} \right)$ A1

[2 marks]

Total [10 marks]

Examiners report

Parts (a), (b) and (d) were generally well done. In (a) (ii), some candidates calculated $1 – {\text{P}}(X \le 3)$.

Parts (a), (b) and (d) were generally well done.

Question

The number of complaints per day received by customer service at a department store follows a Poisson distribution with a mean of $0.6$.

On a randomly chosen day, find the probability that

(i) there are no complaints;

(ii) there are at least three complaints.

[3]

In a randomly chosen five-day week, find the probability that there are no complaints.

[2]

On a randomly chosen day, find the most likely number of complaints received.

Justify your answer.

[3]

The department store introduces a new policy to improve customer service. The number of complaints received per day now follows a Poisson distribution with mean $\lambda $.

On a randomly chosen day, the probability that there are no complaints is now $0.8$.

Find the value of $\lambda $.

[2]

Answer/Explanation

Markscheme

(i) $P(X = 0) = 0.549{\text{ }}( = {{\text{e}}^{ – 0.6}})$ A1

(ii) $P(X \ge 3) = 1 – P(X \le 2)$ (M1)

$P(X \ge 3) = 0.0231$ A1

[3 marks]

EITHER

using $Y \sim {\text{Po(3)}}$ (M1)

using ${(0.549)^5}$ (M1)

THEN

${\text{P}}(Y = 0) = 0.0498{\text{ }}\left( { = {{\text{e}}^{ – 3}}} \right)$ A1

[2 marks]

${\text{P}}(X = 0)$ (most likely number of complaints received is zero) A1

EITHER

calculating ${\text{P}}(X = 0) = 0.549$ and ${\text{P}}(X = 1) = 0.329$ M1A1

sketching an appropriate (discrete) graph of ${\text{P}}(X = x)$ against $x$ M1A1

finding ${\text{P}}(X = 0) = {e^{ – 0.6}}$ and stating that ${\text{P}}(X = 0) > 0.5$ M1A1

using ${\text{P}}(X = x) = {\text{P}}(X = x – 1) \times \frac{\mu }{x}$ where $\mu < 1$ M1A1

[3 marks]

${\text{P}}(X = 0) = 0.8{\text{ (}} \Rightarrow {e^{ – \lambda }} = 0.8)$ (A1)

$\lambda = 0.223\left( { = \ln \frac{5}{4}, = – \ln \frac{4}{5}} \right)$ A1

[2 marks]

Total [10 marks]

Examiners report

Parts (a), (b) and (d) were generally well done. In (a) (ii), some candidates calculated $1 – {\text{P}}(X \le 3)$.

Parts (a), (b) and (d) were generally well done.

Question

The number of complaints per day received by customer service at a department store follows a Poisson distribution with a mean of $0.6$.

On a randomly chosen day, find the probability that

(i) there are no complaints;

(ii) there are at least three complaints.

[3]

In a randomly chosen five-day week, find the probability that there are no complaints.

[2]

On a randomly chosen day, find the most likely number of complaints received.

Justify your answer.

[3]

The department store introduces a new policy to improve customer service. The number of complaints received per day now follows a Poisson distribution with mean $\lambda $.

On a randomly chosen day, the probability that there are no complaints is now $0.8$.

Find the value of $\lambda $.

[2]

Answer/Explanation

Markscheme

(i) $P(X = 0) = 0.549{\text{ }}( = {{\text{e}}^{ – 0.6}})$ A1

(ii) $P(X \ge 3) = 1 – P(X \le 2)$ (M1)

$P(X \ge 3) = 0.0231$ A1

[3 marks]

EITHER

using $Y \sim {\text{Po(3)}}$ (M1)

using ${(0.549)^5}$ (M1)

THEN

${\text{P}}(Y = 0) = 0.0498{\text{ }}\left( { = {{\text{e}}^{ – 3}}} \right)$ A1

[2 marks]

${\text{P}}(X = 0)$ (most likely number of complaints received is zero) A1

EITHER

calculating ${\text{P}}(X = 0) = 0.549$ and ${\text{P}}(X = 1) = 0.329$ M1A1

sketching an appropriate (discrete) graph of ${\text{P}}(X = x)$ against $x$ M1A1

finding ${\text{P}}(X = 0) = {e^{ – 0.6}}$ and stating that ${\text{P}}(X = 0) > 0.5$ M1A1

using ${\text{P}}(X = x) = {\text{P}}(X = x – 1) \times \frac{\mu }{x}$ where $\mu < 1$ M1A1

[3 marks]

${\text{P}}(X = 0) = 0.8{\text{ (}} \Rightarrow {e^{ – \lambda }} = 0.8)$ (A1)

$\lambda = 0.223\left( { = \ln \frac{5}{4}, = – \ln \frac{4}{5}} \right)$ A1

[2 marks]

Total [10 marks]

Examiners report

Parts (a), (b) and (d) were generally well done. In (a) (ii), some candidates calculated $1 – {\text{P}}(X \le 3)$.

Parts (a), (b) and (d) were generally well done.

Question

The random variable $X$ follows a Poisson distribution with mean $m \ne 0$.

Given that $2{\text{P}}(X = 4) = {\text{P}}(X = 5)$, show that $m = 10$.

[3]

Given that $X \le 11$, find the probability that $X = 6$.

[4]

Answer/Explanation

Markscheme

$2\frac{{{{\text{e}}^{ – m}}{m^4}}}{{4!}} = \frac{{{{\text{e}}^{ – m}}{m^5}}}{{5!}}$ M1A1

$\frac{2}{{4!}} = \frac{m}{{5!}}\;\;\;$or other simplification M1

Note: accept a labelled graph showing clearly the solution to the equation. Do not accept simple verification that $m = 10$ is a solution.

$ \Rightarrow m = 10$ AG

[3 marks]

${\text{P}}(X = 6|X \le 11) = \frac{{{\text{P}}(X = 6)}}{{{\text{P}}(X \le 11)}}$ (M1) (A1)

$ = \frac{{0.063055 \ldots }}{{0.696776 \ldots }}$ (A1)

$ = 0.0905$ A1

[4 marks]

Total [7 marks]

Examiners report

Most candidates successfully finished part (a) with two fundamental errors occurring regularly. Either e was granted powers of 4 and 5 or an attempt to show that the value of $m$ was 10 was made by evaluation.

Part (b) was challenging for many candidates that showed that the idea of conditional probability was poorly understood. There were many incorrect solutions where often candidates only found ${\text{P}}(X = 6)$.

Question

The random variable $X$ follows a Poisson distribution with mean $m \ne 0$.

Given that $2{\text{P}}(X = 4) = {\text{P}}(X = 5)$, show that $m = 10$.

[3]

Given that $X \le 11$, find the probability that $X = 6$.

[4]

Answer/Explanation

Markscheme

$2\frac{{{{\text{e}}^{ – m}}{m^4}}}{{4!}} = \frac{{{{\text{e}}^{ – m}}{m^5}}}{{5!}}$ M1A1

$\frac{2}{{4!}} = \frac{m}{{5!}}\;\;\;$or other simplification M1

Note: accept a labelled graph showing clearly the solution to the equation. Do not accept simple verification that $m = 10$ is a solution.

$ \Rightarrow m = 10$ AG

[3 marks]

${\text{P}}(X = 6|X \le 11) = \frac{{{\text{P}}(X = 6)}}{{{\text{P}}(X \le 11)}}$ (M1) (A1)

$ = \frac{{0.063055 \ldots }}{{0.696776 \ldots }}$ (A1)

$ = 0.0905$ A1

[4 marks]

Total [7 marks]

Examiners report

Question

Emma acquires a new cell phone for her birthday and receives texts from her friends. It is assumed that the daily number of texts Emma receives follows a Poisson distribution with mean $m = 5$.

(i) Find the probability that on a certain day Emma receives more than $7$ texts.

(ii) Determine the expected number of days in a week on which Emma receives more than $7$ texts.

[4]

Find the probability that Emma receives fewer than $30$ texts during a week.

[3]

Answer/Explanation

Markscheme

(i) $X \sim Po(5)$

${\text{P}}(X \ge 8) = 0.133$ (M1)A1

(ii) $7 \times 0.133 \ldots $ M1

$ \approx 0.934{\text{ days}}$ A1

Note: Accept “$1$ day”.

[4 marks]

$7 \times 5 = 35\;\;\;\left( {Y \sim Po(35)} \right)$ (A1)

${\text{P}}(Y \le 29) = 0.177$ (M1)A1

[3 marks]

Total [7 marks]

Examiners report

[N/A]

Question

Emma acquires a new cell phone for her birthday and receives texts from her friends. It is assumed that the daily number of texts Emma receives follows a Poisson distribution with mean $m = 5$.

(i) Find the probability that on a certain day Emma receives more than $7$ texts.

(ii) Determine the expected number of days in a week on which Emma receives more than $7$ texts.

[4]

Find the probability that Emma receives fewer than $30$ texts during a week.

[3]

Answer/Explanation

Markscheme

(i) $X \sim Po(5)$

${\text{P}}(X \ge 8) = 0.133$ (M1)A1

(ii) $7 \times 0.133 \ldots $ M1

$ \approx 0.934{\text{ days}}$ A1

Note: Accept “$1$ day”.

[4 marks]

$7 \times 5 = 35\;\;\;\left( {Y \sim Po(35)} \right)$ (A1)

${\text{P}}(Y \le 29) = 0.177$ (M1)A1

[3 marks]

Total [7 marks]

Examiners report

[N/A]

Question

A survey is conducted in a large office building. It is found that $30\% $ of the office workers weigh less than $62$ kg and that $25\% $ of the office workers weigh more than $98$ kg.

The weights of the office workers may be modelled by a normal distribution with mean $\mu $ and standard deviation $\sigma $.

(i) Determine two simultaneous linear equations satisfied by $\mu $ and $\sigma $.

(ii) Find the values of $\mu $ and $\sigma $.

[6]

Find the probability that an office worker weighs more than $100$ kg.

[1]

There are elevators in the office building that take the office workers to their offices.

Given that there are $10$ workers in a particular elevator,

find the probability that at least four of the workers weigh more than $100$ kg.

[2]

Given that there are $10$ workers in an elevator and at least one weighs more than $100$ kg,

find the probability that there are fewer than four workers exceeding $100$ kg.

[3]

The arrival of the elevators at the ground floor between $08:00$ and $09:00$ can be modelled by a Poisson distribution. Elevators arrive on average every $36$ seconds.

Find the probability that in any half hour period between $08:00$ and $09:00$ more than $60$ elevators arrive at the ground floor.

[3]

An elevator can take a maximum of $10$ workers. Given that $400$ workers arrive in a half hour period independently of each other,

find the probability that there are sufficient elevators to take them to their offices.

[3]

Answer/Explanation

Markscheme

Note: In Section B, accept answers that correctly round to 2 sf.

(i) let $W$ be the weight of a worker and $W \sim {\text{N}}(\mu ,{\text{ }}{\sigma ^2})$

${\text{P}}\left( {Z < \frac{{62 – \mu }}{\alpha }} \right) = 0.3$ and ${\text{P}}\left( {Z < \frac{{98 – \mu }}{\sigma }} \right) = 0.75$ (M1)

Note: Award M1 for a correctly shaded and labelled diagram.

$\frac{{62 – \mu }}{\sigma } = {\Phi ^{ – 1}}(0.3)\;\;\;( = – 0.524 \ldots )\;\;\;$and

$\frac{{98 – \mu }}{\sigma } = {\Phi ^{ – 1}}(0.75)\;\;\;( = 0.674 \ldots )$

or linear equivalents A1A1

Note: Condone equations containing the GDC inverse normal command.

(ii) attempting to solve simultaneously (M1)

$\mu = 77.7,{\text{ }}\sigma = 30.0$ A1A1

[6 marks]

Note: In Section B, accept answers that correctly round to 2 sf.

${\text{P}}(W > 100) = 0.229$ A1

[1 mark]

Note: In Section B, accept answers that correctly round to 2 sf.

let $X$ represent the number of workers over $100$ kg in a lift of ten passengers

$X \sim {\text{B}}(10,{\text{ }}0.229 \ldots )$ (M1)

${\text{P}}(X \ge 4) = 0.178$ A1

[2 marks]

Note: In Section B, accept answers that correctly round to 2 sf.

${\text{P}}(X < 4|X \ge 1) = \frac{{{\text{P}}(1 \le X \le 3)}}{{{\text{P}}(X \ge 1)}}$ M1(A1)

Note: Award the M1 for a clear indication of a conditional probability.

$ = 0.808$ A1

[3 marks]

Note: In Section B, accept answers that correctly round to 2 sf.

$L \sim {\text{Po}}(50)$ (M1)

${\text{P}}(L > 60) = 1 – {\text{P}}(L \le 60)$ (M1)

$ = 0.0722$ A1

[3 marks]

Note: In Section B, accept answers that correctly round to 2 sf.

$400$ workers require at least $40$ elevators (A1)

${\text{P}}(L \ge 40) = 1 – {\text{P}}(L \le 39)$ (M1)

$ = 0.935$ A1

[3 marks]

Total [18 marks]

Examiners report

[N/A]

Question

A survey is conducted in a large office building. It is found that $30\% $ of the office workers weigh less than $62$ kg and that $25\% $ of the office workers weigh more than $98$ kg.

The weights of the office workers may be modelled by a normal distribution with mean $\mu $ and standard deviation $\sigma $.

(i) Determine two simultaneous linear equations satisfied by $\mu $ and $\sigma $.

(ii) Find the values of $\mu $ and $\sigma $.

[6]

Find the probability that an office worker weighs more than $100$ kg.

[1]

There are elevators in the office building that take the office workers to their offices.

Given that there are $10$ workers in a particular elevator,

find the probability that at least four of the workers weigh more than $100$ kg.

[2]

Given that there are $10$ workers in an elevator and at least one weighs more than $100$ kg,

find the probability that there are fewer than four workers exceeding $100$ kg.

[3]

The arrival of the elevators at the ground floor between $08:00$ and $09:00$ can be modelled by a Poisson distribution. Elevators arrive on average every $36$ seconds.

Find the probability that in any half hour period between $08:00$ and $09:00$ more than $60$ elevators arrive at the ground floor.

[3]

An elevator can take a maximum of $10$ workers. Given that $400$ workers arrive in a half hour period independently of each other,

find the probability that there are sufficient elevators to take them to their offices.

[3]

Answer/Explanation

Markscheme

Note: In Section B, accept answers that correctly round to 2 sf.

(i) let $W$ be the weight of a worker and $W \sim {\text{N}}(\mu ,{\text{ }}{\sigma ^2})$

${\text{P}}\left( {Z < \frac{{62 – \mu }}{\alpha }} \right) = 0.3$ and ${\text{P}}\left( {Z < \frac{{98 – \mu }}{\sigma }} \right) = 0.75$ (M1)

Note: Award M1 for a correctly shaded and labelled diagram.

$\frac{{62 – \mu }}{\sigma } = {\Phi ^{ – 1}}(0.3)\;\;\;( = – 0.524 \ldots )\;\;\;$and

$\frac{{98 – \mu }}{\sigma } = {\Phi ^{ – 1}}(0.75)\;\;\;( = 0.674 \ldots )$

or linear equivalents A1A1

Note: Condone equations containing the GDC inverse normal command.

(ii) attempting to solve simultaneously (M1)

$\mu = 77.7,{\text{ }}\sigma = 30.0$ A1A1

[6 marks]

Note: In Section B, accept answers that correctly round to 2 sf.

${\text{P}}(W > 100) = 0.229$ A1

[1 mark]

Note: In Section B, accept answers that correctly round to 2 sf.

let $X$ represent the number of workers over $100$ kg in a lift of ten passengers

$X \sim {\text{B}}(10,{\text{ }}0.229 \ldots )$ (M1)

${\text{P}}(X \ge 4) = 0.178$ A1

[2 marks]

Note: In Section B, accept answers that correctly round to 2 sf.

${\text{P}}(X < 4|X \ge 1) = \frac{{{\text{P}}(1 \le X \le 3)}}{{{\text{P}}(X \ge 1)}}$ M1(A1)

Note: Award the M1 for a clear indication of a conditional probability.

$ = 0.808$ A1

[3 marks]

Note: In Section B, accept answers that correctly round to 2 sf.

$L \sim {\text{Po}}(50)$ (M1)

${\text{P}}(L > 60) = 1 – {\text{P}}(L \le 60)$ (M1)

$ = 0.0722$ A1

[3 marks]

Note: In Section B, accept answers that correctly round to 2 sf.

$400$ workers require at least $40$ elevators (A1)

${\text{P}}(L \ge 40) = 1 – {\text{P}}(L \le 39)$ (M1)

$ = 0.935$ A1

[3 marks]

Total [18 marks]

Examiners report

[N/A]

Question

Students sign up at a desk for an activity during the course of an afternoon. The arrival of each student is independent of the arrival of any other student and the number of students arriving per hour can be modelled as a Poisson distribution with a mean of $\lambda $.

The desk is open for 4 hours. If exactly 5 people arrive to sign up for the activity during that time find the probability that exactly 3 of them arrived during the first hour.

Answer/Explanation

Markscheme

P(3 in the first hour) $ = \frac{{{\lambda ^3}{e^{ – \lambda }}}}{{3!}}$ A1

number to arrive in the four hours follows $Po(4\lambda )$ M1

P(5 arrive in total) $ = \frac{{{{(4\lambda )}^5}{e^{ – 4\lambda }}}}{{5!}}$ A1

attempt to find P(2 arrive in the next three hours) M1

$ = \frac{{{{(3\lambda )}^2}{e^{ – 3\lambda }}}}{{2!}}$ A1

use of conditional probability formula M1

P(3 in the first hour given 5 in total) $ = \frac{{\frac{{{\lambda ^3}{e^{ – \lambda }}}}{{3!}} \times \frac{{{{(3\lambda )}^2}{e^{ – 3\lambda }}}}{{2!}}}}{{\frac{{{{(4\lambda )}^5}{e^{ – 4\lambda }}}}{{5!}}}}$ A1

$\frac{{\left( {\frac{9}{{2!3!}}} \right)}}{{\left( {\frac{{{4^5}}}{{5!}}} \right)}} = \frac{{45}}{{512}} = 0.0879$ A1

[8 marks]

Examiners report

A more difficult question, but it was still surprising how many candidates were unable to make a good start with it. Many were using $\lambda = \frac{5}{4}$ and consequently unable to progress very far. Many students failed to recognise that a conditional probability should be used.

Question

Six balls numbered 1, 2, 2, 3, 3, 3 are placed in a bag. Balls are taken one at a time from the bag at random and the number noted. Throughout the question a ball is always replaced before the next ball is taken.

Three balls are taken from the bag. Find the probability that

A single ball is taken from the bag. Let $X$ denote the value shown on the ball.

Find ${\text{E}}(X)$.

[2]

the total of the three numbers is 5;

[3]

b.i.

the median of the three numbers is 1.

[3]

b.ii.

Ten balls are taken from the bag. Find the probability that less than four of the balls are numbered 2.

[3]

Find the least number of balls that must be taken from the bag for the probability of taking out at least one ball numbered 2 to be greater than 0.95.

[3]

Another bag also contains balls numbered 1 , 2 or 3.

Eight balls are to be taken from this bag at random. It is calculated that the expected number of balls numbered 1 is 4.8 , and the variance of the number of balls numbered 2 is 1.5.

Find the least possible number of balls numbered 3 in this bag.

[8]

Answer/Explanation

Markscheme

${\text{E}}(X) = 1 \times \frac{1}{6} + 2 \times \frac{2}{6} + 3 \times \frac{3}{6} = \frac{{14}}{6}{\text{ }}\left( { = \frac{7}{3} = 2.33} \right)$ (M1)A1

[2 marks]

$3 \times {\text{P}}(113) + 3 \times {\text{P}}(122)$ (M1)

$3 \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{2} + 3 \times \frac{1}{6} \times \frac{1}{3} \times \frac{1}{3} = \frac{7}{{72}}{\text{ }}( = 0.0972)$ A1

Note: Award M1 for attempt to find at least four of the cases.

[3 marks]

b.i.

recognising 111 as a possibility $\left( {{\text{implied by }}\frac{1}{{216}}} \right)$ (M1)

recognising 112 and 113 as possibilities $\left( {{\text{implied by }}\frac{2}{{216}}{\text{ and }}\frac{3}{{216}}} \right)$ (M1)

seeing the three arrangements of 112 and 113 (M1)

${\text{P}}(111) + 3 \times {\text{P}}(112) + 3 \times {\text{P}}(113)$

$ = \frac{1}{{216}} + \frac{6}{{216}} + \frac{9}{{216}} = \frac{{16}}{{216}}{\text{ }}\left( { = \frac{2}{{27}} = 0.0741} \right)$ A1

[3 marks]

b.ii.

let the number of twos be $X,{\text{ }}X \sim B\left( {10,{\text{ }}\frac{1}{3}} \right)$ (M1)

${\text{P}}(X < 4) = {\text{P}}(X \leqslant 3) = 0.559$ (M1)A1

[3 marks]

let $n$ be the number of balls drawn

${\text{P}}(X \geqslant 1) = 1 – {\text{P}}(X = 0)$ M1

$ = 1 – {\left( {\frac{2}{3}} \right)^n} > 0.95$ M1

${\left( {\frac{2}{3}} \right)^n} > 0.05$

$n = 8$ A1

[3 marks]

$8{p_1} = 4.8 \Rightarrow {p_1} = \frac{3}{5}$ (M1)A1

$8{p_2}(1 – {p_2}) = 1.5$ (M1)

$p_2^2 – {p_2} – 0.1875 = 0$ (M1)

${p_2} = \frac{1}{4}{\text{ }}\left( {{\text{or }}\frac{3}{4}} \right)$ A1

reject $\frac{3}{4}$ as it gives a total greater than one

${\text{P}}(1{\text{ or }}2) = \frac{{17}}{{20}}{\text{ or P}}(3) = \frac{3}{{20}}$ (A1)

recognising LCM as 20 so min total number is 20 (M1)

the least possible number of 3’s is 3 A1

[8 marks]

Examiners report

Part (a) was generally well done, although many candidates lost their way after that.

Candidates had difficulty recognising all the different cases in part (b).

b.i.

Candidates had difficulty recognising all the different cases in part (b).

b.ii.

Parts (c) and (d) should have been more standard questions, but many were unable to tackle them.

Part (e) was poorly answered in general.

Question

Three balls are taken from the bag. Find the probability that

A single ball is taken from the bag. Let $X$ denote the value shown on the ball.

Find ${\text{E}}(X)$.

[2]

the total of the three numbers is 5;

[3]

b.i.

the median of the three numbers is 1.

[3]

b.ii.

Ten balls are taken from the bag. Find the probability that less than four of the balls are numbered 2.

[3]

Find the least number of balls that must be taken from the bag for the probability of taking out at least one ball numbered 2 to be greater than 0.95.

[3]

Another bag also contains balls numbered 1 , 2 or 3.

Eight balls are to be taken from this bag at random. It is calculated that the expected number of balls numbered 1 is 4.8 , and the variance of the number of balls numbered 2 is 1.5.

Find the least possible number of balls numbered 3 in this bag.

[8]

Answer/Explanation

Markscheme

${\text{E}}(X) = 1 \times \frac{1}{6} + 2 \times \frac{2}{6} + 3 \times \frac{3}{6} = \frac{{14}}{6}{\text{ }}\left( { = \frac{7}{3} = 2.33} \right)$ (M1)A1

[2 marks]

$3 \times {\text{P}}(113) + 3 \times {\text{P}}(122)$ (M1)

$3 \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{2} + 3 \times \frac{1}{6} \times \frac{1}{3} \times \frac{1}{3} = \frac{7}{{72}}{\text{ }}( = 0.0972)$ A1

Note: Award M1 for attempt to find at least four of the cases.

[3 marks]

b.i.

recognising 111 as a possibility $\left( {{\text{implied by }}\frac{1}{{216}}} \right)$ (M1)

recognising 112 and 113 as possibilities $\left( {{\text{implied by }}\frac{2}{{216}}{\text{ and }}\frac{3}{{216}}} \right)$ (M1)

seeing the three arrangements of 112 and 113 (M1)

${\text{P}}(111) + 3 \times {\text{P}}(112) + 3 \times {\text{P}}(113)$

$ = \frac{1}{{216}} + \frac{6}{{216}} + \frac{9}{{216}} = \frac{{16}}{{216}}{\text{ }}\left( { = \frac{2}{{27}} = 0.0741} \right)$ A1

[3 marks]

b.ii.

let the number of twos be $X,{\text{ }}X \sim B\left( {10,{\text{ }}\frac{1}{3}} \right)$ (M1)

${\text{P}}(X < 4) = {\text{P}}(X \leqslant 3) = 0.559$ (M1)A1

[3 marks]

let $n$ be the number of balls drawn

${\text{P}}(X \geqslant 1) = 1 – {\text{P}}(X = 0)$ M1

$ = 1 – {\left( {\frac{2}{3}} \right)^n} > 0.95$ M1

${\left( {\frac{2}{3}} \right)^n} > 0.05$

$n = 8$ A1

[3 marks]

$8{p_1} = 4.8 \Rightarrow {p_1} = \frac{3}{5}$ (M1)A1

$8{p_2}(1 – {p_2}) = 1.5$ (M1)

$p_2^2 – {p_2} – 0.1875 = 0$ (M1)

${p_2} = \frac{1}{4}{\text{ }}\left( {{\text{or }}\frac{3}{4}} \right)$ A1

reject $\frac{3}{4}$ as it gives a total greater than one

${\text{P}}(1{\text{ or }}2) = \frac{{17}}{{20}}{\text{ or P}}(3) = \frac{3}{{20}}$ (A1)

recognising LCM as 20 so min total number is 20 (M1)

the least possible number of 3’s is 3 A1

[8 marks]

Examiners report

Part (a) was generally well done, although many candidates lost their way after that.

Candidates had difficulty recognising all the different cases in part (b).

b.i.

Candidates had difficulty recognising all the different cases in part (b).

b.ii.

Parts (c) and (d) should have been more standard questions, but many were unable to tackle them.

Part (e) was poorly answered in general.

Question

Three balls are taken from the bag. Find the probability that

A single ball is taken from the bag. Let $X$ denote the value shown on the ball.

Find ${\text{E}}(X)$.

[2]

the total of the three numbers is 5;

[3]

b.i.

the median of the three numbers is 1.

[3]

b.ii.

Ten balls are taken from the bag. Find the probability that less than four of the balls are numbered 2.

[3]

Find the least number of balls that must be taken from the bag for the probability of taking out at least one ball numbered 2 to be greater than 0.95.

[3]

Another bag also contains balls numbered 1 , 2 or 3.

Eight balls are to be taken from this bag at random. It is calculated that the expected number of balls numbered 1 is 4.8 , and the variance of the number of balls numbered 2 is 1.5.

Find the least possible number of balls numbered 3 in this bag.

[8]

Answer/Explanation

Markscheme

${\text{E}}(X) = 1 \times \frac{1}{6} + 2 \times \frac{2}{6} + 3 \times \frac{3}{6} = \frac{{14}}{6}{\text{ }}\left( { = \frac{7}{3} = 2.33} \right)$ (M1)A1

[2 marks]

$3 \times {\text{P}}(113) + 3 \times {\text{P}}(122)$ (M1)

$3 \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{2} + 3 \times \frac{1}{6} \times \frac{1}{3} \times \frac{1}{3} = \frac{7}{{72}}{\text{ }}( = 0.0972)$ A1

Note: Award M1 for attempt to find at least four of the cases.

[3 marks]

b.i.

recognising 111 as a possibility $\left( {{\text{implied by }}\frac{1}{{216}}} \right)$ (M1)

recognising 112 and 113 as possibilities $\left( {{\text{implied by }}\frac{2}{{216}}{\text{ and }}\frac{3}{{216}}} \right)$ (M1)

seeing the three arrangements of 112 and 113 (M1)

${\text{P}}(111) + 3 \times {\text{P}}(112) + 3 \times {\text{P}}(113)$

$ = \frac{1}{{216}} + \frac{6}{{216}} + \frac{9}{{216}} = \frac{{16}}{{216}}{\text{ }}\left( { = \frac{2}{{27}} = 0.0741} \right)$ A1

[3 marks]

b.ii.

let the number of twos be $X,{\text{ }}X \sim B\left( {10,{\text{ }}\frac{1}{3}} \right)$ (M1)

${\text{P}}(X < 4) = {\text{P}}(X \leqslant 3) = 0.559$ (M1)A1

[3 marks]

let $n$ be the number of balls drawn

${\text{P}}(X \geqslant 1) = 1 – {\text{P}}(X = 0)$ M1

$ = 1 – {\left( {\frac{2}{3}} \right)^n} > 0.95$ M1

${\left( {\frac{2}{3}} \right)^n} > 0.05$

$n = 8$ A1

[3 marks]

$8{p_1} = 4.8 \Rightarrow {p_1} = \frac{3}{5}$ (M1)A1

$8{p_2}(1 – {p_2}) = 1.5$ (M1)

$p_2^2 – {p_2} – 0.1875 = 0$ (M1)

${p_2} = \frac{1}{4}{\text{ }}\left( {{\text{or }}\frac{3}{4}} \right)$ A1

reject $\frac{3}{4}$ as it gives a total greater than one

${\text{P}}(1{\text{ or }}2) = \frac{{17}}{{20}}{\text{ or P}}(3) = \frac{3}{{20}}$ (A1)

recognising LCM as 20 so min total number is 20 (M1)

the least possible number of 3’s is 3 A1

[8 marks]

Examiners report

Part (a) was generally well done, although many candidates lost their way after that.

Candidates had difficulty recognising all the different cases in part (b).

b.i.

Candidates had difficulty recognising all the different cases in part (b).

b.ii.

Parts (c) and (d) should have been more standard questions, but many were unable to tackle them.

Part (e) was poorly answered in general.

Question

A company produces rectangular sheets of glass of area 5 square metres. During manufacturing these glass sheets flaws occur at the rate of 0.5 per 5 square metres. It is assumed that the number of flaws per glass sheet follows a Poisson distribution.

Glass sheets with no flaws earn a profit of $5. Glass sheets with at least one flaw incur a loss of $3.

This company also produces larger glass sheets of area 20 square metres. The rate of occurrence of flaws remains at 0.5 per 5 square metres.

A larger glass sheet is chosen at random.

Find the probability that a randomly chosen glass sheet contains at least one flaw.

[3]

Find the expected profit, $P$ dollars, per glass sheet.

[3]

Find the probability that it contains no flaws.

[2]

Answer/Explanation

Markscheme

${\text{X}} \sim {\text{Po}}(0.5)$ (A1)

${\text{P}}(X \geqslant 1) = 0.393{\text{ }}( = 1 – {{\text{e}}^{ – 0.5}})$ (M1)A1

[3 marks]

${\text{P}}(X = 0) = 0.607 \ldots $ (A1)

${\text{E}}(P) = (0.607 \ldots \times 5) – (0.393 \ldots \times 3)$ (M1)

the expected profit is $1.85 per glass sheet A1

[3 marks]

$Y \sim {\text{Po}}(2)$ (M1)

${\text{P}}(Y = 0) = 0.135{\text{ }}( = {{\text{e}}^{ – 2}})$ A1

[2 marks]

Examiners report

Part (a) was reasonably well done. Some candidates calculated ${\text{P}}(X = 1)$.

Part (b) was not as well done as expected with a surprising number of candidates calculating $5{\text{P}}(X = 0) + 3{\text{P}}(X \geqslant 1)$ rather than $5{\text{P}}(X = 0) – 3{\text{P}}(X \geqslant 1)$.

Part (c) was very well done.

Question

Glass sheets with no flaws earn a profit of $5. Glass sheets with at least one flaw incur a loss of $3.

This company also produces larger glass sheets of area 20 square metres. The rate of occurrence of flaws remains at 0.5 per 5 square metres.

A larger glass sheet is chosen at random.

Find the probability that a randomly chosen glass sheet contains at least one flaw.

[3]

Find the expected profit, $P$ dollars, per glass sheet.

[3]

Find the probability that it contains no flaws.

[2]

Answer/Explanation

Markscheme

${\text{X}} \sim {\text{Po}}(0.5)$ (A1)

${\text{P}}(X \geqslant 1) = 0.393{\text{ }}( = 1 – {{\text{e}}^{ – 0.5}})$ (M1)A1

[3 marks]

${\text{P}}(X = 0) = 0.607 \ldots $ (A1)

${\text{E}}(P) = (0.607 \ldots \times 5) – (0.393 \ldots \times 3)$ (M1)

the expected profit is $1.85 per glass sheet A1

[3 marks]

$Y \sim {\text{Po}}(2)$ (M1)

${\text{P}}(Y = 0) = 0.135{\text{ }}( = {{\text{e}}^{ – 2}})$ A1

[2 marks]

Examiners report

Part (a) was reasonably well done. Some candidates calculated ${\text{P}}(X = 1)$.

Part (c) was very well done.

Question

Glass sheets with no flaws earn a profit of $5. Glass sheets with at least one flaw incur a loss of $3.

This company also produces larger glass sheets of area 20 square metres. The rate of occurrence of flaws remains at 0.5 per 5 square metres.

A larger glass sheet is chosen at random.

Find the probability that a randomly chosen glass sheet contains at least one flaw.

[3]

Find the expected profit, $P$ dollars, per glass sheet.

[3]

Find the probability that it contains no flaws.

[2]

Answer/Explanation

Markscheme

${\text{X}} \sim {\text{Po}}(0.5)$ (A1)

${\text{P}}(X \geqslant 1) = 0.393{\text{ }}( = 1 – {{\text{e}}^{ – 0.5}})$ (M1)A1

[3 marks]

${\text{P}}(X = 0) = 0.607 \ldots $ (A1)

${\text{E}}(P) = (0.607 \ldots \times 5) – (0.393 \ldots \times 3)$ (M1)

the expected profit is $1.85 per glass sheet A1

[3 marks]

$Y \sim {\text{Po}}(2)$ (M1)

${\text{P}}(Y = 0) = 0.135{\text{ }}( = {{\text{e}}^{ – 2}})$ A1

[2 marks]

Examiners report

Part (a) was reasonably well done. Some candidates calculated ${\text{P}}(X = 1)$.

Part (c) was very well done.

Question

A discrete random variable $X$ follows a Poisson distribution ${\text{Po}}(\mu )$.

Show that ${\text{P}}(X = x + 1) = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x),{\text{ }}x \in \mathbb{N}$.

[3]

Given that ${\text{P}}(X = 2) = 0.241667$ and ${\text{P}}(X = 3) = 0.112777$, use part (a) to find the value of $\mu $.

[3]

Answer/Explanation

Markscheme

METHOD 1

${\text{P}}(X = x + 1) = \frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ – \mu }}$ A1

$ = \frac{\mu }{{x + 1}} \times \frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ – \mu }}$ M1A1

$ = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x)$ AG

METHOD 2

$\frac{\mu }{{x + 1}} \times {\text{P}}(X = x) = \frac{\mu }{{x + 1}} \times \frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ – \mu }}$ A1

$ = \frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ – \mu }}$ M1A1

$ = {\text{P}}(X = x + 1)$ AG

METHOD 3

$\frac{{{\text{P}}(X = x + 1)}}{{{\text{P}}(X = x)}} = \frac{{\frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ – \mu }}}}{{\frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ – \mu }}}}$ (M1)

$ = \frac{{{\mu ^{x + 1}}}}{{{\mu ^x}}} \times \frac{{x!}}{{(x + 1)!}}$ A1

$ = \frac{\mu }{{x + 1}}$ A1

and so ${\text{P}}(X = x + 1) = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x)$ AG

[3 marks]

${\text{P}}(X = 3) = \frac{\mu }{3} \bullet {\text{P}}(X = 2){\text{ }}\left( {0.112777 = \frac{\mu }{3} \bullet 0.241667} \right)$ A1

attempting to solve for $\mu $ (M1)

$\mu = 1.40$ A1

[3 marks]

Examiners report

[N/A]

Question

A discrete random variable $X$ follows a Poisson distribution ${\text{Po}}(\mu )$.

Show that ${\text{P}}(X = x + 1) = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x),{\text{ }}x \in \mathbb{N}$.

[3]

Given that ${\text{P}}(X = 2) = 0.241667$ and ${\text{P}}(X = 3) = 0.112777$, use part (a) to find the value of $\mu $.

[3]

Answer/Explanation

Markscheme

METHOD 1

${\text{P}}(X = x + 1) = \frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ – \mu }}$ A1

$ = \frac{\mu }{{x + 1}} \times \frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ – \mu }}$ M1A1

$ = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x)$ AG

METHOD 2

$\frac{\mu }{{x + 1}} \times {\text{P}}(X = x) = \frac{\mu }{{x + 1}} \times \frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ – \mu }}$ A1

$ = \frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ – \mu }}$ M1A1

$ = {\text{P}}(X = x + 1)$ AG

METHOD 3

$\frac{{{\text{P}}(X = x + 1)}}{{{\text{P}}(X = x)}} = \frac{{\frac{{{\mu ^{x + 1}}}}{{(x + 1)!}}{{\text{e}}^{ – \mu }}}}{{\frac{{{\mu ^x}}}{{x!}}{{\text{e}}^{ – \mu }}}}$ (M1)

$ = \frac{{{\mu ^{x + 1}}}}{{{\mu ^x}}} \times \frac{{x!}}{{(x + 1)!}}$ A1

$ = \frac{\mu }{{x + 1}}$ A1

and so ${\text{P}}(X = x + 1) = \frac{\mu }{{x + 1}} \times {\text{P}}(X = x)$ AG

[3 marks]

${\text{P}}(X = 3) = \frac{\mu }{3} \bullet {\text{P}}(X = 2){\text{ }}\left( {0.112777 = \frac{\mu }{3} \bullet 0.241667} \right)$ A1

attempting to solve for $\mu $ (M1)

$\mu = 1.40$ A1

[3 marks]

Examiners report

[N/A]

Question

The number of bananas that Lucca eats during any particular day follows a Poisson distribution with mean 0.2.

Find the probability that Lucca eats at least one banana in a particular day.

[2]

Find the expected number of weeks in the year in which Lucca eats no bananas.

[4]

Answer/Explanation

Markscheme

let $X$ be the number of bananas eaten in one day

$X \sim {\text{Po}}(0.2)$

${\text{P}}(X \geqslant 1) = 1 – {\text{P}}(X = 0)$ (M1)

$ = 0.181{\text{ }}( = 1 – {{\text{e}}^{ – 0.2}})$ A1

[2 marks]

EITHER

let $Y$ be the number of bananas eaten in one week

${\text{Y}} \sim {\text{Po}}(1.4)$ (A1)

${\text{P}}(Y = 0) = 0.246596 \ldots {\text{ }}( = {{\text{e}}^{ – 1.4}})$ (A1)

let $Z$ be the number of days in one week at least one banana is eaten

$Z \sim {\text{B}}(7,{\text{ }}0.181 \ldots )$ (A1)

${\text{P}}(Z = 0) = 0.246596 \ldots $ (A1)

THEN

$52 \times 0.246596 \ldots $ (M1)

$ = 12.8{\text{ }}( = 52{{\text{e}}^{ – 1.4}})$ A1

[4 marks]

Examiners report

[N/A]

Question

The number of bananas that Lucca eats during any particular day follows a Poisson distribution with mean 0.2.

Find the probability that Lucca eats at least one banana in a particular day.

[2]

Find the expected number of weeks in the year in which Lucca eats no bananas.

[4]

Answer/Explanation

Markscheme

let $X$ be the number of bananas eaten in one day

$X \sim {\text{Po}}(0.2)$

${\text{P}}(X \geqslant 1) = 1 – {\text{P}}(X = 0)$ (M1)

$ = 0.181{\text{ }}( = 1 – {{\text{e}}^{ – 0.2}})$ A1

[2 marks]

EITHER

let $Y$ be the number of bananas eaten in one week

${\text{Y}} \sim {\text{Po}}(1.4)$ (A1)

${\text{P}}(Y = 0) = 0.246596 \ldots {\text{ }}( = {{\text{e}}^{ – 1.4}})$ (A1)

let $Z$ be the number of days in one week at least one banana is eaten

$Z \sim {\text{B}}(7,{\text{ }}0.181 \ldots )$ (A1)

${\text{P}}(Z = 0) = 0.246596 \ldots $ (A1)

THEN

$52 \times 0.246596 \ldots $ (M1)

$ = 12.8{\text{ }}( = 52{{\text{e}}^{ – 1.4}})$ A1

[4 marks]

Examiners report

[N/A]

Question

The mean number of squirrels in a certain area is known to be 3.2 squirrels per hectare of woodland. Within this area, there is a 56 hectare woodland nature reserve. It is known that there are currently at least 168 squirrels in this reserve.

Assuming the population of squirrels follow a Poisson distribution, calculate the probability that there are more than 190 squirrels in the reserve.

Answer/Explanation

Markscheme

X is number of squirrels in reserve
X ∼ Po(179.2) A1

Note: Award A1 if 179.2 or 56 × 3.2 seen or implicit in future calculations.

recognising conditional probability M1

P(X > 190 | X ≥ 168)

$ = \frac{{{\text{P}}\left( {X > 190} \right)}}{{{\text{P}}\left( {X \geqslant 168} \right)}} = \left( {\frac{{0.19827 \ldots }}{{0.80817 \ldots }}} \right)$ (A1)(A1)

= 0.245 A1

[5 marks]

Examiners report

[N/A]

Question

The number of taxis arriving at Cardiff Central railway station can be modelled by a Poisson distribution. During busy periods of the day, taxis arrive at a mean rate of 5.3 taxis every 10 minutes. Let T represent a random 10 minute busy period.

Find the probability that exactly 4 taxis arrive during T.

[2]

a.i.

Find the most likely number of taxis that would arrive during T.

[2]

a.ii.

Given that more than 5 taxis arrive during T, find the probability that exactly 7 taxis arrive during T.

[3]

a.iii.

During quiet periods of the day, taxis arrive at a mean rate of 1.3 taxis every 10 minutes.

Find the probability that during a period of 15 minutes, of which the first 10 minutes is busy and the next 5 minutes is quiet, that exactly 2 taxis arrive.

[6]

Answer/Explanation

Markscheme

$X \sim {\text{Po}}\left( {5.3} \right)$

${\text{P}}\left( {X = 4} \right) = {{\text{e}}^{ – 5.3}}\frac{{{{5.3}^4}}}{{4{\text{!}}}}$ (M1)

= 0.164 A1

[2 marks]

a.i.

METHOD 1

listing probabilities (table or graph) M1

mode X = 5 (with probability 0.174) A1

Note: Award M0A0 for 5 (taxis) or mode = 5 with no justification.

METHOD 2

mode is the integer part of mean R1

E(X) = 5.3 ⇒ mode = 5 A1

Note: Do not allow R0A1.

[2 marks]

a.ii.

attempt at conditional probability (M1)

$\frac{{{\text{P}}\left( {X = 7} \right)}}{{{\text{P}}\left( {X \geqslant 6} \right)}}$ or equivalent $\left( { = \frac{{0.1163 \ldots }}{{0.4365 \ldots }}} \right)$ A1

= 0.267 A1

[3 marks]

a.iii.

METHOD 1

the possible arrivals are (2,0), (1,1), (0,2) (A1)

$Y \sim {\text{Po}}\left( {0.65} \right)$ A1

attempt to compute, using sum and product rule, (M1)

0.070106… × 0.52204… + 0.026455… × 0.33932… + 0.0049916… × 0.11028… (A1)(A1)

Note: Award A1 for one correct product and A1 for two other correct products.

= 0.0461 A1

[6 marks]

METHOD 2

recognising a sum of 2 independent Poisson variables eg Z = X + Y R1

$\lambda = 5.3 + \frac{{1.3}}{2}$

P(Z = 2) = 0.0461 (M1)A3

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

Question

Find the probability that exactly 4 taxis arrive during T.

[2]

a.i.

Find the most likely number of taxis that would arrive during T.

[2]

a.ii.

Given that more than 5 taxis arrive during T, find the probability that exactly 7 taxis arrive during T.

[3]

a.iii.

During quiet periods of the day, taxis arrive at a mean rate of 1.3 taxis every 10 minutes.

Find the probability that during a period of 15 minutes, of which the first 10 minutes is busy and the next 5 minutes is quiet, that exactly 2 taxis arrive.

[6]

Answer/Explanation

Markscheme

$X \sim {\text{Po}}\left( {5.3} \right)$

${\text{P}}\left( {X = 4} \right) = {{\text{e}}^{ – 5.3}}\frac{{{{5.3}^4}}}{{4{\text{!}}}}$ (M1)

= 0.164 A1

[2 marks]

a.i.

METHOD 1

listing probabilities (table or graph) M1

mode X = 5 (with probability 0.174) A1

Note: Award M0A0 for 5 (taxis) or mode = 5 with no justification.

METHOD 2

mode is the integer part of mean R1

E(X) = 5.3 ⇒ mode = 5 A1

Note: Do not allow R0A1.

[2 marks]

a.ii.

attempt at conditional probability (M1)

$\frac{{{\text{P}}\left( {X = 7} \right)}}{{{\text{P}}\left( {X \geqslant 6} \right)}}$ or equivalent $\left( { = \frac{{0.1163 \ldots }}{{0.4365 \ldots }}} \right)$ A1

= 0.267 A1

[3 marks]

a.iii.

METHOD 1

the possible arrivals are (2,0), (1,1), (0,2) (A1)

$Y \sim {\text{Po}}\left( {0.65} \right)$ A1

attempt to compute, using sum and product rule, (M1)

0.070106… × 0.52204… + 0.026455… × 0.33932… + 0.0049916… × 0.11028… (A1)(A1)

Note: Award A1 for one correct product and A1 for two other correct products.

= 0.0461 A1

[6 marks]

METHOD 2

recognising a sum of 2 independent Poisson variables eg Z = X + Y R1

$\lambda = 5.3 + \frac{{1.3}}{2}$

P(Z = 2) = 0.0461 (M1)A3

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

Question

Find the probability that exactly 4 taxis arrive during T.

[2]

a.i.

Find the most likely number of taxis that would arrive during T.

[2]

a.ii.

Given that more than 5 taxis arrive during T, find the probability that exactly 7 taxis arrive during T.

[3]

a.iii.

During quiet periods of the day, taxis arrive at a mean rate of 1.3 taxis every 10 minutes.

Find the probability that during a period of 15 minutes, of which the first 10 minutes is busy and the next 5 minutes is quiet, that exactly 2 taxis arrive.

[6]

Answer/Explanation

Markscheme

$X \sim {\text{Po}}\left( {5.3} \right)$

${\text{P}}\left( {X = 4} \right) = {{\text{e}}^{ – 5.3}}\frac{{{{5.3}^4}}}{{4{\text{!}}}}$ (M1)

= 0.164 A1

[2 marks]

a.i.

METHOD 1

listing probabilities (table or graph) M1

mode X = 5 (with probability 0.174) A1

Note: Award M0A0 for 5 (taxis) or mode = 5 with no justification.

METHOD 2

mode is the integer part of mean R1

E(X) = 5.3 ⇒ mode = 5 A1

Note: Do not allow R0A1.

[2 marks]

a.ii.

attempt at conditional probability (M1)

$\frac{{{\text{P}}\left( {X = 7} \right)}}{{{\text{P}}\left( {X \geqslant 6} \right)}}$ or equivalent $\left( { = \frac{{0.1163 \ldots }}{{0.4365 \ldots }}} \right)$ A1

= 0.267 A1

[3 marks]

a.iii.

METHOD 1

the possible arrivals are (2,0), (1,1), (0,2) (A1)

$Y \sim {\text{Po}}\left( {0.65} \right)$ A1

attempt to compute, using sum and product rule, (M1)

0.070106… × 0.52204… + 0.026455… × 0.33932… + 0.0049916… × 0.11028… (A1)(A1)

Note: Award A1 for one correct product and A1 for two other correct products.

= 0.0461 A1

[6 marks]

METHOD 2

recognising a sum of 2 independent Poisson variables eg Z = X + Y R1

$\lambda = 5.3 + \frac{{1.3}}{2}$

P(Z = 2) = 0.0461 (M1)A3

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

Question

Find the probability that exactly 4 taxis arrive during T.

[2]

a.i.

Find the most likely number of taxis that would arrive during T.

[2]

a.ii.

Given that more than 5 taxis arrive during T, find the probability that exactly 7 taxis arrive during T.

[3]

a.iii.

During quiet periods of the day, taxis arrive at a mean rate of 1.3 taxis every 10 minutes.

Find the probability that during a period of 15 minutes, of which the first 10 minutes is busy and the next 5 minutes is quiet, that exactly 2 taxis arrive.

[6]

Answer/Explanation

Markscheme

$X \sim {\text{Po}}\left( {5.3} \right)$

${\text{P}}\left( {X = 4} \right) = {{\text{e}}^{ – 5.3}}\frac{{{{5.3}^4}}}{{4{\text{!}}}}$ (M1)

= 0.164 A1

[2 marks]

a.i.

METHOD 1

listing probabilities (table or graph) M1

mode X = 5 (with probability 0.174) A1

Note: Award M0A0 for 5 (taxis) or mode = 5 with no justification.

METHOD 2

mode is the integer part of mean R1

E(X) = 5.3 ⇒ mode = 5 A1

Note: Do not allow R0A1.

[2 marks]

a.ii.

attempt at conditional probability (M1)

$\frac{{{\text{P}}\left( {X = 7} \right)}}{{{\text{P}}\left( {X \geqslant 6} \right)}}$ or equivalent $\left( { = \frac{{0.1163 \ldots }}{{0.4365 \ldots }}} \right)$ A1

= 0.267 A1

[3 marks]

a.iii.

METHOD 1

the possible arrivals are (2,0), (1,1), (0,2) (A1)

$Y \sim {\text{Po}}\left( {0.65} \right)$ A1

attempt to compute, using sum and product rule, (M1)

0.070106… × 0.52204… + 0.026455… × 0.33932… + 0.0049916… × 0.11028… (A1)(A1)

Note: Award A1 for one correct product and A1 for two other correct products.

= 0.0461 A1

[6 marks]

METHOD 2

recognising a sum of 2 independent Poisson variables eg Z = X + Y R1

$\lambda = 5.3 + \frac{{1.3}}{2}$

P(Z = 2) = 0.0461 (M1)A3

[6 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

a.iii.

[N/A]

Question

The weights, in kg, of male birds of a certain species are modelled by a normal distribution with mean $\mu $ and standard deviation $\sigma $ .

Given that 70 % of the birds weigh more than 2.1 kg and 25 % of the birds weigh more than 2.5 kg, calculate the value of $\mu $ and the value of $\sigma $ .

[4]

A random sample of ten of these birds is obtained. Let X denote the number of birds in the sample weighing more than 2.5 kg.

(i) Calculate ${\text{E}}(X)$ .

(ii) Calculate the probability that exactly five of these birds weigh more than 2.5 kg.

(iii) Determine the most likely value of X .

[5]

The number of eggs, Y , laid by female birds of this species during the nesting season is modelled by a Poisson distribution with mean $\lambda $ . You are given that ${\text{P}}(Y \geqslant 2) = 0.80085$ , correct to 5 decimal places.

(i) Determine the value of $\lambda $ .

(ii) Calculate the probability that two randomly chosen birds lay a total of

two eggs between them.

(iii) Given that the two birds lay a total of two eggs between them, calculate the probability that they each lay one egg.

[8]

Answer/Explanation

Markscheme

we are given that

$2.1 = \mu – 0.5244\sigma $

$2.5 = \mu + 0.6745\sigma $ M1A1

$\mu = 2.27{\text{ , }}\sigma = 0.334$ A1A1

[4 marks]

(i) let X denote the number of birds weighing more than 2.5 kg

then X is B(10, 0.25) A1

${\text{E}}(X) = 2.5$ A1

(ii) 0.0584 A1

(iii) to find the most likely value of X , consider

${p_0} = 0.0563 \ldots ,{\text{ }}{p_1} = 0.1877 \ldots ,{\text{ }}{p_2} = 0.2815 \ldots ,{\text{ }}{p_3} = 0.2502 \ldots $ M1

therefore, most likely value = 2 A1

[5 marks]

(i) we solve $1 – {\text{P}}(Y \leqslant 1) = 0.80085$ using the GDC M1

$\lambda = 3.00$ A1

(ii) let ${X_1}{\text{, }}{X_2}$ denote the number of eggs laid by each bird

${\text{P}}({X_1} + {X_2} = 2) = {\text{P}}({X_1} = 0){\text{P}}({X_2} = 1) + {\text{P}}({X_1} = 1){\text{P}}({X_2} = 1) + {\text{P}}({X_1} = 2){\text{P}}({X_2} = 0)$ M1A1

$ = {{\text{e}}^{ – 3}} \times {{\text{e}}^{ – 3}} \times \frac{9}{2} + {({{\text{e}}^{ – 3}} \times 3)^2} + {{\text{e}}^{ – 3}} \times \frac{9}{2} \times {{\text{e}}^{ – 3}} = 0.0446$ A1

(iii) ${\text{P}}({X_1} = 1,{\text{ }}{X_2} = 1|{X_1} + {X_2} = 2) = \frac{{{\text{P}}({X_1} = 1,{\text{ }}{X_2} = 1)}}{{{\text{P}}({X_1} + {X_2} = 2)}}$ M1A1

$ = 0.5$ A1

[8 marks]

Examiners report

[N/A]