Question
The fish in a lake have weights that are normally distributed with a mean of 1.3 kg and a standard deviation of 0.2 kg.
a.Determine the probability that a fish which is caught weighs less than 1.4 kg.[1]
b.John catches 6 fish. Calculate the probability that at least 4 of the fish weigh more than 1.4 kg.[3]
c.Determine the probability that a fish which is caught weighs less than 1 kg, given that it weighs less than 1.4 kg.[2]
▶️Answer/Explanation
Markscheme
\({\text{P}}(x < 1.4) = 0.691\,\,\,\,\,\)(accept 0.692) A1
[1 mark]
METHOD 1
\(y \sim {\text{B(6, 0.3085…)}}\) (M1)
\({\text{P}}(Y \geqslant 4) = 1 – {\text{P}}(Y \leqslant 3)\) (M1)
\( = 0.0775\,\,\,\,\,\)(accept 0.0778 if 3sf approximation from (a) used) A1
METHOD 2
\(X \sim {\text{B(6, 0.6914…)}}\) (M1)
\({\text{P}}(X \leqslant 2)\) (M1)
\( = 0.0775\,\,\,\,\,\)(accept 0.0778 if 3sf approximation from (a) used) A1
[3 marks]
\({\text{P}}(x < 1|x < 1.4) = \frac{{{\text{P}}(x < 1)}}{{{\text{P}}(x < 1.4)}}\) M1
\( = \frac{{0.06680…}}{{0.6914…}}\)
\( = 0.0966\,\,\,\,\,\)(accept 0.0967) A1
[2 marks]
Question
At the start of each week, Eric and Marina pick a night at random on which they will watch a movie.
If they choose a Saturday night, the probability that they watch a French movie is \(\frac{7}{9}\) and if they choose any other night the probability that they watch a French movie is \(\frac{4}{9}\).
a.Find the probability that they watch a French movie.[3]
b.Given that last week they watched a French movie, find the probability that it was on a Saturday night.[2]
▶️Answer/Explanation
Markscheme
\({\text{P}}(F) = \left( {\frac{1}{7} \times \frac{7}{9}} \right) + \left( {\frac{6}{7} \times \frac{4}{9}} \right)\) (M1)(A1)
Note: Award M1 for the sum of two products.
\( = \frac{{31}}{{63}}{\text{ }}( = 0.4920 \ldots )\) A1
[3 marks]
Use of \({\text{P}}(S|F) = \frac{{{\text{P}}(S \cap F)}}{{{\text{P}}(F)}}\) to obtain \({\text{P}}(S|F) = \frac{{\frac{1}{7} \times \frac{7}{9}}}{{\frac{{31}}{{63}}}}\). M1
Note: Award M1 only if the numerator results from the product of two probabilities.
\( = \frac{7}{{31}}{\text{ }}( = 0.2258 \ldots )\) A1
[2 marks]
Question
The number of birds seen on a power line on any day can be modelled by a Poisson distribution with mean 5.84.
a.Find the probability that during a certain seven-day week, more than 40 birds have been seen on the power line.[2]
b.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]
Question
The random variable \(X\) follows a Poisson distribution with mean \(m \ne 0\).
a.Given that \(2{\text{P}}(X = 4) = {\text{P}}(X = 5)\), show that \(m = 10\).[3]
b.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]
Question
Farmer Suzie grows turnips and the weights of her turnips are normally distributed with a mean of \(122g\) and standard deviation of \(14.7g\).
a.(i) Calculate the percentage of Suzie’s turnips that weigh between \(110g\) and \(130g\).
(ii) Suzie has \(100\) turnips to take to market. Find the expected number weighing more than \(130g\).
(iii) Find the probability that at least \(30\) of the \(100g\) turnips weigh more than \(130g\).[6]
Farmer Ray also grows turnips and the weights of his turnips are normally distributed with a mean of \(144g\). Ray only takes to market turnips that weigh more than \(130g\). Over a period of time, Ray finds he has to reject \(1\) in \(15\) turnips due to their being underweight.
(i) Find the standard deviation of the weights of Ray’s turnips.
(ii) Ray has \(200\) turnips to take to market. Find the expected number weighing more than \(150g\).[6]
▶️Answer/Explanation
Markscheme
(i) \(P(110 < X < 130) = 0.49969 \ldots = 0.500 = 50.0\% \) (M1)A1
Note: Accept \(50\)
Note: Award M1A0 for \(0.50\) (\(0.500\))
(ii) \(P(X > 130) = (1 – 0.707 \ldots ) = 0.293 \ldots \) M1
expected number of turnips \( = 29.3\) A1
Note: Accept \(29\).
(iii) no of turnips weighing more than \(130\) is \(Y \sim B(100,{\text{ }}0.293)\) M1
\(P(Y \ge 30) = 0.478\) A1
[6 marks]
(i) \(X \sim N(144,{\text{ }}{\sigma ^2})\)
\(P(X \le 130) = \frac{1}{{15}} = 0.0667\) (M1)
\(P\left( {Z \le \frac{{130 – 144}}{\sigma }} \right) = 0.0667\)
\(\frac{{14}}{\sigma } = 1.501\) (A1)
\(\sigma = 9.33{\text{ g}}\) A1
(ii) \(P(X > 150|X > 130) = \frac{{P(X > 150)}}{{P(X > 130)}}\) M1
\( = \frac{{0.26008 \ldots }}{{1 – 0.06667}} = 0.279\) A1
expected number of turnips \( = 55.7\) A1
[6 marks]
Total [12 marks]
Examiners report
[N/A]
[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 \(\sigma \).
a.(i) Determine two simultaneous linear equations satisfied by \(\mu \) and \(\sigma \).
(ii) Find the values of \(\mu \) and \(\sigma \).[6]
Given that there are \(10\) workers in a particular elevator,
c.find the probability that at least four of the workers weigh more than \(100\) kg.[2]
d.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]
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]
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]
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]