Question
Taylor is playing a computer game in which they shoot at spaceships and battleships. The number of spaceships they hit per minute can be modelled by a Poisson distribution with mean 4.2. The number of battleships they hit per minute can be modelled by a Poisson distribution with a mean of 2.3. Any single hit occurs independently of all others.
(a) Find the probability Taylor hits
(i) at most 10 spaceships in 2 minutes.
(ii) a total of more than 10 spaceships and battleships in one minute.
Every spaceship that is hit earns Taylor 3 points and every battleship 5 points. Let T be the total points earned in one minute.
(b) Find
(i) E(T)
(ii) Var(T)
(c) State one reason why the distribution of T cannot be Poisson. Taylor intends to play the game for one hour.
(d) Use the central limit theorem to find the probability that Taylor’s mean score per minute is greater than 25.
▶️Answer/Explanation
Detail Solution
Part (a)(i): Probability Taylor hits at most 10 spaceships in 2 minutes
The number of spaceships hit per minute follows a Poisson distribution with a mean of 4.2. Since we’re looking at a 2-minute period, we need the rate over 2 minutes. For a Poisson process, the parameter scales with time, so:
Mean spaceships in 2 minutes = 4.2 × 2 = 8.4.
Let \( S \) be the number of spaceships hit in 2 minutes. Thus, \( S \sim \text{Poisson}(8.4) \). We need \( P(S \leq 10) \), the probability of hitting at most 10 spaceships.
Using the Poisson cumulative distribution function:
\[ P(S \leq 10) = \sum_{k=0}^{10} e^{-8.4} \frac{8.4^k}{k!} \]
This is the sum of probabilities from 0 to 10 hits. Calculating this exactly requires summing 11 terms, which can be tedious by hand, but Poisson probabilities are well-tabulated or computable. Using a calculator or statistical software for \( \text{Poisson}(8.4) \):
\( P(S \leq 10) \approx 0.7673 \) (rounded to 4 decimal places).
This makes sense intuitively: the mean is 8.4, so 10 is slightly above the mean, and we expect a significant portion of the distribution to lie at or below 10.
Part (a)(ii): Probability of hitting more than 10 spaceships and battleships total in one minute
Let \( X \) be the number of spaceships hit in 1 minute, \( X \sim \text{Poisson}(4.2) \), and \( Y \) be the number of battleships hit in 1 minute, \( Y \sim \text{Poisson}(2.3) \). Since hits are independent, the total number of hits \( Z = X + Y \) is the sum of two independent Poisson variables. The sum of independent Poisson random variables is also Poisson, with the mean being the sum of the individual means:
Mean of \( Z = X + Y = 4.2 + 2.3 = 6.5 \).
Thus, \( Z \sim \text{Poisson}(6.5) \). We need \( P(Z > 10) \), which is:
\[ P(Z > 10) = 1 – P(Z \leq 10) \]
Compute \( P(Z \leq 10) \) for \( \text{Poisson}(6.5) \):
\[ P(Z \leq 10) = \sum_{k=0}^{10} e^{-6.5} \frac{6.5^k}{k!} \]
Using a Poisson calculator or tables:
\( P(Z \leq 10) \approx 0.9326 \).
So:
\[ P(Z > 10) = 1 – 0.9326 = 0.0674 \]
This probability is small, which aligns with 10 being well above the mean of 6.5, placing it in the right tail of the distribution.
Part (b)(i): Expected value of total points in one minute, \( E(T) \)
Each spaceship hit earns 3 points, and each battleship hit earns 5 points. Define \( T \) as the total points in 1 minute:
\[ T = 3X + 5Y \]
where \( X \sim \text{Poisson}(4.2) \) and \( Y \sim \text{Poisson}(2.3) \), and they are independent. The expected value of a linear combination is:
\[ E(T) = E(3X + 5Y) = 3E(X) + 5E(Y) \]
For Poisson variables, the mean is the parameter:
\( E(X) = 4.2 \),
\( E(Y) = 2.3 \).
So:
\[ E(T) = 3 \times 4.2 + 5 \times 2.3 = 12.6 + 11.5 = 24 \]
Thus, \( E(T) = 24 \) points per minute.
Part (b)(ii): Variance of total points, \( \text{Var}(T) \)
Since \( X \) and \( Y \) are independent, the variance of a linear combination \( T = 3X + 5Y \) is:
\[ \text{Var}(T) = \text{Var}(3X + 5Y) = 3^2 \text{Var}(X) + 5^2 \text{Var}(Y) \]
For a Poisson random variable, the variance equals the mean:
\( \text{Var}(X) = 4.2 \),
\( \text{Var}(Y) = 2.3 \).
So:
\[ \text{Var}(T) = 9 \times 4.2 + 25 \times 2.3 = 37.8 + 57.5 = 95.3 \]
Thus, \( \text{Var}(T) = 95.3 \).
Part (c): Reason why \( T \) cannot be Poisson
A Poisson distribution models the number of events in a fixed interval, taking non-negative integer values (0, 1, 2, …), and each event contributes equally to the count. Here, \( T = 3X + 5Y \) represents points, where \( X \) and \( Y \) are Poisson, but the points per hit (3 or 5) differ:
\( T \) takes values that are multiples of 3 and 5 (e.g., 0, 3, 5, 6, 8, 9, 10, …), not all non-negative integers (e.g., 1, 2, 4, 7 are impossible).
A Poisson variable’s values increment by 1, but \( T \)’s possible values depend on combinations of 3 and 5, skipping some integers.
Thus, one reason is: *The distribution of \( T \) cannot be Poisson because it does not take all non-negative integer values, only those that are sums of multiples of 3 and 5.
Part(d) Central Limit Theorem
let $ T_{i}$ be the points earned in the -th minute.
$E(T_{i})=24.1$ and $Var(T_{i})$=95.3
Let $\bar {T}$ be the mean score per minute
By the central limit theorem, $\bar{T} $ is approximately normally distributed with
Mean:
variance $ Var( \bar{T}) $=$\frac{Var(T_{i})} {60}$ = $\frac{95.3}{60} \approx 1.5883 $
Standard Deviation $ \sqrt{1.5883} \approx 1.2603 $
we want $ P(\bar{T}>25) $
$ P(\bar{T}>25)=P(Z>\frac{25-24.1}{1.2603})=P(Z>\frac{0.9}{1.2603}) \approx P(Z>0.7141) $
using a standard normal distribution table or calculator
$P(Z>0.7141) \approx 1-0.7624=0.2376$
Therefore,the probability that Taylor’s mean score per minute is greater than 25 is approximately 0.2376
————Markscheme—————–
(a) (i) let S be the number of spaceships hit and B the number of battleships
mean = 8.4
P(S ≤ 10) = 0.774 (0.774301…)
(ii) attempt to add two means
4.2 + 2.3 = 6.5
P(S + B > 10) = P(S + B ≥ 11)
0.0668 (0.0668387…)
(b) (i) E(T) = 3 × 4.2 + 5 × 2.3 = 24.1
(ii) Var(T) = 3² × 4.2 + 5² × 2.3 = 95.3
(c) any valid reason
for example:
mean is not equal to variance OR T cannot take all integer values
(d) distribution of mean score is $N(24.1, \frac{95.3}{60}) (N(24.1, 1.58833…))$
$P(\overline{T}>25) = 0.238 (0.237576…)$
Question 2. [Maximum mark: 16]
It is known that the weights of male Persian cats are normally distributed with mean 6.1 kg
and variance 0.52 kg2.
a. Sketch a diagram showing the above information. [2]
b. Find the proportion of male Persian cats weighing between 5.5 kg and 6.5 kg. [2]
A group of 80 male Persian cats are drawn from this population.
c. Determine the expected number of cats in this group that have a weight of less than 5.3 kg. [3]
The male cats are now joined by 80 female Persian cats. The female cats are drawn from a population whose weights are normally distributed with mean 4.5 kg and standard deviation 0.45 kg.
Ten female cats are chosen at random.
d. (i)Find the probability that exactly one of them weighs over 4.62 kg.
Let N be the number of cats weighing over 4.62 kg.
d. (ii) Find the variance of N . [5]
A cat is selected at random from all 160 cats.
e. Find the probability that the cat was female, given that its weight was over 4.7 kg. [4]
▶️Answer/Explanation
(a) 
(b) \(X ~ N (6.1, 0.52 ) P (5.5 < X < 6.5)\) OR labelled sketch of region \(= 0.673 (0.673074…)\)
(c) \((P (X < 5.3) =0.0547992… 0.0547992… \times 80 = 4.38 (4.38393…)\)
(d) (i) \(Y ~ N (4.5, 0.452 ) (P ( Y > 4.62) ) 0.394862…\) use of binomial seen or implied using \(B(10, 0.394862…) 0.0430 (0.0429664…)\)
(ii)\(np (1- p) = 2.39 (2.38946…)\)
(e) \(P (F ∩(W> 4.7) = 0.5 × 0.3284 =(0.1642 )\)
use of tree diagram OR use of
\(P\((F|W>4.7)=\frac{F\cap (W>4.7)}{P(W>4.7)}\frac{0.5\times 0.3284}{0.5\times 0.9974+0.5\times 0.3284}\)
\(= 0.248 (0.247669…)\)
Question
The aircraft for a particular flight has 72 seats. The airline’s records show that historically
for this flight only 90% of the people who purchase a ticket arrive to board the flight.
They assume this trend will continue and decide to sell extra tickets and hope that no
more than 72 passengers will arrive.
The number of passengers that arrive to board this flight is assumed to follow a binomial
distribution with a probability of 0.9.
(a) The airline sells 74 tickets for this flight. Find the probability that more than 72
passengers arrive to board the flight.
(b) (i) Write down the expected number of passengers who will arrive to board the flight if 72 tickets are sold.
(ii) Find the maximum number of tickets that could be sold if the expected number
of passengers who arrive to board the flight must be less than or equal to 72.
Each passenger pays $150 for a ticket. If too many passengers arrive, then the airline will
give $300 in compensation to each passenger that cannot board.
(c) Find, to the nearest integer, the expected increase or decrease in the money made
by the airline if they decide to sell 74 tickets rather than 72.
▶️Answer/Explanation
Ans:
Question
A machine fills containers with grass seed. Each container is supposed to weigh \(28\) kg. However the weights vary with a standard deviation of \(0.54\) kg. A random sample of \(24\) bags is taken to check that the mean weight is \(28\) kg.
A.a.Assuming the series for \({{\rm{e}}^x}\) , find the first five terms of the Maclaurin series for\[\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ – {x^2}}}{2}}} {\rm{ .}}\][3]
A.b.(i) Use your answer to (a) to find an approximate expression for the cumulative distributive function of \({\rm{N}}(0,1)\) .
(ii) Hence find an approximate value for \({\rm{P}}( – 0.5 \le Z \le 0.5)\) , where \(Z \sim {\rm{N}}(0,1)\) .[6]
B.a.State and justify an appropriate test procedure giving the null and alternate hypotheses.[5]
B.b.What is the critical region for the sample mean if the probability of a Type I error is to be \(3.5\%\)?[7]
B.c.If the mean weight of the bags is actually \(28\).1 kg, what would be the probability of a Type II error?[2]
▶️Answer/Explanation
Markscheme
\({{\rm{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \ldots \)
\({{\rm{e}}^{\frac{{ – {x^2}}}{2}}} = 1 + \left( { – \frac{{{x^2}}}{2}} \right) + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^2}}}{{2!}} + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^3}}}{{3!}} + \frac{{{{\left( { – \frac{{{x^2}}}{2}} \right)}^4}}}{{4!}} + \ldots \) M1A1
\(\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ – {x^2}}}{2}}} = \frac{1}{{\sqrt {2\pi } }}\left( {1 – \frac{{{x^2}}}{2} + \frac{{{x^4}}}{8} – \frac{{{x^6}}}{{48}} + \frac{{{x^8}}}{{384}}} \right)\) A1
[3 marks]
(i) \(\frac{1}{{\sqrt {2\pi } }}\int_0^x {1 – \frac{{{t^2}}}{2}} + \frac{{{t^4}}}{8} – \frac{{{t^6}}}{{48}} + \frac{{{t^8}}}{{384}}{\rm{d}}t\) M1
\( = \frac{1}{{\sqrt {2\pi } }}\left( {x – \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} – \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}}} \right)\) A1
\({\rm{P}}(Z \le x) = 0.5 + \frac{1}{{\sqrt {2\pi } }}\left( {x – \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} – \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}} – \ldots } \right)\) R1A1
(ii) \({\rm{P}}( – 0.5 \le Z \le 0.5) = \frac{2}{{\sqrt {2\pi } }}\left( {0.5 – \frac{{{{0.5}^3}}}{6} + \frac{{{{0.5}^5}}}{{40}} – \frac{{{{0.5}^7}}}{{336}} + \frac{{{{0.5}^9}}}{{3456}} – \ldots } \right)\) M1
\( = 0.38292 = 0.383\) A1
[6 marks]
this is a two tailed test of the sample mean \(\overline x \)
we use the central limit theorem to justify assuming that R1
\(\overline X \sim {\rm{N}}\left( {28,\frac{{{{0.54}^2}}}{{24}}} \right)\) R1A1
\({{\rm{H}}_0}:\mu = 28\) A1
\({{\rm{H}}_1}:\mu \ne 28\) A1
[5 marks]
since \({\text{P(Type I error)}} = 0.035\) , critical value \(2.108\) (M1)A1
and (\(\overline x \le 28 – 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) or \(\overline x \ge 28 + 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) ) (M1)(A1)(A1)
\(\overline x \le 27.7676\) or \(\overline x \ge 28.2324\)
so \(\overline x \le 27.8\) or \(\overline x \ge 28.2\) A1A1
[7 marks]
if \(\mu = 28.1\)
\(\overline X \sim {\rm{N}}\left( {28.1,\frac{{{{0.54}^2}}}{{24}}} \right)\) R1
\({\text{P(Type II error)}} = {\rm{P}}(27.7676 < \overline x < 28.2324)\)
\( = 0.884\) A1
Note: Depending on the degree of accuracy used for the critical region the answer for part (c) can be anywhere from \(0.8146\) to \(0.879\).
[2 marks]
Question
a.The continuous random variable \(X\) takes values only in the interval [\(a\), \(b\)] and \(F\) denotes its cumulative distribution function. Using integration by parts, show that:\[E(X) = b – \int_a^b {F(x){\rm{d}}x}. \][4]
{f(y) = \cos y,}&{0 \leqslant y \leqslant \frac{\pi }{2}} \\
{f(y) = 0,}&{{\text{elsewhere}}{\text{.}}}
\end{array}\]
(i) Obtain an expression for the cumulative distribution function of \(Y\) , valid for \(0 \le y \le \frac{\pi }{2}\) . Use the result in (a) to determine \(E(Y)\) .
(ii) The random variable \(U\) is defined by \(U = {Y^n}\) , where \(n \in {\mathbb{Z}^ + }\) . Obtain an expression for the cumulative distribution function of \(U\) valid for \(0 \le u \le {\left( {\frac{\pi }{2}} \right)^n}\) .
(iii) The medians of \(U\) and \(Y\) are denoted respectively by \({m_u}\) and \({m_y}\) . Show that \({m_u} = m_y^n\) .[14]
▶️Answer/Explanation
Markscheme
\(E(X) = \int_a^b {xf(x){\rm{d}}x} \) M1
\( = \left[ {xF(x)} \right]_a^b – \int_a^b {F(x){\rm{d}}x} \) A1
\( = bF(b) – aF(a) – \int_a^b {F(x){\rm{d}}x} \) A1
\( = b – \int_a^b {F(x){\rm{d}}x} \) because \(F(a) = 0\) and \(F(b) = 1\) A1
[4 marks]
(i) let \(G\) denote the cumulative distribution function of \(Y\)
\(G(y) = \int_0^y {\cos t{\rm{d}}t} \) M1
\( = \left[ {\sin t} \right]_0^y\) (A1)
\( = \sin y\) A1
\(E(Y) = \frac{\pi }{2} – \int_0^{\frac{\pi }{2}} {\sin y{\rm{d}}y} \) M1
\( = \frac{\pi }{2} + \left[ {\cos y} \right]_0^{\frac{\pi }{2}}\) A1
\( = \frac{\pi }{2} – 1\) A1
(ii) CDF of \(U = P(U \le u)\) M1
\( = P({Y^n} \le u)\) A1
\( = P({Y^{}} \le {u^{\frac{1}{n}}})\) A1
\( = G({u^{\frac{1}{n}}})\) (A1)
\( = \sin \left( {{u^{\frac{1}{n}}}} \right)\) A1
(iii) \({m_y}\) satisfies the equation \(\sin {m_y} = \frac{1}{2}\) A1
\({m_u}\) satisfies the equation \(\sin \left( {m_u^{\frac{1}{n}}} \right) = \frac{1}{2}\) A1
therefore \({m_y} = m_u^{\frac{1}{n}}\) A1
\({m_u} = m_y^n\) AG
[14 marks]
Question
a.A random variable \(X\) has probability density function \(f\) given by:\[f(x) = \left\{ {\begin{array}{*{20}{l}}
{\lambda {e^{ – \lambda x}},}&{{\text{for }}x \geqslant 0{\text{ where }}\lambda > 0} \\
{0,}&{{\text{for }}x < 0.}
\end{array}} \right.\]
(i) Find an expression for \({\rm{P}}(X > a)\) , where \(a > 0\) .
A chicken crosses a road. It is known that cars pass the chicken’s crossing route, with intervals between cars measured in seconds, according to the random variable \(X\) , with \(\lambda = 0.03\) . The chicken, which takes \(10\) seconds to cross the road, starts to cross just as one car passes.
(ii) Find the probability that the chicken will reach the other side of the road before the next car arrives.
Later, the chicken crosses the road again just after a car has passed.
(iii) Show that the probability that the chicken completes both crossings is greater than \(0.5\).[6]
b.A rifleman shoots at a circular target. The distance in centimetres from the centre of the target at which the bullet hits, can be modelled by \(X\) with \(\lambda = 0.4\) . The rifleman scores \(10\) points if \(X \le 1\) , \(5\) points if \(1 < X \le 5\) , \(1\) point if \(5 < X \le 10\) and no points if \(X > 10\) .
(i) Find the expected score when one bullet is fired at the target.
A second rifleman, whose shooting can also be modelled by \(X\) , wishes to find his value of \(\lambda \) .
(ii) Given that his expected score is \(6.5\), find his value of \(\lambda \) .[10]
▶️Answer/Explanation
Markscheme
(i) \({\rm{P}}(X > a) = \int_a^\infty {\lambda {e^{ – \lambda x}}{\rm{d}}x} \) M1
\(\left[ { – {e^{ – \lambda x}}} \right]_a^\infty \) A1
\( = {e^{ – \lambda a}}\) A1
(ii) \({\rm{P}}(X > 10) = {e^{ – 0.3}}( = 0.74 \ldots )\) (M1)A1
(iii) probability of a safe double crossing \( = {e^{ – 0.6}}\) \(( = {0.74^2})\) \( = 0.55\) A1
which is greater than \(0.5\) AG
[6 marks]
(i) \({\rm{P}}(X \le 1) = 0.3296 \ldots \) (A1)
\({\rm{P}}(1 \le X \le 5) = 0.5349 \ldots \) (A1)
\({\rm{P}}(5 \le X \le 10) = 0.1170 \ldots \) (A1)
\({\rm{E(score)}} = 10 \times 0.3296 \ldots + 5 \times 0.5349 \ldots + 1 \times 0.1170 \ldots \) M1A1
\( = 6.09\) A1
Note: Accept probabilities in exponential form until the final decimal answer.
(ii) \({\rm{E(score)}}\) for X with unknown parameter can be expressed as \(10 \times (1 – {e^{ – \lambda }}) + 5 \times ({e^{ – \lambda }} – {e^{ – 5\lambda }}) + ({e^{ – 5\lambda }} – {e^{ – 10\lambda }})\) (M1)(A1)
attempt to solve \({\rm{E(score)}} = 6.5\) (M1)
obtain \(\lambda = 0.473\) A1
[10 marks]
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