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
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
The random variable \(X\) has probability density function given by
\[f(x) = \left\{ {\begin{array}{*{20}{l}}
{x{{\text{e}}^{ – x}},}&{{\text{for }}x \geqslant 0,} \\
{0,}&{{\text{otherwise}}}
\end{array}} \right..\]
A sample of size 50 is taken from the distribution of \(X\).
a.Use l’Hôpital’s rule to show that \(\mathop {\lim }\limits_{x \to \infty } \frac{{{x^3}}}{{{{\text{e}}^x}}} = 0\).[3]
b.(i) Find \({\text{E}}({X^2})\).
(ii) Show that \({\text{Var}}(X) = 2\).[10]
c.State the central limit theorem.[2]
d.Find the probability that the sample mean is less than 2.3.[2]
▶️Answer/Explanation
Markscheme
attempt to apply l’Hôpital’s rule M1
\(\mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2}}}{{{{\text{e}}^x}}}\) A1
then \(\mathop {\lim }\limits_{x \to \infty } \frac{{6x}}{{{{\text{e}}^x}}}\)
then \(\mathop {\lim }\limits_{x \to \infty } \frac{6}{{{{\text{e}}^x}}}\) A1
\( = 0\) AG
[3 marks]
(i) \({\text{E}}({X^2}) = \mathop {\lim }\limits_{R \to \infty } \int\limits_0^R {{x^3}{{\text{e}}^{ – x}}{\text{d}}x} \) M1
attempt at integration by parts M1
the integral \( = [ – {x^3}{{\text{e}}^{ – x}}]_0^R + \mathop \smallint \limits_0^R 3{x^2}{{\text{e}}^{ – x}}{\text{d}}x\) A1A1
\( = [ – {x^3}{{\text{e}}^{ – x}}]_0^R + [ – 3{x^2}{{\text{e}}^{ – x}}]_0^R + \int\limits_0^R {6x{{\text{e}}^{ – x}}{\text{d}}x} \) M1
\( = [ – {x^3}{{\text{e}}^{ – x}}]_0^R + [ – 3{x^2}{{\text{e}}^{ – x}}]_0^R + [ – 6x{{\text{e}}^{ – x}}]_0^R + \int\limits_0^R {6{{\text{e}}^{ – x}}{\text{d}}x} \) A1
\( = [ – {x^3}{{\text{e}}^{ – x}}]_0^R + [ – 3{x^2}{{\text{e}}^{ – x}}]_0^R + [ – 6x{{\text{e}}^{ – x}}]_0^R + [ – 6{{\text{e}}^{ – x}}]_0^R\) A1
\( = 6\) when \(R \to \infty \) R1
(ii) \({\text{E}}(X) = 2\) A1
\({\text{Var}}(X) = {\text{E}}({X^2}) – {\left( {{\text{E}}(X)} \right)^2} = 6 – {2^2}\) M1
\( = 2\) AG
[10 marks]
if a random sample of size \(n\) is taken from any distribution \(X\), with \({\text{E}}(X) = \mu \) and \({\text{Var}}(X) = {\sigma ^2}\), then, for large n, A1
the sample mean \(\bar X\) has approximate distribution \({\text{N}}\left( {\mu ,{\text{ }}\frac{{{\sigma ^2}}}{n}} \right)\) A1
[2 marks]
\(\bar X \sim {\text{N}}\left( {2,{\text{ }}\frac{2}{{50}} = {{(0.2)}^2}} \right)\) (A1)
\({\text{P}}(\bar X < 2.3) = \left( {{\text{P}}(Z < 1.5)} \right) = 0.933\) A1
[2 marks]
Question
In a large population of sheep, their weights are normally distributed with mean \(\mu \) kg and standard deviation \(\sigma \) kg. A random sample of \(100\) sheep is taken from the population.
The mean weight of the sample is \(\bar X\) kg.
a.State the distribution of \(\bar X\) , giving its mean and standard deviation.[2]
b.The sample values are summarized as \(\sum {x = 3782} \) and \(\sum {{x^2} = 155341} \) where \(x\) kg is the weight of a sheep.
(i) Find unbiased estimates for \(\mu \) and \({\sigma ^2}\).
(ii) Find a \(95\%\) confidence interval for \(\mu \).[6]
c.Test, at the \(1\%\) level of significance, the null hypothesis \(\mu = 35\) against the alternative hypothesis that \(\mu > 35\).[5]
▶️Answer/Explanation
Markscheme
\(\bar X \sim N\left( {\mu ,{\text{ }}\frac{{{\sigma ^2}}}{{100}}} \right)\) A1A1
Note: Award A1 for \(N\), A1 for the parameters.
(i) \(\bar x = \frac{{\sum x }}{n} = \frac{{3782}}{{100}} = 37.8\) A1
\(s_{n – 1}^2 = \frac{{155341}}{{99}} – \frac{{{{3782}^2}}}{{9900}} = 124\) M1A1
(ii) \(95\% CI = 37.82 \pm 1.98\sqrt {\frac{{124.3006}}{{100}}} \) (M1)(A1)
\( = (35.6,{\text{ }}40.0)\) A1
METHOD 1
one tailed t-test A1
testing \(37.82\) A1
\(99\) degrees of freedom
reject if \(t > 2.36\) A1
t-value being tested is \(2.5294\) A1
since \(2.5294 > 2.36\) we reject the null hypothesis and accept the alternative hypothesis R1
METHOD 2
one tailed t-test (A1)
\(p = 0.00650\) A3
since \(p{\text{ – value}} < 0.01\) we reject the null hypothesis and accept the alternative hypothesis R1