Edexcel S4 — Question 6 14 marks

Exam BoardEdexcel
ModuleS4 (Statistics 4)
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoment generating functions
TypeCompare estimator properties
DifficultyStandard +0.3 This is a standard S4 question on unbiased estimators and variance that follows a predictable template. Parts (a)-(d) involve routine application of Poisson distribution properties (E[X]=Var[X]=λ×area) and basic variance rules. Part (e) requires comparing variances by solving an inequality, which is straightforward algebra. While it requires multiple steps and understanding of statistical estimation, it demands no novel insight—just systematic application of well-practiced techniques from the S4 syllabus.
Spec5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02i Poisson distribution: random events model5.02k Calculate Poisson probabilities5.02m Poisson: mean = variance = lambda

Faults occur in a roll of material at a rate of \(\lambda\) per m\(^2\). To estimate \(\lambda\), three pieces of material of sizes 3 m\(^2\), 7 m\(^2\) and 10 m\(^2\) are selected and the number of faults \(X_1\), \(X_2\) and \(X_3\) respectively are recorded. The estimator \(\hat{\lambda}\), where $$\hat{\lambda} = k(X_1 + X_2 + X_3)$$ is an unbiased estimator of \(\lambda\).
  1. Write down the distributions of \(X_1\), \(X_2\) and \(X_3\) and find the value of \(k\). [4]
  2. Find Var(\(\hat{\lambda}\)). [3]
A random sample of \(n\) pieces of this material, each of size 4 m\(^2\), was taken. The number of faults on each piece, \(Y\), was recorded.
  1. Show that \(\frac{1}{4}\bar{Y}\) is an unbiased estimator of \(\lambda\). [2]
  2. Find Var(\(\frac{1}{4}\bar{Y}\)). [3]
  3. Find the minimum value of \(n\) for which \(\frac{1}{4}\bar{Y}\) becomes a better estimator of \(\lambda\) than \(\hat{\lambda}\). [2]

Question 6:
6
6

Total

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1. George owns a garage and he records the mileage of cars, x thousands of miles, between
services. The results from a random sample of 10 cars are summarised below.
!x = 113.4 !x2 = 1414.08
The mileage of cars between services is normally distributed and George believes that the
standard deviation is 2.4 thousand miles.
Stating your hypotheses clearly, test, at the 5% level of significance, whether or not these
data support George’s belief.
(7)
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AnswerMarks Guidance
EngineerA B
January17 19
July19 18
SiteSample size (n)
Sample mean (x)Standard deviation (s)
A7 8.43
B13 14.31
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4. A random sample of size 2, X and X , is taken from the random variable X which has a
1 2
continuous uniform distribution over the interval [–a, 2a], a > 0
X + X
(a) Show that X= 1 2 is a biased estimator of a and find the bias.
2
(3)
The random variable Y = kX is an unbiased estimator of a.
(b) Write down the value of the constant k.
(1)
(c) Find Var(Y).
(4)
The random variable M is the maximum of X and X
1 2
The probability density function, m(x), of M is given by
⎧2(x + a)
⎪ −a ! x ! 2a
m(x) = ⎨ 9a2
⎩ 0 otherwise
(d) Show that M is an unbiased estimator of a.
(4)
3
Given that E(M2) = a2
2
(e) find Var(M).
(1)
(f) State, giving a reason, whether you would use Y or M as an estimator of a.
(2)
A random sample of two values of X are 5 and –1
(g) Use your answer to part (f) to estimate a.
(1)
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AnswerMarks Guidance
!1.5 2
Power0.59 0.75
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AnswerMarks Guidance
!1.5 2
Power0.59 0.75
Sc i
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(Total 17 marks)
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6. The carbon content, measured in suitable units, of steel is normally distributed. Two
independent random samples of steel were taken from a refining plant at different times
and their carbon content recorded. The results are given below.
Sample A: 1.5 0.9 1.3 1.2
Sample B: 0.4 0.6 0.8 0.3 0.5 0.4
(a) Stating your hypotheses clearly, carry out a suitable test, at the 10% level of
significance, to show that both samples can be assumed to have come from populations
with a common variance &2.
(7)
(b) Showing your working clearly, find the 99% confidence interval for &2 based on both
samples.
(6)
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Question 6:
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Total
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1. George owns a garage and he records the mileage of cars, x thousands of miles, between
services. The results from a random sample of 10 cars are summarised below.
!x = 113.4 !x2 = 1414.08
The mileage of cars between services is normally distributed and George believes that the
standard deviation is 2.4 thousand miles.
Stating your hypotheses clearly, test, at the 5% level of significance, whether or not these
data support George’s belief.
(7)
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Engineer | A | B | C | D | E | F | G | H
January | 17 | 19 | 22 | 26 | 15 | 28 | 18 | 21
July | 19 | 18 | 25 | 24 | 17 | 25 | 16 | 19
Site | Sample size (n) | –
Sample mean (x) | Standard deviation (s)
A | 7 | 8.43 | 4.24
B | 13 | 14.31 | 4.37
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4. A random sample of size 2, X and X , is taken from the random variable X which has a
1 2
continuous uniform distribution over the interval [–a, 2a], a > 0
X + X
(a) Show that X= 1 2 is a biased estimator of a and find the bias.
2
(3)
The random variable Y = kX is an unbiased estimator of a.
(b) Write down the value of the constant k.
(1)
(c) Find Var(Y).
(4)
The random variable M is the maximum of X and X
1 2
The probability density function, m(x), of M is given by
⎧2(x + a)
⎪ −a ! x ! 2a
m(x) = ⎨ 9a2
⎪
⎩ 0 otherwise
(d) Show that M is an unbiased estimator of a.
(4)
3
Given that E(M2) = a2
2
(e) find Var(M).
(1)
(f) State, giving a reason, whether you would use Y or M as an estimator of a.
(2)
A random sample of two values of X are 5 and –1
(g) Use your answer to part (f) to estimate a.
(1)
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! | 1.5 | 2 | 2.5 | 3 | 3.5 | 4
Power | 0.59 | 0.75 | 0.86 | r | 0.96 | 0.97
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! | 1.5 | 2 | 2.5 | 3 | 3.5 | 4
Power | 0.59 | 0.75 | 0.86 | r | 0.96 | 0.97
S | c | i | e | n | ti | s | t’ | s | t | e | st
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6. The carbon content, measured in suitable units, of steel is normally distributed. Two
independent random samples of steel were taken from a refining plant at different times
and their carbon content recorded. The results are given below.
Sample A: 1.5 0.9 1.3 1.2
Sample B: 0.4 0.6 0.8 0.3 0.5 0.4
(a) Stating your hypotheses clearly, carry out a suitable test, at the 10% level of
significance, to show that both samples can be assumed to have come from populations
with a common variance &2.
(7)
(b) Showing your working clearly, find the 99% confidence interval for &2 based on both
samples.
(6)
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Faults occur in a roll of material at a rate of $\lambda$ per m$^2$. To estimate $\lambda$, three pieces of material of sizes 3 m$^2$, 7 m$^2$ and 10 m$^2$ are selected and the number of faults $X_1$, $X_2$ and $X_3$ respectively are recorded.

The estimator $\hat{\lambda}$, where
$$\hat{\lambda} = k(X_1 + X_2 + X_3)$$
is an unbiased estimator of $\lambda$.

\begin{enumerate}[label=(\alph*)]
\item Write down the distributions of $X_1$, $X_2$ and $X_3$ and find the value of $k$.
[4]

\item Find Var($\hat{\lambda}$).
[3]
\end{enumerate}

A random sample of $n$ pieces of this material, each of size 4 m$^2$, was taken. The number of faults on each piece, $Y$, was recorded.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumii}{2}
\item Show that $\frac{1}{4}\bar{Y}$ is an unbiased estimator of $\lambda$.
[2]

\item Find Var($\frac{1}{4}\bar{Y}$).
[3]

\item Find the minimum value of $n$ for which $\frac{1}{4}\bar{Y}$ becomes a better estimator of $\lambda$ than $\hat{\lambda}$.
[2]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S4  Q6 [14]}}