| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Estimator properties and bias |
| Difficulty | Standard +0.3 This is a straightforward S3 question testing standard results: (a) is a bookwork proof of unbiasedness requiring simple expectation algebra, (b) applies standard formulas for sample mean and variance, and (c) is a routine normal distribution confidence interval calculation to find sample size. All parts follow textbook methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(\bar{X}) = \frac{1}{n}E(X_1 + \ldots + X_n) = \frac{1}{n}(E(X_1) + \ldots + E(X_n)) = \frac{1}{n}(\mu + \ldots + \mu) = \frac{n\mu}{n} = \mu\) | B1cso | Require \(E(\bar{X}) = \mu\) with at least 1 correct intermediate step and no incorrect working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\bar{x} = \frac{1}{5}(197+203+205+201+195) = 200.2\) g | B1 | B1 for 200.2 or \(\frac{1001}{5}\) |
| \(s^2 = \frac{1}{n-1}\left(\sum x^2 - n\bar{x}^2\right) = \frac{1}{4}(200469 - 5(200.2^2)) = 17.2\) | M1, A1 | M1 for use of correct formula; accept \(\frac{1}{4}S_{xx} = \frac{1}{4} \times 68.8\); A1 awrt 17.2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2 \times 1.25 \geq \text{Width of CI}\), i.e. \(2.5 \geq \frac{2 \times 1.96 \times 4.8}{\sqrt{n}}\) | M1B1 | M1 for use of any equivalent expression; accept equality; accept their \(s\) instead of 4.8; B1 for 1.96 seen with s.e. |
| \(\sqrt{n} \geq \frac{2 \times 1.96 \times 4.8}{2.5} = 7.5264\), so \(n \geq 56.6(5)\) | A1 | 1st A1 for 56.6(5) |
| Minimum sample size is 57 | A1 | Must follow from correct working e.g. \(\sqrt{n} \leq 7.5264\) resulting in \(n=57\) award A0 |
# Question 6:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(\bar{X}) = \frac{1}{n}E(X_1 + \ldots + X_n) = \frac{1}{n}(E(X_1) + \ldots + E(X_n)) = \frac{1}{n}(\mu + \ldots + \mu) = \frac{n\mu}{n} = \mu$ | B1cso | Require $E(\bar{X}) = \mu$ with at least 1 correct intermediate step and no incorrect working |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{x} = \frac{1}{5}(197+203+205+201+195) = 200.2$ g | B1 | B1 for 200.2 or $\frac{1001}{5}$ |
| $s^2 = \frac{1}{n-1}\left(\sum x^2 - n\bar{x}^2\right) = \frac{1}{4}(200469 - 5(200.2^2)) = 17.2$ | M1, A1 | M1 for use of correct formula; accept $\frac{1}{4}S_{xx} = \frac{1}{4} \times 68.8$; A1 awrt 17.2 |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $2 \times 1.25 \geq \text{Width of CI}$, i.e. $2.5 \geq \frac{2 \times 1.96 \times 4.8}{\sqrt{n}}$ | M1B1 | M1 for use of any equivalent expression; accept equality; accept their $s$ instead of 4.8; B1 for 1.96 seen with s.e. |
| $\sqrt{n} \geq \frac{2 \times 1.96 \times 4.8}{2.5} = 7.5264$, so $n \geq 56.6(5)$ | A1 | 1st A1 for 56.6(5) |
| Minimum sample size is 57 | A1 | Must follow from correct working e.g. $\sqrt{n} \leq 7.5264$ resulting in $n=57$ award A0 |6. A random sample $X _ { 1 } , X _ { 2 } , \ldots , X _ { n }$ is taken from a population with mean $\mu$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\bar { X } = \frac { 1 } { n } \left( X _ { 1 } + X _ { 2 } + \ldots + X _ { n } \right)$ is an unbiased estimator of the population mean $\mu$.
A company produces small jars of coffee.
Five jars of coffee were taken at random and weighed.
The weights, in grams, were as follows
$$\begin{array} { l l l l l }
197 & 203 & 205 & 201 & 195
\end{array}$$
\item Calculate unbiased estimates of the population mean and variance of the weights of the jars produced by the company.
It is known from previous results that the weights are normally distributed with standard deviation 4.8 g .
The manager is going to take a second random sample. He wishes to ensure that there is at least a $95 \%$ probability that the estimate of the population mean is within 1.25 g of its true value.
\item Find the minimum sample size required.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2014 Q6 [8]}}