Edexcel S2 2010 January — Question 7 11 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2010
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentral limit theorem
TypeDeriving sampling distribution
DifficultyModerate -0.5 This is a straightforward S2 question on sampling distributions with a small discrete population. Part (a) requires basic mean/variance calculation from a simple probability distribution (1p with probability 0.25, 2p with probability 0.75). Parts (b) and (c) involve systematic enumeration of all possible samples of size 3 and calculating their means - tedious but conceptually simple with no novel insight required. Easier than average A-level as it's mostly mechanical computation rather than problem-solving.
Spec5.05a Sample mean distribution: central limit theorem

  1. A bag contains a large number of coins. It contains only \(1 p\) and \(2 p\) coins in the ratio \(1 : 3\)
    1. Find the mean \(\mu\) and the variance \(\sigma ^ { 2 }\) of the values of this population of coins.
    A random sample of size 3 is taken from the bag.
  2. List all the possible samples.
  3. Find the sampling distribution of the mean value of the samples.

AnswerMarks Guidance
(a)x 1p
\(P(X = x)\)\(\frac{1}{4}\) \(\frac{3}{4}\)
\(\mu = 1 \times \frac{1}{4} + 2 \times \frac{3}{4} = \frac{7}{4}\) or \(1\frac{3}{4}\) or \(1.75\)B1 \(\sigma^2 = 1^2 \times \frac{1}{4} + 2^2 \times \frac{3}{4} - \left(\frac{7}{4}\right)^2 = \frac{3}{16}\) or \(0.1875\)
(b) \((1,1,1)\), \((1,1,2)\) any order, \((1,2,2)\) any order, \((2,2,2)\); \((1,2,1)\), \((2,1,1)\), \((2,1,2)\), \((2,2,1)\)B1 B1 May be implied by \(3 \times (1,1,2)\) and \(3 \times (1,2,2)\)
(c)\(\bar{x}\) 1
\(P(\bar{X} = \bar{x})\)\(\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}\) \(3 \times \frac{1}{4} \times \frac{1}{4} \times \frac{3}{4} = \frac{9}{64}\)
B1 M1 A1 M1 A1 A1(6 marks)
Total [11]
**(a)** | x | 1p | 2p |
|---|---|---|
| $P(X = x)$ | $\frac{1}{4}$ | $\frac{3}{4}$ |

$\mu = 1 \times \frac{1}{4} + 2 \times \frac{3}{4} = \frac{7}{4}$ or $1\frac{3}{4}$ or $1.75$ | B1 | $\sigma^2 = 1^2 \times \frac{1}{4} + 2^2 \times \frac{3}{4} - \left(\frac{7}{4}\right)^2 = \frac{3}{16}$ or $0.1875$ | M1 A1 | (3 marks)

**(b)** $(1,1,1)$, $(1,1,2)$ any order, $(1,2,2)$ any order, $(2,2,2)$; $(1,2,1)$, $(2,1,1)$, $(2,1,2)$, $(2,2,1)$ | B1 B1 | May be implied by $3 \times (1,1,2)$ and $3 \times (1,2,2)$ | (2 marks)

**(c)** | $\bar{x}$ | 1 | $\frac{4}{3}$ | $\frac{5}{3}$ | 2 |
|---|---|---|---|---|
| $P(\bar{X} = \bar{x})$ | $\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}$ | $3 \times \frac{1}{4} \times \frac{1}{4} \times \frac{3}{4} = \frac{9}{64}$ | $3 \times \frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{27}{64}$ | $\frac{3}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{27}{64}$ |

| B1 M1 A1 M1 A1 A1 | (6 marks)

| **Total [11]**
\begin{enumerate}
  \item A bag contains a large number of coins. It contains only $1 p$ and $2 p$ coins in the ratio $1 : 3$\\
(a) Find the mean $\mu$ and the variance $\sigma ^ { 2 }$ of the values of this population of coins.
\end{enumerate}

A random sample of size 3 is taken from the bag.\\
(b) List all the possible samples.\\
(c) Find the sampling distribution of the mean value of the samples.\\

\hfill \mbox{\textit{Edexcel S2 2010 Q7 [11]}}