| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Discrete uniform distribution sample mean |
| Difficulty | Moderate -0.3 This is a straightforward application of the Central Limit Theorem with a discrete uniform distribution. Part (i) is routine variance calculation using the formula, part (ii) requires standard CLT application with n=300 (large sample), and part (iii) asks for textbook justification. All steps are mechanical with no novel insight required, making it slightly easier than average for S2 level. |
| Spec | 2.04f Find normal probabilities: Z transformation5.02b Expectation and variance: discrete random variables5.05a Sample mean distribution: central limit theorem |
| \(x\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( X = x )\) | 0.25 | 0.25 | 0.25 | 0.25 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.25(1 + 4 + 9) - 1.5^2 = 1.25\) AG | B1 [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1.4 - 1.5}{\sqrt{\frac{5}{4} \div 300}}\) \((= -1.549)\) | M1 | \(\frac{1.4 - \frac{1}{600} - 1.5}{\sqrt{\frac{5}{4} \div 300}}\) \((= -1.523)\) |
| \(\Phi("-1.549") = 1 - \Phi("1.549")\) | M1 | \(\Phi("-1.523") = 1 - \Phi("1.523")\) |
| \(= 0.0607\) (3 sf) | A1 [3] | \(= 0.0639\) (3 sf) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Large sample or large \(n\); \(\bar{X}\) (approx) normally distributed; or Central Limit Theorem | B1 [1] |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.25(1 + 4 + 9) - 1.5^2 = 1.25$ **AG** | B1 [1] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1.4 - 1.5}{\sqrt{\frac{5}{4} \div 300}}$ $(= -1.549)$ | M1 | $\frac{1.4 - \frac{1}{600} - 1.5}{\sqrt{\frac{5}{4} \div 300}}$ $(= -1.523)$ |
| $\Phi("-1.549") = 1 - \Phi("1.549")$ | M1 | $\Phi("-1.523") = 1 - \Phi("1.523")$ |
| $= 0.0607$ (3 sf) | A1 [3] | $= 0.0639$ (3 sf) |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Large sample or large $n$; $\bar{X}$ (approx) normally distributed; or Central Limit Theorem | B1 [1] | |
---5 The score on one throw of a 4 -sided die is denoted by the random variable $X$ with probability distribution as shown in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( X = x )$ & 0.25 & 0.25 & 0.25 & 0.25 \\
\hline
\end{tabular}
\end{center}
(i) Show that $\operatorname { Var } ( X ) = 1.25$.
The die is thrown 300 times. The score on each throw is noted and the mean, $\bar { X }$, of the 300 scores is found.\\
(ii) Use a normal distribution to find $\mathrm { P } ( \bar { X } < 1.4 )$.\\
(iii) Justify the use of the normal distribution in part (ii).
\hfill \mbox{\textit{CAIE S2 2014 Q5 [5]}}