| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Custom discrete distribution sample mean |
| Difficulty | Standard +0.3 This is a straightforward application of the Central Limit Theorem with a discrete distribution. Part (a) is routine variance calculation, part (b) applies CLT with normal tables, part (c) tests conceptual understanding, and part (d) is a standard one-tailed hypothesis test. All steps follow textbook procedures with no novel problem-solving required, making it slightly easier than average for A-level Further Maths Statistics. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| \(x\) | 0 | 1 | 2 | 3 | 4 |
| \(\mathrm { P } ( X = x )\) | 0.1 | 0.2 | 0.4 | 0.2 | 0.1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(E(X) = 2\) | B1 | |
| \(0.2 \times 1 + 0.4 \times 2^2 + 0.2 \times 3^2 + 0.1 \times 4^2 - 2^2 (= 1.2)\) | B1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{a-2}{\sqrt{1.2 \div 200}} = \phi^{-1}(0.9)\) | M1 | M1 for LHS, M1 for RHS |
| (M1 for LHS, M1 for RHS) | M1 | |
| \(a = 2 + \sqrt{1.2 \div 200} \times 1.282\) | M1 | |
| \(2.10\) (3 sf) | A1 |
| Answer | Marks |
|---|---|
| Yes, because \(X\) is not normally distributed. | B1 |
| Answer | Marks |
|---|---|
| \(H_0\): pop mean \(= 2\); \(H_1\): pop mean \(< 2\) | B1 |
| \(\frac{1.86 - 2}{\sqrt{1.2 \div 200}}\) | M1 |
| \(1.807\) | A1 |
| comp \(z = 1.645\) | M1 |
| There is evidence that the spinner is biased so that mean is less than 2 | A1 |
## Question 4:
**Part 4(a):**
| $E(X) = 2$ | B1 | |
| $0.2 \times 1 + 0.4 \times 2^2 + 0.2 \times 3^2 + 0.1 \times 4^2 - 2^2 (= 1.2)$ | B1 | AG |
**Part 4(b):**
| $\frac{a-2}{\sqrt{1.2 \div 200}} = \phi^{-1}(0.9)$ | M1 | M1 for LHS, M1 for RHS |
| (M1 for LHS, M1 for RHS) | M1 | |
| $a = 2 + \sqrt{1.2 \div 200} \times 1.282$ | M1 | |
| $2.10$ (3 sf) | A1 | |
**Part 4(c):**
| Yes, because $X$ is not normally distributed. | B1 | |
**Part 4(d):**
| $H_0$: pop mean $= 2$; $H_1$: pop mean $< 2$ | B1 | |
| $\frac{1.86 - 2}{\sqrt{1.2 \div 200}}$ | M1 | |
| $1.807$ | A1 | |
| comp $z = 1.645$ | M1 | |
| There is evidence that the spinner is biased so that mean is less than 2 | A1 | |
---4 The score on one spin of a 5 -sided spinner is denoted by the random variable $X$ with probability distribution as shown in the table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = x )$ & 0.1 & 0.2 & 0.4 & 0.2 & 0.1 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Show that $\operatorname { Var } ( X ) = 1.2$.\\
The spinner is spun 200 times. The score on each spin is noted and the mean, $\bar { X }$, of the 200 scores is found.
\item Given that $\mathrm { P } ( \bar { X } > a ) = 0.1$, find the value of $a$.
\item Explain whether it was necessary to use the Central Limit theorem in your answer to part (b).
\item Johann has another, similar, spinner. He suspects that it is biased so that the mean score is less than 2 . He spins his spinner 200 times and finds that the mean of the 200 scores is 1.86 .
Given that the variance of the score on one spin of this spinner is also 1.2 , test Johann's suspicion at the 5\% significance level.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2020 Q4 [12]}}