| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Justifying CLT for sampling distribution |
| Difficulty | Moderate -0.5 This is a straightforward CLT application requiring calculation of a sampling distribution probability and explaining why normality isn't needed due to large sample size (n=120). The conceptual understanding required is standard S2 material with no novel insight needed, making it slightly easier than average. |
| Spec | 5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{6}{\sqrt{120}}\) oe seen | B1 | Or \(6^2/120\) oe seen |
| \(\frac{30-29}{\left(\frac{6}{\sqrt{120}}\right)} = 1.826\) | M1 | \(\pm\) Allow without \(\sqrt{120}\). No sd/var mix |
| \(P(z > \text{`}1.826\text{'}) = 1 - \Phi(\text{`}1.826\text{'}) = 0.034\) (2 sf) | M1, A1 [4] | Correct tail consistent with their working; 0.0339 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| No; \(n\) is large \((\geqslant 30)\) | B1 | 1st B1 for either comment |
| Sample mean is (approx) normally distributed or The CLT applies oe | B1 [2] | 2nd B1 for 'No' with 2nd comment (No mark for 'No' alone) |
## Question 2:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{6}{\sqrt{120}}$ oe seen | B1 | Or $6^2/120$ oe seen |
| $\frac{30-29}{\left(\frac{6}{\sqrt{120}}\right)} = 1.826$ | M1 | $\pm$ Allow without $\sqrt{120}$. No sd/var mix |
| $P(z > \text{`}1.826\text{'}) = 1 - \Phi(\text{`}1.826\text{'}) = 0.034$ (2 sf) | M1, A1 [4] | Correct tail consistent with their working; 0.0339 |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| No; $n$ is large $(\geqslant 30)$ | B1 | 1st B1 for either comment |
| Sample mean is (approx) normally distributed or The CLT applies oe | B1 [2] | 2nd B1 for 'No' with 2nd comment (No mark for 'No' alone) |
---2 The mean and standard deviation of the time spent by people in a certain library are 29 minutes and 6 minutes respectively.\\
(i) Find the probability that the mean time spent in the library by a random sample of 120 people is more than 30 minutes.\\
(ii) Explain whether it was necessary to assume that the time spent by people in the library is normally distributed in the solution to part (i).
\hfill \mbox{\textit{CAIE S2 2015 Q2 [6]}}