| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics B AS (Further Statistics B AS) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Justifying CLT for confidence intervals |
| Difficulty | Standard +0.3 Part (i) is a standard confidence interval calculation using given statistics. Part (ii) tests understanding of CLT justification (large n=70), which is a key conceptual point but straightforward to state. Part (iii) requires working backwards from a confidence interval to find variance, which is slightly less routine but still mechanical algebra. Overall slightly easier than average due to computational nature and clear structure. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05d Confidence intervals: using normal distribution |
| Sample mean | Sample variance |
| 118.86 | 86.57 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (i) | CI is given by |
| Answer | Marks |
|---|---|
| 116.68(cid:100)(cid:80)(cid:100)121.04 | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.4 |
| Answer | Marks |
|---|---|
| 1.1 | For general form |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (ii) | The sample is large |
| Answer | Marks |
|---|---|
| are approximately Normally distributed | E1 |
| Answer | Marks |
|---|---|
| [2] | 2.4 |
| 2.4 | E |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (iii) | s2 |
| Answer | Marks |
|---|---|
| s2 (cid:32)104.96 | M1 |
| Answer | Marks |
|---|---|
| [2] | I3.1b |
| 1.1 | M |
Question 6:
6 | (i) | CI is given by
86.57
118.86(cid:114)1.96(cid:117)
70
116.68(cid:100)(cid:80)(cid:100)121.04 | M1
B1
M1
A1
[4] | 3.4
1.1a
1.1
1.1 | For general form
For 1.96
86.57
For o.e.
N70
6 | (ii) | The sample is large
the Central Limit Theorem states that sample means
are approximately Normally distributed | E1
E1
[2] | 2.4
2.4 | E
For large sample
For mention of use of CLT
6 | (iii) | s2
CI width = 2(cid:117)1.96(cid:117) (cid:32)4.8
70
s2 (cid:32)104.96 | M1
C
A1
[2] | I3.1b
1.1 | M
Or equation based on half CI
width6 The table below shows the mean and variance of the test scores of a random samples of 70 girls who are starting an A level Mathematics course.
\begin{center}
\begin{tabular}{ | c | c | }
\hline
Sample mean & Sample variance \\
\hline
118.86 & 86.57 \\
\hline
\end{tabular}
\end{center}
(i) Showing your working, find a $95 \%$ confidence interval for the population mean.\\
(ii) Explain why you can construct the interval in part (i) despite no information about the distribution of the parent population being given.\\
(iii) The same random sample of girls repeats the test. The mean improvement in score is 0.9 . The $95 \%$ confidence interval for the improvement is $[ - 1.5,3.3 ]$. What is the sample variance for the improvement in score?
\hfill \mbox{\textit{OCR MEI Further Statistics B AS Q6 [8]}}