| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics B AS (Further Statistics B AS) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Justifying CLT for sampling distribution |
| Difficulty | Standard +0.3 This is a straightforward application of the Central Limit Theorem requiring students to (a) apply the CLT formula with given parameters (n=50, μ=4.96, σ²=0.15) and (b) explain why CLT allows this despite unknown population distribution. The calculation is routine and the justification is standard bookwork, making it slightly easier than average but still requiring understanding of when CLT applies. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | Distribution is N(4.96, 0.15/50) |
| P(Mean > 5.00) = 0.233 (0.23260…) | B1 |
| Answer | Marks |
|---|---|
| [3] | For N(4.96, … |
| Answer | Marks |
|---|---|
| BC | Alternative is to calculate |
total > 250 with N(50×4.96,
| Answer | Marks |
|---|---|
| (b) | Because by the central limit theorem, for large |
| Answer | Marks |
|---|---|
| approximately Normal | B1 |
| Answer | Marks |
|---|---|
| [3] | For CLT |
| Answer | Marks |
|---|---|
| For full answer. | Must mention SAMPLE |
Question 1:
1 | (a) | Distribution is N(4.96, 0.15/50)
P(Mean > 5.00) = 0.233 (0.23260…) | B1
M1
A1
[3] | For N(4.96, …
For … 0.15/50)
BC | Alternative is to calculate
total > 250 with N(50×4.96,
0.15×50) or N(248,7.5)
(b) | Because by the central limit theorem, for large
values of n, the distribution of the sample mean is
approximately Normal | B1
B1
B1
[3] | For CLT
For large values of n
For full answer. | Must mention SAMPLE
mean1 It is known that the red blood cell count of adults in a particular country, measured in suitable units, has mean 4.96 and variance 0.15.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that the mean red blood cell count of a random sample of 50 adults from this country is at least 5.00.
\item Explain how you can find the probability in part (a) despite the fact that you do not know the distribution of red blood cell counts.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics B AS 2019 Q1 [6]}}