| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Random Variables |
| Type | Linear combinations of independent variables |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths statistics question requiring understanding of discrete uniform distributions, variance of sample means, and normal approximation. Part (a) requires careful case analysis for even/odd n, part (b) applies standard variance formulas (Var(X̄) = Var(X)/100), and part (c) uses CLT with continuity correction. While it tests multiple concepts and requires careful algebraic manipulation, the techniques are standard for Further Statistics and follow predictable patterns without requiring novel insight. |
| Spec | 5.02e Discrete uniform distribution5.03a Continuous random variables: pdf and cdf5.03b Solve problems: using pdf5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (a) | 1 |
| Answer | Marks |
|---|---|
| 2(𝑛−24) | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | For 𝑛−24 seen |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (b) | Var(X) = 1 ((𝑛−24)2−1) |
| Answer | Marks |
|---|---|
| 1200 | B1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 2.1 | For correct use of formula |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (c) | 75+25 |
| Answer | Marks |
|---|---|
| P(Mean < 47.995) = 0.0866 | B1 |
| Answer | Marks |
|---|---|
| [5] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | For substituting n = 75 into their variance from part (b) |
Question 11:
11 | (a) | 1
For n even, the probability =
2
Number of values of X is 𝑛−24
𝑛+25 𝑛−25
For n odd, the number that are less than is
2 2
𝑛−25
For n odd, the probability =
2(𝑛−24) | B1
M1
M1
A1
[4] | 1.1
3.1a
3.1a
1.1 | For 𝑛−24 seen
𝑛−25
For seen
2
If odd and even answers reversed allow B0M1M1A0
11 | (b) | Var(X) = 1 ((𝑛−24)2−1)
12
Var of mean of 100 values = 1 ((𝑛−24)2−1)
1200
= ( 1 (𝑛2−48𝑛+575))
1200 | B1
B1
[2] | 3.1a
2.1 | For correct use of formula
For division of their Var by 100
11 | (c) | 75+25
Mean = = 50
2
((75−24)2−1) 2600 13
Variance = = = = 2.1666…
1200 1200 6
13
By CLT distribution is approx. N(50, )
6
For P(Mean < 48) use a continuity correction to find
P(Mean < 47.995)
P(Mean < 47.995) = 0.0866 | B1
M1
M1
M1
A1
[5] | 1.1
3.1a
2.2a
3.4
1.1 | For substituting n = 75 into their variance from part (b)
2600 13
NB Standard error = √ = √ = 1.4719… scores M1
1200 6
For distribution with correct mean but their variance
For correct continuity correction
Allow max 4 marks for P(Mean < 48) = 0.0871 (without
correct continuity correction)
Allow equivalent method eg using total rather than mean11 The discrete random variable $X$ has a uniform distribution over the set of all integers between 25 and $n$ inclusive, where $n$ is a positive integer with $n > 25$.
\begin{enumerate}[label=(\alph*)]
\item Determine $\mathrm { P } \left( \mathrm { X } < \frac { \mathrm { n } + 25 } { 2 } \right)$ in each of the following cases.
\begin{itemize}
\item $n$ is even
\item $n$ is odd
\item Determine an expression in terms of $n$ for the variance of the mean of 100 independent values of $X$.
\item Given that $n = 75$, calculate an estimate of the probability that the mean of 100 independent values of $X$ is less than 48 .
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2024 Q11 [11]}}