| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Multiple independent binomial calculations |
| Difficulty | Moderate -0.8 This is a straightforward application of standard binomial distribution formulas with no conceptual challenges. Part (a) uses complement rule, (b) is direct formula recall (np(1-p)), (c) applies variance addition property, and (d) requires basic understanding of sampling independence. All calculations are routine with no problem-solving insight required. |
| Spec | 5.02d Binomial: mean np and variance np(1-p)5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | B(20, 0.05) |
| = 1−0.7358 = 0.2642 | B1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 1.1 | For stating correct binomial distribution/calculation |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (b) | Var(X) [= 20×0.05×0.95] = 0.95 |
| [1] | 1.1 | |
| 3 | (c) | Var(Y) = 30×0.07×0.93 [= 1.953] |
| Answer | Marks |
|---|---|
| SD (X + Y) = 1.70 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | FT their Var(X) + their Var(Y), provided their Var(Y) is |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (d) | e.g. Because there could be a fault that affected only some |
| Answer | Marks | Guidance |
|---|---|---|
| sample comes from several batches | B1 | |
| [1] | 2.2b | For a suitable comment that suggests that one batch may not |
Question 3:
3 | (a) | B(20, 0.05)
= 1−0.7358 = 0.2642 | B1
B1
[2] | 3.3
1.1 | For stating correct binomial distribution/calculation
For 0.2642 Allow 0.264
3 | (b) | Var(X) [= 20×0.05×0.95] = 0.95 | B1
[1] | 1.1
3 | (c) | Var(Y) = 30×0.07×0.93 [= 1.953]
Var (X + Y) = 0.95 + 1.953 [=2.903]
SD (X + Y) = 1.70 | M1
M1
A1
[3] | 3.1b
1.1
1.1 | FT their Var(X) + their Var(Y), provided their Var(Y) is
identified.
cao (1.70381…)
3 | (d) | e.g. Because there could be a fault that affected only some
of the batches and it is more likely to be found if the
sample comes from several batches | B1
[1] | 2.2b | For a suitable comment that suggests that one batch may not
be representative.3 At a pottery which manufactures mugs, it is known that $5 \%$ of mugs are faulty. The mugs are produced in batches of 20 . Faults are modelled as occurring randomly and independently. The number of faulty mugs in a batch is denoted by the random variable $X$.
\begin{enumerate}[label=(\alph*)]
\item Determine $\mathrm { P } ( X \geqslant 2 )$.
\item Find $\operatorname { Var } ( X )$.
Independently of the mugs, the pottery also manufactures cups, and it is known that $7 \%$ of cups are faulty. The cups are produced in batches of 30 . Faults are modelled as occurring randomly and independently. The number of faulty cups in a batch is denoted by the random variable $Y$.
\item Determine the standard deviation of $X + Y$.
When 10 batches of cups have been produced, a sample of 15 cups is tested to ensure that the handles of the cups are properly attached.
\item Explain why it might not be sensible to select a sample of 15 cups from the same batch.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2023 Q3 [7]}}