| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2009 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Non-normal population sample mean (CLT) |
| Difficulty | Standard +0.3 This is a straightforward application of binomial distributions with normal approximation. Part (i) requires basic expectation and variance formulas for independent binomials. Part (ii) involves standard normal approximation with continuity correction and inverse normal lookup. The question is slightly above average difficulty due to the two-variable setup and need to justify approximations, but follows a well-practiced S3 template with no novel insight required. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions5.05a Sample mean distribution: central limit theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use \(B \sim B(29, 0.3)\), \(G \sim B(26, 0.2)\) | M1 | |
| \(E(F) = 29 \times 0.3 + 26 \times 0.2 = 13.9\) | M1A1 | |
| \(\text{Var}(F) = 29 \times 0.3 \times 0.7 + 26 \times 0.2 \times 0.8 = 10.25\) | M1A1 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(B\): \(np = 8.7\), \(nq = 20.3\); \(G\): \(np = 5.2\), \(nq = 20.8\). All exceed 5, so normal approximation valid for each | B2 | Must check numerically. B1 for checking one distribution |
| \(F \sim N(13.9, 10.25)\) (approximately) (Requires \(P(F \leq n) = 0.99\)) | \(\text{M1}\sqrt{}\) | Use normal. May be implied |
| \([n + 0.5 - 13.9]/\sqrt{10.25} = 2.326\), their 10.25 | M1B1 | Standardise. M0 if variance has divisors |
| A1 | cc | |
| \(n = 20.85\) | M1 | Solving similar |
| Need to have 21 spares available | A1 8 | No cc, lose last A1 (\(n=22\)). Wrong cc, lose A1A1 |
| SR Using \(B(55, 0.2527)\): B1; M1(\(N(13.9, 10.39)\)); M1B1M1A0 (Max 5/8) |
# Question 6:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use $B \sim B(29, 0.3)$, $G \sim B(26, 0.2)$ | M1 | |
| $E(F) = 29 \times 0.3 + 26 \times 0.2 = 13.9$ | M1A1 | |
| $\text{Var}(F) = 29 \times 0.3 \times 0.7 + 26 \times 0.2 \times 0.8 = 10.25$ | M1A1 **5** | |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $B$: $np = 8.7$, $nq = 20.3$; $G$: $np = 5.2$, $nq = 20.8$. All exceed 5, so normal approximation valid for each | B2 | Must check numerically. B1 for checking one distribution |
| $F \sim N(13.9, 10.25)$ (approximately) (Requires $P(F \leq n) = 0.99$) | $\text{M1}\sqrt{}$ | Use normal. May be implied |
| $[n + 0.5 - 13.9]/\sqrt{10.25} = 2.326$, their 10.25 | M1B1 | Standardise. M0 if variance has divisors |
| | A1 | cc |
| $n = 20.85$ | M1 | Solving similar |
| Need to have 21 spares available | A1 **8** | No cc, lose last A1 ($n=22$). Wrong cc, lose A1A1 |
| SR Using $B(55, 0.2527)$: B1; M1($N(13.9, 10.39)$); M1B1M1A0 (Max 5/8) | | |
---6 A mathematics examination is taken by 29 boys and 26 girls. Experience has shown that the probability that any boy forgets to bring a calculator to the examination is 0.3 , and that any girl forgets is 0.2 . Whether or not any student forgets to bring a calculator is independent of all other students. The numbers of boys and girls who forget to bring a calculator are denoted by $B$ and $G$ respectively, and $F = B + G$.\\
(i) Find $\mathrm { E } ( F )$ and $\operatorname { Var } ( F )$.\\
(ii) Using suitable approximations to the distributions of $B$ and $G$, which should be justified, find the smallest number of spare calculators that should be available in order to be at least $99 \%$ certain that all 55 students will have a calculator.
\hfill \mbox{\textit{OCR S3 2009 Q6 [13]}}