| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Combined events across distributions |
| Difficulty | Standard +0.3 This is a straightforward application of normal distribution techniques with standard procedures: (i) basic probability calculation using tables, (ii) combining independent events with clear structure, (iii) sketching distributions, and (iv) reverse normal calculation. All parts use routine methods taught in S2 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.2960\) (4 s.f.) or 0.296 (to 3 s.f.) | M1 for standardizing, A1 for -0.538 and 0.231, M1 for prob. with tables and correct structure, A1 CAO (min 3 s.f., to include use of difference column) | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{One over, one under}) = 0.8220 \times 0.7257 + 0.1780 \times 0.2743 = 0.645\) | B1 for 0.8220 or 0.1780, B1 for 0.2743 or 0.7257, M1 for either product, M1 for sum of both products, A1 CAO | 5 |
| (iii) [Graph showing two normal curves] | G1 for shape, G1 for means, shown explicitly or by scale, G1 for lower max height in young adults, G1 for greater variance in young adults | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sigma = 17.0\) | B1 for 1.175 seen, M1 for equation in \(\sigma\) with z-value, M1 for correct handling of LH tail, A1 cao | 4 |
**(i)** $X \sim N(56, 6.5^2)$
$P(52.5 < X < 57.5) = P\left(\frac{52.5 - 56}{6.5} < Z < \frac{57.5 - 56}{6.5}\right)$
$= P(-0.538 < Z < 0.231)$
$= \Phi(0.231) - (1 - \Phi(0.538))$
$= 0.5914 - (1 - 0.7046)$
$= 0.5914 - 0.2954$
$= 0.2960$ (4 s.f.) or 0.296 (to 3 s.f.) | M1 for standardizing, A1 for -0.538 and 0.231, M1 for prob. with tables and correct structure, A1 CAO (min 3 s.f., to include use of difference column) | 4
**(ii)** $P(5\text{-year-old} < 62) = P\left(Z < \frac{62 - 56}{6.5}\right) = \Phi(0.923) = 0.8220$
$P(\text{young adult} < 62) = P\left(Z < \frac{62 - 68}{10}\right) = \Phi(-0.6) = 1 - 0.7257 = 0.2743$
$P(\text{One over, one under}) = 0.8220 \times 0.7257 + 0.1780 \times 0.2743 = 0.645$ | B1 for 0.8220 or 0.1780, B1 for 0.2743 or 0.7257, M1 for either product, M1 for sum of both products, A1 CAO | 5
**(iii)** [Graph showing two normal curves] | G1 for shape, G1 for means, shown explicitly or by scale, G1 for lower max height in young adults, G1 for greater variance in young adults | 4
**(iv)** $Y \sim N(82, \sigma^2)$
From tables $\Phi^{-1}(0.88) = 1.175$
$\frac{62 - 82}{\sigma} = -1.175$
$-20 = -1.175\sigma$
$\sigma = 17.0$ | B1 for 1.175 seen, M1 for equation in $\sigma$ with z-value, M1 for correct handling of LH tail, A1 cao | 4
---3 In a large population, the diastolic blood pressure (DBP) of 5-year-old children is Normally distributed with mean 56 and standard deviation 6.5.\\
(i) Find the probability that the DBP of a randomly selected 5-year-old child is between 52.5 and 57.5.
The DBP of young adults is Normally distributed with mean 68 and standard deviation 10.\\
(ii) A 5-year-old child and a young adult are selected at random. Find the probability that the DBP of one of them is over 62 and the other is under 62.\\
(iii) Sketch both distributions on a single diagram.\\
(iv) For another age group, the DBP is Normally distributed with mean 82. The DBP of $12 \%$ of people in this age group is below 62. Find the standard deviation for this age group.
\hfill \mbox{\textit{OCR MEI S2 2008 Q3 [17]}}