| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2008 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Direct binomial from normal probability |
| Difficulty | Moderate -0.3 This is a straightforward application of standard normal distribution techniques and basic hypothesis testing. Part (i) is a routine z-score calculation, (ii) applies binomial distribution using the probability from (i), (iii) uses inverse normal tables, and (iv-v) follow a standard one-tailed z-test procedure. All steps are textbook exercises requiring no problem-solving insight, though the multi-part structure and hypothesis test elevate it slightly above pure recall. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.05a Hypothesis testing language: null, alternative, p-value, significance2.05e Hypothesis test for normal mean: known variance |
| Answer | Marks | Guidance |
|---|---|---|
| \(X \sim N(27500, 4000^2)\) | — | |
| \(P(X>25000) = P\left(Z>\frac{25000-27500}{4000}\right)\) | M1 for standardising | |
| \(= P(Z>-0.625)\) | A1 for \(-0.625\), M1 dep for correct tail | |
| \(= \Phi(0.625) = 0.7340\) (3 s.f.) | A1 CAO (must include use of differences) | 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{7 of 10 last more than 25000}) = \binom{10}{7}\times 0.7340^7\times 0.2660^3 = 0.2592\) | M1 for coefficient, M1 for \(0.7340^7\times 0.2660^3\), A1 FT (min 2sf) | 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| From tables \(\Phi^{-1}(0.99) = 2.326\) | B1 for 2.326 seen | |
| \(\frac{k-27500}{4000} = -2.326\) | M1 for equation in \(k\) and negative z-value | |
| \(x = 27500 - 2.326\times 4000 = 18200\) | A1 CAO for awrt 18200 | 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): \(\mu = 27500\); \(H_1\): \(\mu > 27500\) | B1 for use of 27500, B1 for both correct | |
| Where \(\mu\) denotes the mean lifetime of the new tyres | B1 for definition of \(\mu\) | 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Test statistic \(= \frac{28630-27500}{4000/\sqrt{15}} = \frac{1130}{1032.8} = 1.094\) | M1 must include \(\sqrt{15}\), A1 FT | |
| 5% level 1-tailed critical value of \(z = 1.645\) | B1 for 1.645 | |
| \(1.094 < 1.645\) so not significant | M1 dep for sensible comparison leading to a conclusion | |
| There is insufficient evidence to conclude that the new tyres last longer | A1 for conclusion in words in context | 5 marks total |
# Question 3:
## Part (i)
$X \sim N(27500, 4000^2)$ | — |
$P(X>25000) = P\left(Z>\frac{25000-27500}{4000}\right)$ | M1 for standardising |
$= P(Z>-0.625)$ | A1 for $-0.625$, M1 dep for correct tail |
$= \Phi(0.625) = 0.7340$ (3 s.f.) | A1 CAO (must include use of differences) | 4 marks total
## Part (ii)
$P(\text{7 of 10 last more than 25000}) = \binom{10}{7}\times 0.7340^7\times 0.2660^3 = 0.2592$ | M1 for coefficient, M1 for $0.7340^7\times 0.2660^3$, A1 FT (min 2sf) | 3 marks total
## Part (iii)
From tables $\Phi^{-1}(0.99) = 2.326$ | B1 for 2.326 seen |
$\frac{k-27500}{4000} = -2.326$ | M1 for equation in $k$ and negative z-value |
$x = 27500 - 2.326\times 4000 = 18200$ | A1 CAO for awrt 18200 | 3 marks total
## Part (iv)
$H_0$: $\mu = 27500$; $H_1$: $\mu > 27500$ | B1 for use of 27500, B1 for both correct |
Where $\mu$ denotes the mean lifetime of the new tyres | B1 for definition of $\mu$ | 3 marks total
## Part (v)
Test statistic $= \frac{28630-27500}{4000/\sqrt{15}} = \frac{1130}{1032.8} = 1.094$ | M1 must include $\sqrt{15}$, A1 FT |
5% level 1-tailed critical value of $z = 1.645$ | B1 for 1.645 |
$1.094 < 1.645$ so not significant | M1 dep for sensible comparison leading to a conclusion |
There is insufficient evidence to conclude that the new tyres last longer | A1 for conclusion in words in context | 5 marks total
---3 A company has a fleet of identical vans. Company policy is to replace all of the tyres on a van as soon as any one of them is worn out. The random variable $X$ represents the number of miles driven before the tyres on a van are replaced. $X$ is Normally distributed with mean 27500 and standard deviation 4000.\\
(i) Find $\mathrm { P } ( X > 25000 )$.\\
(ii) 10 vans in the fleet are selected at random. Find the probability that the tyres on exactly 7 of them last for more than 25000 miles.\\
(iii) The tyres of $99 \%$ of vans last for more than $k$ miles. Find the value of $k$.
A tyre supplier claims that a different type of tyre will have a greater mean lifetime. A random sample of 15 vans is fitted with these tyres. For each van, the number of miles driven before the tyres are replaced is recorded. A hypothesis test is carried out to investigate the claim. You may assume that these lifetimes are also Normally distributed with standard deviation 4000.\\
(iv) Write down suitable null and alternative hypotheses for the test.\\
(v) For the 15 vans, it is found that the mean lifetime of the tyres is 28630 miles. Carry out the test at the $5 \%$ level.
\hfill \mbox{\textit{OCR MEI S2 2008 Q3 [18]}}