| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Normal Distribution |
| Type | Probability calculation plus find unknown boundary |
| Difficulty | Standard +0.3 This is a straightforward application of standard Normal distribution techniques: z-score calculations for probabilities, inverse Normal for finding k, and a basic two-tailed hypothesis test with known variance. All procedures are routine S2 content with no novel problem-solving required, though the multi-part structure and hypothesis test execution place it slightly above average difficulty. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (A) \(P(X < 325) = P\!\left(Z < \frac{325-355}{52}\right) = P(Z < -0.577) = 1 - \Phi(0.577) = 1 - 0.7181 = 0.2819\) | M1 standardising, M1 correct structure, A1 CAO | Ignore spurious continuity corrections & allow reversal of numerator. Allow answers which round to 0.282 |
| (B) \(P(300 < X < 400) = P\!\left(\frac{300-355}{52} < Z < \frac{400-355}{52}\right) = P(-1.058 < Z < 0.865)\) | M1 standardising both | Penalise spurious continuity corrections |
| \(= \Phi(0.865) - (1 - \Phi(1.058)) = 0.8065 - (1-0.8549) = 0.6614\) (0.6615 from GDC) | M1 correct structure, A1 CAO | Allow 0.663 if penalised inappropriate table use already. Use of \(sd = \sqrt{52}\) or \(52^2\) can earn M1 for structure only; max 2/6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| From tables \(\Phi^{-1}(0.2) = -0.8416\) | B1 for \(\pm 0.8416\) seen | NOT \(1 - 0.8416\) |
| \(\frac{k - 355}{52} = -0.8416\) | M1 equation in \(k\) | Equation must be equivalent to this. Penalise use of \(+0.8416\) unless numerator reversed. Condone other \(z\) values but use of probabilities e.g. \(\Phi(0.2)=0.5793\) gets M0 A0 |
| \(k = 355 - 0.8416 \times 52 = 311.2\) | A1 CAO | Allow 311 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0: \mu = 355\); \(H_1: \mu \neq 355\). Where \(\mu\) denotes the population mean (reaction time for women) | B1 for use of 355 in hypotheses, B1 both correct, B1 definition of \(\mu\) | Use of 355 in hypotheses and hypotheses given in terms of \(\mu\) not \(p\) or \(x\), etc. unless letter clearly defined as population mean |
| Test statistic \(= \frac{344-355}{52/\sqrt{25}} = \frac{-11}{10.4} = -1.058\) | M1 must include \(\sqrt{25}\), A1 | Allow \(+1.058\) only if later compared with \(+1.96\) |
| 5% level 2-tailed critical value of \(z = 1.96\). \(-1.058 > -1.96\) so not significant. There is not sufficient evidence to reject \(H_0\) | B1 for 1.96, M1 sensible comparison leading to conclusion | Or \(-1.96\) |
| There is insufficient evidence to conclude that women have a different reaction time from men in this experiment | A1 correct conclusion in words in context | Do not accept 'men and women have same reaction time' |
# Question 3 (part i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| (A) $P(X < 325) = P\!\left(Z < \frac{325-355}{52}\right) = P(Z < -0.577) = 1 - \Phi(0.577) = 1 - 0.7181 = 0.2819$ | M1 standardising, M1 correct structure, A1 CAO | Ignore spurious continuity corrections & allow reversal of numerator. Allow answers which round to 0.282 |
| (B) $P(300 < X < 400) = P\!\left(\frac{300-355}{52} < Z < \frac{400-355}{52}\right) = P(-1.058 < Z < 0.865)$ | M1 standardising both | Penalise spurious continuity corrections |
| $= \Phi(0.865) - (1 - \Phi(1.058)) = 0.8065 - (1-0.8549) = 0.6614$ (0.6615 from GDC) | M1 correct structure, A1 CAO | Allow 0.663 if penalised inappropriate table use already. Use of $sd = \sqrt{52}$ or $52^2$ can earn M1 for structure only; max 2/6 |
**Total: 3+3 marks**
# Question 3 (part ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| From tables $\Phi^{-1}(0.2) = -0.8416$ | B1 for $\pm 0.8416$ seen | NOT $1 - 0.8416$ |
| $\frac{k - 355}{52} = -0.8416$ | M1 equation in $k$ | Equation must be equivalent to this. Penalise use of $+0.8416$ unless numerator reversed. Condone other $z$ values but use of probabilities e.g. $\Phi(0.2)=0.5793$ gets M0 A0 |
| $k = 355 - 0.8416 \times 52 = 311.2$ | A1 CAO | Allow 311 |
**Total: 3 marks**
# Question 3 (part iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0: \mu = 355$; $H_1: \mu \neq 355$. Where $\mu$ denotes the population mean (reaction time for women) | B1 for use of 355 in hypotheses, B1 both correct, B1 definition of $\mu$ | Use of 355 in hypotheses and hypotheses given in terms of $\mu$ not $p$ or $x$, etc. unless letter clearly defined as population mean |
| Test statistic $= \frac{344-355}{52/\sqrt{25}} = \frac{-11}{10.4} = -1.058$ | M1 must include $\sqrt{25}$, A1 | Allow $+1.058$ only if later compared with $+1.96$ |
| 5% level 2-tailed critical value of $z = 1.96$. $-1.058 > -1.96$ so not significant. There is not sufficient evidence to reject $H_0$ | B1 for 1.96, M1 sensible comparison leading to conclusion | Or $-1.96$ |
| There is insufficient evidence to conclude that women have a different reaction time from men in this experiment | A1 correct conclusion in words in context | Do not accept 'men and women have same reaction time' |
**Total: 8 marks**3 The random variable $X$ represents the reaction times, in milliseconds, of men in a driving simulator. $X$ is Normally distributed with mean 355 and standard deviation 52.
\begin{enumerate}[label=(\roman*)]
\item Find\\
(A) $\mathrm { P } ( X < 325 )$,\\
(B) $\mathrm { P } ( 300 < X < 400 )$.
\item Find the value of $k$ for which $\mathrm { P } ( X < k ) = 0.2$.
It is thought that women may have a different mean reaction time from men. In order to test this, a random sample of 25 women is selected. The mean reaction time of these women in the driving simulator is 344 milliseconds. You may assume that women's reaction times are also Normally distributed with standard deviation 52 milliseconds. A hypothesis test is carried out to investigate whether women have a different mean reaction time from men.
\item Carry out the test at the $5 \%$ significance level.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2011 Q3 [17]}}