| Exam Board | OCR MEI |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | June |
| Marks | 19 |
| 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 multi-part normal distribution question covering standard techniques: z-score calculations, independent events (raising probability to power 5), inverse normal for percentiles, and a one-tailed hypothesis test with known variance. All parts are routine S2 material requiring only direct application of formulas with no novel problem-solving, making it slightly easier than average A-level difficulty. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.05e Hypothesis test for normal mean: known variance |
| Answer | Marks | Guidance |
|---|---|---|
| Part (i) | (A) \(P(X < 65) = P\left(Z < \frac{65-63}{5.2}\right) = P(Z < 0.3846) = \Phi(0.3846) = 0.6497\) | M1 for standardizing; M1 for structure; A1 CAO (min 3 s.f.) NB When a candidate's answers suggest that (s)he appears to have neglected to use the difference column of the Normal distribution tables penalise the first occurrence only |
| (B) \(P(60 < X < 65) = P\left(\frac{60-63}{5.2} < Z < \frac{65-63}{5.2}\right)\) | M1 for standardizing both | |
| \(= P(-0.5769 < Z < 0.3846) = \Phi(0.3846) - (1 - \Phi(0.5769)) = 0.6497 - (1 - 0.7181) = 0.3678\) | M1 for correct structure; A1 CAO 3s.f. | 3 |
| Part (ii) | \(P(\text{All 5 between 60 and 65}) = 0.3678^5 = 0.00673\) | M1 A1 FT (min 2sf) |
| Part (iii) | From tables \(\Phi^{-1}(0.95) = 1.645\) | B1 for ±1.645 seen; M1 for correct equation in \(k\) |
| \(\frac{k-63}{5.2} = -1.645\) | ||
| \(x = 63 - 5.2 \times 1.645 = 54.45\) mins | A1 CAO | 3 |
| Part (iv) | \(H_0: \mu = 63\) minutes; \(H_1: \mu < 63\) minutes. Where \(\mu\) denotes the population mean time on the new course. | B1 for use of 63; B1 for both correct; B1 for definition of \(\mu\) |
| Test statistic = \(\frac{61.7-63}{5.2/\sqrt{15}} = \frac{-1.3}{1.3426} = -0.968\) | M1 must include \(\sqrt{15}\); A1 | |
| 5% level 1 tailed critical value of \(z = 1.645\) | B1 for ±1.645 | |
| \(-0.968 > -1.645\) so not significant. There is not sufficient evidence to reject \(H_0\) | M1 for sensible comparison leading to a conclusion | |
| There is insufficient evidence to conclude that the new course results in lower times. | A1 FT for correct conclusion in words in context | 5 |
| TOTAL | 19 |
| **Part (i)** | **(A)** $P(X < 65) = P\left(Z < \frac{65-63}{5.2}\right) = P(Z < 0.3846) = \Phi(0.3846) = 0.6497$ | M1 for standardizing; M1 for structure; A1 CAO (min 3 s.f.) NB When a candidate's answers suggest that (s)he appears to have neglected to use the difference column of the Normal distribution tables penalise the first occurrence only | 3 |
| | **(B)** $P(60 < X < 65) = P\left(\frac{60-63}{5.2} < Z < \frac{65-63}{5.2}\right)$ | M1 for standardizing both | |
| | $= P(-0.5769 < Z < 0.3846) = \Phi(0.3846) - (1 - \Phi(0.5769)) = 0.6497 - (1 - 0.7181) = 0.3678$ | M1 for correct structure; A1 CAO 3s.f. | 3 |
| **Part (ii)** | $P(\text{All 5 between 60 and 65}) = 0.3678^5 = 0.00673$ | M1 A1 FT (min 2sf) | 2 |
| **Part (iii)** | From tables $\Phi^{-1}(0.95) = 1.645$ | B1 for ±1.645 seen; M1 for correct equation in $k$ | |
| | $\frac{k-63}{5.2} = -1.645$ | | |
| | $x = 63 - 5.2 \times 1.645 = 54.45$ mins | A1 CAO | 3 |
| **Part (iv)** | $H_0: \mu = 63$ minutes; $H_1: \mu < 63$ minutes. Where $\mu$ denotes the population mean time on the new course. | B1 for use of 63; B1 for both correct; B1 for definition of $\mu$ | |
| | Test statistic = $\frac{61.7-63}{5.2/\sqrt{15}} = \frac{-1.3}{1.3426} = -0.968$ | M1 must include $\sqrt{15}$; A1 | |
| | 5% level 1 tailed critical value of $z = 1.645$ | B1 for ±1.645 | |
| | $-0.968 > -1.645$ so not significant. There is not sufficient evidence to reject $H_0$ | M1 for sensible comparison leading to a conclusion | |
| | There is insufficient evidence to conclude that the new course results in lower times. | A1 FT for correct conclusion in words in context | 5 |
| | | **TOTAL** | **19** |
---3 In a men's cycling time trial, the times are modelled by the random variable $X$ minutes which is Normally distributed with mean 63 and standard deviation 5.2.
\begin{enumerate}[label=(\roman*)]
\item Find
$$\begin{aligned}
& \text { (A) } \mathrm { P } ( X < 65 ) \text {, } \\
& \text { (B) } \mathrm { P } ( 60 < X < 65 ) \text {. }
\end{aligned}$$
\item Find the probability that 5 riders selected at random all record times between 60 and 65 minutes.
\item A competitor aims to be in the fastest $5 \%$ of entrants (i.e. those with the lowest times). Find the maximum time that he can take.
It is suggested that holding the time trial on a new course may result in lower times. To investigate this, a random sample of 15 competitors is selected. These 15 competitors do the time trial on the new course. The mean time taken by these riders is 61.7 minutes. You may assume that times are Normally distributed and the standard deviation is still 5.2 minutes. A hypothesis test is carried out to investigate whether times on the new course are lower.
\item Write down suitable null and alternative hypotheses for the test. Carry out the test at the 5\% significance level.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S2 2010 Q3 [19]}}