| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Multiple stage process probability |
| Difficulty | Standard +0.3 This is a straightforward application of standard normal distribution techniques and linear combinations. Parts (i)-(iii) require routine standardization and combining normal variables using given formulas (mean and variance of linear combinations). Part (iv) is a standard confidence interval calculation from sample data. Part (v) requires basic statistical reasoning. All techniques are directly taught in S3 with no novel problem-solving required, making this slightly easier than average for A-level statistics. |
| Spec | 5.04b Linear combinations: of normal distributions5.05d Confidence intervals: using normal distribution |
| Mean |
| |||
| Swimming | 11.07 | 2.36 | ||
| Cycling | 57.33 | 8.76 | ||
| Running | 24.23 | 3.75 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S \sim N(11.07,\ 2.36^2)\), \(C \sim N(57.33,\ 8.76^2)\), \(R \sim N(24.23,\ 3.75^2)\) | When a candidate's answers suggest neglect of difference columns of Normal distribution tables, penalise first occurrence only. | |
| \(P(10 < S < 13) = P\!\left(\frac{10-11.07}{2.36} < Z < \frac{13-11.07}{2.36}\right)\) | M1 | For standardising. Award once, here or elsewhere. |
| \(= P(-0.4534 < Z < 0.8178)\) | A1 | |
| \(= 0.7931 - (1 - 0.6748) = 0.4679\) | A1 | Cao. Accept 0.468(0), 0.4681, 0.4682, but not 0.4683. |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Want \(P(R > S + 10)\) i.e. \(P(R - S > 10)\) | M1 | Allow \(S - R\) provided subsequent work is consistent |
| \(R - S \sim N(24.23 - 11.07,\ 3.75^2 + 2.36^2) = N(13.16,\ 19.6321)\) | B1 B1 | Mean. Variance. Accept sd \(= \sqrt{19.6321} = 4.4308\ldots\) |
| \(P(\text{this} > 10) = P\!\left(Z > \frac{10 - 13.16}{\sqrt{19.6321}}\right) = P(Z > -0.7132)\) | ||
| \(= 0.7621\) | A1 | cao |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Want \(P(S + R > \tfrac{2}{3}C)\) i.e. \(P(S + R - \tfrac{2}{3}C > 0)\) | M1 | Allow \(\frac{2}{3}L - (S+R)\) provided subsequent work is consistent |
| \(S + R - \tfrac{2}{3}C \sim N\!\left(11.07 + 24.23 - \tfrac{2}{3}\times 57.33,\ \right.\) \(\left. 2.36^2 + 3.75^2 + \left(\tfrac{2}{3}\times 8.76\right)^2\right) = N(-2.92,\ 53.7377)\) | B1 B1 | Mean. Variance. Accept sd \(= \sqrt{53.7377} = 7.3306\ldots\) |
| \(P(\text{this} > 0) = P\!\left(Z > \frac{0-(-2.92)}{\sqrt{53.7377}}\right) = P(Z > 0.3983)\) | ||
| \(= 1 - 0.6548 = 0.3452\) | A1 | cao |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\bar{x} = 98.484\), \(s_{n-1} = 10.1594\) | B1 | Do not allow \(s_n = 9.7269\) |
| CI given by \(98.484 \pm\) | M1 | ft c's \(\bar{x}\) |
| \(2.201\) | B1 | From \(t_{11}\) |
| \(\times \frac{10.1594}{\sqrt{12}}\) | M1 | ft c's \(s_{n-1}\) |
| \(= 98.484 \pm 6.455 = (92.03,\ 104.94)\) | A1 | cao. Must be expressed as an interval. Require 1 or 2 dp; condone 3dp. |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Normality is unlikely to be reasonable – times could well be (positively) skewed. | E1 | Discussion required. Accept any reasonable point. Accept "reasonable" provided adequate explanation is given. |
| Independence is unlikely to be reasonable – e.g. a competitor who is fast in one stage may well be fast in all three. | E1 | Discussion required. Accept any reasonable point. This is independence between stages for a particular competitor, not between competitors. |
| [2] |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S \sim N(11.07,\ 2.36^2)$, $C \sim N(57.33,\ 8.76^2)$, $R \sim N(24.23,\ 3.75^2)$ | | When a candidate's answers suggest neglect of difference columns of Normal distribution tables, penalise first occurrence only. |
| $P(10 < S < 13) = P\!\left(\frac{10-11.07}{2.36} < Z < \frac{13-11.07}{2.36}\right)$ | M1 | For standardising. Award once, here or elsewhere. |
| $= P(-0.4534 < Z < 0.8178)$ | A1 | |
| $= 0.7931 - (1 - 0.6748) = 0.4679$ | A1 | Cao. Accept 0.468(0), 0.4681, 0.4682, but not 0.4683. |
| **[3]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Want $P(R > S + 10)$ i.e. $P(R - S > 10)$ | M1 | Allow $S - R$ provided subsequent work is consistent |
| $R - S \sim N(24.23 - 11.07,\ 3.75^2 + 2.36^2) = N(13.16,\ 19.6321)$ | B1 B1 | Mean. Variance. Accept sd $= \sqrt{19.6321} = 4.4308\ldots$ |
| $P(\text{this} > 10) = P\!\left(Z > \frac{10 - 13.16}{\sqrt{19.6321}}\right) = P(Z > -0.7132)$ | | |
| $= 0.7621$ | A1 | cao |
| **[4]** | | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Want $P(S + R > \tfrac{2}{3}C)$ i.e. $P(S + R - \tfrac{2}{3}C > 0)$ | M1 | Allow $\frac{2}{3}L - (S+R)$ provided subsequent work is consistent |
| $S + R - \tfrac{2}{3}C \sim N\!\left(11.07 + 24.23 - \tfrac{2}{3}\times 57.33,\ \right.$ $\left. 2.36^2 + 3.75^2 + \left(\tfrac{2}{3}\times 8.76\right)^2\right) = N(-2.92,\ 53.7377)$ | B1 B1 | Mean. Variance. Accept sd $= \sqrt{53.7377} = 7.3306\ldots$ |
| $P(\text{this} > 0) = P\!\left(Z > \frac{0-(-2.92)}{\sqrt{53.7377}}\right) = P(Z > 0.3983)$ | | |
| $= 1 - 0.6548 = 0.3452$ | A1 | cao |
| **[4]** | | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} = 98.484$, $s_{n-1} = 10.1594$ | B1 | Do not allow $s_n = 9.7269$ |
| CI given by $98.484 \pm$ | M1 | ft c's $\bar{x}$ |
| $2.201$ | B1 | From $t_{11}$ |
| $\times \frac{10.1594}{\sqrt{12}}$ | M1 | ft c's $s_{n-1}$ |
| $= 98.484 \pm 6.455 = (92.03,\ 104.94)$ | A1 | cao. Must be expressed as an interval. Require 1 or 2 dp; condone 3dp. |
| **[5]** | | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Normality is unlikely to be reasonable – times could well be (positively) skewed. | E1 | Discussion required. Accept any reasonable point. Accept "reasonable" provided adequate explanation is given. |
| Independence is unlikely to be reasonable – e.g. a competitor who is fast in one stage may well be fast in all three. | E1 | Discussion required. Accept any reasonable point. This is independence between stages for a particular competitor, not between competitors. |
| **[2]** | | |
---3 The triathlon is a sports event in which competitors take part in three stages, swimming, cycling and running, one straight after the other. The winner is the competitor with the shortest overall time. In this question the times for the separate stages are assumed to be Normally distributed and independent of each other.
For a particular triathlon event in which there was a very large number of competitors, the mean and standard deviation of the times, measured in minutes, for each stage were as follows.
\begin{center}
\begin{tabular}{ | l | c | c | }
\hline
& Mean & \begin{tabular}{ c }
Standard \\
deviation \\
\end{tabular} \\
\hline
Swimming & 11.07 & 2.36 \\
\hline
Cycling & 57.33 & 8.76 \\
\hline
Running & 24.23 & 3.75 \\
\hline
\end{tabular}
\end{center}
(i) For a randomly chosen competitor, find the probability that the swimming time is between 10 and 13 minutes.\\
(ii) For a randomly chosen competitor, find the probability that the running time exceeds the swimming time by more than 10 minutes.\\
(iii) For a randomly chosen competitor, find the probability that the swimming and running times combined exceed $\frac { 2 } { 3 }$ of the cycling time.\\
(iv) In a different triathlon event the total times, in minutes, for a random sample of 12 competitors were as follows.
$$\begin{array} { l l l l l l l l l l l l }
103.59 & 99.04 & 85.03 & 81.34 & 106.79 & 89.14 & 98.55 & 98.22 & 108.87 & 116.29 & 102.51 & 92.44
\end{array}$$
Find a 95\% confidence interval for the mean time of all competitors in this event.\\
(v) Discuss briefly whether the assumptions of Normality and independence for the stages of triathlon events are reasonable.
\hfill \mbox{\textit{OCR MEI S3 2012 Q3 [18]}}