| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2016 |
| 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 +0.8 This question requires understanding of linear combinations of normal variables across four multi-step parts, including working with means of samples, sums of independent normals, and comparing linear combinations. While the individual calculations are systematic once set up, part (iii) requires forming an inequality between two sums, and part (iv) involves comparing means which requires careful algebraic manipulation. The conceptual demand of correctly identifying distributions and their parameters across multiple scenarios elevates this above a standard question, though it remains within reach for well-prepared S3 students. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Type of round | Mean | Standard deviation |
| Starter | 200 | 15 |
| Middle | 220 | 25 |
| Final | 250 | 20 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F \sim N(250, 20^2)\) | ||
| \(P(\bar{F}_4 > 260) = P\left(Z > \frac{260-250}{10}\right)\) | M1 | standardisation including division by \(\sqrt{n}\) |
| \(= P(Z > 1)\) | M1 | correct tail (probability < 0.5) |
| \(= 0.1587\) | A1 | cao (to 3 or 4 sf) |
| Answer | Marks | Guidance |
|---|---|---|
| \((F_1+F_2+F_3+F_4) = F' \sim N(1000, 1600)\) | ||
| \((M_1+M_2+\ldots+M_{12}) = M' \sim N(2640, 7500)\) | ||
| \((S_1+S_2+S_3+S_4) = S' \sim N(800, 900)\) | M1 | for variances: at least one of \(4 \times 15^2\) etc. seen; allow \(4\times15 + 12\times25 + 4\times20\), but not \(4^2\) etc. |
| \(\rightarrow T \sim N(4440, 100^2)\) | A1 | for 10,000 (or 2.778 in minutes) |
| B1 | for 4440 (or 74 in minutes) | |
| \(P(T < 4500) = P\left(Z < \frac{4500-4440}{100}\right) = P(Z < 0.6)\) | M1 | correct tail (probability > 0.5) and \(\sqrt{\text{their variance}}\) |
| \(= 0.7258\) | A1 | art 0.726 (given to 3 or 4 sf) |
| Answer | Marks | Guidance |
|---|---|---|
| Looking for \(P\big((M' - 3.5S') > 0\big)\) | M1 | interpret the question correctly; e.g. \(12M - 3.5\times4S\) or \(12M > 3.5\times4S\) seen |
| \([M' \sim N(2640, 7500)]\) | ||
| \(3.5S' \sim N(2800, 11025)\) | M1 | |
| \((M' - 3.5S') \sim N(-160, 18525)\) | B1, A1 | mean and variance |
| \(= P\left(Z > \frac{160}{\sqrt{18525}}\right) = P(Z > 1.1755) = 0.1199\) | A1 | cao (0.1198 to 0.120) |
| Answer | Marks | Guidance |
|---|---|---|
| Looking for \(P\big((\bar{F}_4 - \bar{M}_{12}) > 25\big)\) | M1 | interpret the question correctly; e.g. \(P(\bar{F}_4 > \bar{M}_{12} + 25)\) seen |
| \(\bar{M}_{12} \sim N\left(220, \frac{625}{12}\right)\) | ||
| \(\bar{F}_4 \sim N(250, 100)\) | M1 | variance: at least one of \(\frac{25^2}{12}\) or \(\frac{20^2}{4}\) seen |
| \((\bar{F}_4 - \bar{M}_{12}) \sim N(30, 152.08)\) | A1 | correct variance |
| B1 | correct mean | |
| \(P\left(Z > \frac{25-30}{\sqrt{152.08}}\right) = P(Z > -0.4054) = 0.6574\) | A1 | answer rounds to 0.657 or 0.658 |
# Question 1:
## Part i:
$F \sim N(250, 20^2)$ | | |
$P(\bar{F}_4 > 260) = P\left(Z > \frac{260-250}{10}\right)$ | M1 | standardisation including division by $\sqrt{n}$
$= P(Z > 1)$ | M1 | correct tail (probability < 0.5)
$= 0.1587$ | A1 | cao (to 3 or 4 sf)
## Part ii:
$(F_1+F_2+F_3+F_4) = F' \sim N(1000, 1600)$ | |
$(M_1+M_2+\ldots+M_{12}) = M' \sim N(2640, 7500)$ | |
$(S_1+S_2+S_3+S_4) = S' \sim N(800, 900)$ | M1 | for variances: at least one of $4 \times 15^2$ etc. seen; allow $4\times15 + 12\times25 + 4\times20$, but not $4^2$ etc.
$\rightarrow T \sim N(4440, 100^2)$ | A1 | for 10,000 (or 2.778 in minutes)
| B1 | for 4440 (or 74 in minutes)
$P(T < 4500) = P\left(Z < \frac{4500-4440}{100}\right) = P(Z < 0.6)$ | M1 | correct tail (probability > 0.5) and $\sqrt{\text{their variance}}$
$= 0.7258$ | A1 | art 0.726 (given to 3 or 4 sf)
## Part iii:
Looking for $P\big((M' - 3.5S') > 0\big)$ | M1 | interpret the question correctly; e.g. $12M - 3.5\times4S$ or $12M > 3.5\times4S$ seen
$[M' \sim N(2640, 7500)]$ | |
$3.5S' \sim N(2800, 11025)$ | M1 |
$(M' - 3.5S') \sim N(-160, 18525)$ | B1, A1 | mean and variance
$= P\left(Z > \frac{160}{\sqrt{18525}}\right) = P(Z > 1.1755) = 0.1199$ | A1 | cao (0.1198 to 0.120)
## Part iv:
Looking for $P\big((\bar{F}_4 - \bar{M}_{12}) > 25\big)$ | M1 | interpret the question correctly; e.g. $P(\bar{F}_4 > \bar{M}_{12} + 25)$ seen
$\bar{M}_{12} \sim N\left(220, \frac{625}{12}\right)$ | |
$\bar{F}_4 \sim N(250, 100)$ | M1 | variance: at least one of $\frac{25^2}{12}$ or $\frac{20^2}{4}$ seen
$(\bar{F}_4 - \bar{M}_{12}) \sim N(30, 152.08)$ | A1 | correct variance
| B1 | correct mean
$P\left(Z > \frac{25-30}{\sqrt{152.08}}\right) = P(Z > -0.4054) = 0.6574$ | A1 | answer rounds to 0.657 or 0.658
---1 A game consists of 20 rounds. Each round is denoted as either a starter, middle or final round. The times taken for each round are independently and Normally distributed with the following parameters (given in seconds).
\begin{center}
\begin{tabular}{ | l | c | c | }
\hline
Type of round & Mean & Standard deviation \\
\hline
Starter & 200 & 15 \\
\hline
Middle & 220 & 25 \\
\hline
Final & 250 & 20 \\
\hline
\end{tabular}
\end{center}
The game consists of 4 starter, 12 middle and 4 final rounds. Find the probability that\\
(i) the mean time per round for the 4 final rounds will exceed 260 seconds,\\
(ii) all 20 rounds will be completed in a total time of 75 minutes or less,\\
(iii) the 12 middle rounds will take at least 3.5 times as long in total as the 4 starter rounds,\\
(iv) the mean time per round for the 12 middle rounds will be at least 25 seconds less than the mean time per round for the 4 final rounds.
\hfill \mbox{\textit{OCR MEI S3 2016 Q1 [18]}}