| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Single sum threshold probability |
| Difficulty | Standard +0.3 This is a straightforward application of normal distribution properties with standard transformations. Parts (i)-(ii) involve basic standardization and Z-score lookups. Part (iii) uses the sum of independent normals (a standard result). Part (iv) requires finding a percentile. All techniques are routine for S3 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.03c Calculate mean/variance: by integration5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X < 170) = P\left(Z < \frac{170-180}{12} = -0.8333\right) = 1 - 0.7976 = 0.2024\) | M, A1, A1 | For standardising. Award once, here or elsewhere. |
| Answer | Marks | Guidance |
|---|---|---|
| \(X_1 + X_2 + X_3 + X_4 + X_5 \sim N(900, \sigma^2 = 720[\sigma = 26.8328])\) | B1, B1 | Mean. Variance. Accept sd. |
| \(P(\text{this} < 840) = P\left(Z < \frac{840-900}{26.8328} = -2.236\right) = 1 - 0.9873 = 0.0127\) | A1 | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| \(Y \sim N(50, \sigma = 6)\) | ||
| \(X + Y \sim N(230, \sigma^2 = 180[\sigma = 13.4164])\) | B1, B1 | Mean. Variance. Accept sd. |
| \(P(\text{this} > 240) = P\left(Z > \frac{240-230}{13.4164} = 0.7454\right) = 1 - 0.7720 = 0.2280\) | A1 | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{4}X \sim N\left(45, \sigma^2 = \frac{1}{16} \cdot 144 = 9[\sigma = 3]\right)\) | B1 | Variance. Accept sd. FT incorrect mean. |
| Require \(t\) such that | M | Formulation of requirement. |
| \(0.9 = P(\text{this} < t) = P\left(Z < \frac{t-45}{3}\right) = P(Z < 1.282)\) | 1.282 | |
| \(\therefore t - 45 = 3 \times 1.282 \Rightarrow t = 48.85(48.846)\) | B1, A1 | ft only for incorrect mean |
| Answer | Marks | Guidance |
|---|---|---|
| \(I = 45 + T\) where \(T \sim N(120, \sigma = 10)\) | ||
| \(\therefore I \sim N(165, \sigma = 10)\) | B1 | for unchanged \(\sigma\) (candidates might work with \(P(T < 105)\)) |
| \(P(I < 150) = P\left(Z < \frac{150-165}{10} = -1.5\right) = 1 - 0.9332 = 0.0668\) | A1 | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| \(J = 30 + \frac{3}{5}T\) where \(T \sim N(120, \sigma = 10)\) | ||
| \(\therefore J \sim N\left(102, \sigma^2 = \frac{9}{25} \times 100 = 36[\sigma = 6]\right)\) | B1, B1 | Mean. Variance. Accept sd. |
| \(P(J < 105) = P\left(Z < \frac{105-102}{6} = 0.5\right) = 0.6915\) | A1 | c.a.o. |
## (i)
$P(X < 170) = P\left(Z < \frac{170-180}{12} = -0.8333\right) = 1 - 0.7976 = 0.2024$ | M, A1, A1 | For standardising. Award once, here or elsewhere. | 3 marks
## (ii)
$X_1 + X_2 + X_3 + X_4 + X_5 \sim N(900, \sigma^2 = 720[\sigma = 26.8328])$ | B1, B1 | Mean. Variance. Accept sd. | 3 marks
$P(\text{this} < 840) = P\left(Z < \frac{840-900}{26.8328} = -2.236\right) = 1 - 0.9873 = 0.0127$ | A1 | c.a.o. |
## (iii)
$Y \sim N(50, \sigma = 6)$ | | |
$X + Y \sim N(230, \sigma^2 = 180[\sigma = 13.4164])$ | B1, B1 | Mean. Variance. Accept sd. | 3 marks
$P(\text{this} > 240) = P\left(Z > \frac{240-230}{13.4164} = 0.7454\right) = 1 - 0.7720 = 0.2280$ | A1 | c.a.o. |
## (iv)
$\frac{1}{4}X \sim N\left(45, \sigma^2 = \frac{1}{16} \cdot 144 = 9[\sigma = 3]\right)$ | B1 | Variance. Accept sd. FT incorrect mean. | 4 marks
Require $t$ such that | M | Formulation of requirement. |
$0.9 = P(\text{this} < t) = P\left(Z < \frac{t-45}{3}\right) = P(Z < 1.282)$ | | 1.282 |
$\therefore t - 45 = 3 \times 1.282 \Rightarrow t = 48.85(48.846)$ | B1, A1 | ft only for incorrect mean |
## (v)
$I = 45 + T$ where $T \sim N(120, \sigma = 10)$ | | |
$\therefore I \sim N(165, \sigma = 10)$ | B1 | for unchanged $\sigma$ (candidates might work with $P(T < 105)$) | 2 marks
$P(I < 150) = P\left(Z < \frac{150-165}{10} = -1.5\right) = 1 - 0.9332 = 0.0668$ | A1 | c.a.o. |
## (vi)
$J = 30 + \frac{3}{5}T$ where $T \sim N(120, \sigma = 10)$ | | |
$\therefore J \sim N\left(102, \sigma^2 = \frac{9}{25} \times 100 = 36[\sigma = 6]\right)$ | B1, B1 | Mean. Variance. Accept sd. | 3 marks
$P(J < 105) = P\left(Z < \frac{105-102}{6} = 0.5\right) = 0.6915$ | A1 | c.a.o. |
---Geoffrey is a university lecturer. He has to prepare five questions for an examination. He knows by experience that it takes about 3 hours to prepare a question, and he models the time (in minutes) taken to prepare one by the Normally distributed random variable $X$ with mean 180 and standard deviation 12, independently for all questions.
\begin{enumerate}[label=(\roman*)]
\item One morning, Geoffrey has a gap of 2 hours 50 minutes (170 minutes) between other activities. Find the probability that he can prepare a question in this time. [3]
\item One weekend, Geoffrey can devote 14 hours to preparing the complete examination paper. Find the probability that he can prepare all five questions in this time. [3]
\end{enumerate}
A colleague, Helen, has to check the questions.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item She models the time (in minutes) to check a question by the Normally distributed random variable $Y$ with mean 50 and standard deviation 6, independently for all questions and independently of $X$. Find the probability that the total time for Geoffrey to prepare a question and Helen to check it exceeds 4 hours. [3]
\item When working under pressure of deadlines, Helen models the time to check a question in a different way. She uses the Normally distributed random variable $\frac{1}{2}X$, where $X$ is as above. Find the length of time, as given by this model, which Helen needs to ensure that, with probability 0.9, she has time to check a question. [4]
\end{enumerate}
Ian, an educational researcher, suggests that a better model for the time taken to prepare a question would be a constant $k$ representing "thinking time" plus a random variable $T$ representing the time required to write the question itself, independently for all questions.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{4}
\item Taking $k$ as 45 and $T$ as Normally distributed with mean 120 and standard deviation 10 (all units are minutes), find the probability according to Ian's model that a question can be prepared in less than 2 hours 30 minutes. [2]
\end{enumerate}
Juliet, an administrator, proposes that the examination should be reduced in time and shorter questions should be used.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{5}
\item Juliet suggests that Ian's model should be used for the time taken to prepare such shorter questions but with $k = 30$ and $T$ replaced by $\frac{2}{3}T$. Find the probability as given by this model that a question can be prepared in less than $1\frac{1}{4}$ hours. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S3 2006 Q2 [18]}}