| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2008 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Distribution of scaled variable |
| Difficulty | Standard +0.3 This is a straightforward application of standard normal distribution techniques (scaling, linear combinations, and confidence intervals). All parts follow routine procedures taught in S3 with no novel problem-solving required. The multi-part structure and variety of techniques (transformation, sum of normals, confidence interval) make it slightly above average, but each individual step is mechanical. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X < 300) = P\left(Z < \frac{300-260}{24} = 1.6667\right)\) | M1, A1 | For standardising. Award once, here or elsewhere. |
| \(= 0.9522\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(Y \sim N(260 \times 0.6 = 156,\ 24^2 \times 0.6^2 = 207.36)\) | B1, B1 | Mean. Variance. Accept sd \((= 14.4)\). |
| \(P(Y > 175) = P\left(Z > \frac{175-156}{14.4} = 1.3194\right)\) | ||
| \(= 1 - 0.9063 = 0.0937\) | A1 | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(Y_1 + Y_2 + Y_3 + Y_4 \sim N(624,\ 829.44)\) | B1, B1 | Mean. Ft mean of (ii). Variance. Accept sd \((= 28.8)\). Ft variance of (ii). |
| \(P(\text{this} < 600) = P\left(Z < \frac{600-624}{28.8} = -0.8333\right)\) | ||
| \(= 1 - 0.7976 = 0.2024\) | A1 | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Require \(w\) such that \(0.975 = P(\text{above} > w) = P\left(Z > \frac{w-624}{28.8}\right)\) | M1 | Formulation of requirement. |
| \(-1.96\) | B1 | |
| \(\therefore w - 624 = 28.8 \times -1.96 \Rightarrow w = 567.5(52)\) | A1 | Ft parameters of (iii). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(On \sim N(150,\ \sigma = 18)\); \(X_1 + X_2 + X_3 + On_1 + On_2 \sim N(1080,\ 2376)\) | B1, B1 | Mean. Variance. Accept sd \((= 48.744)\). |
| \(P(\text{this} > 1000) = P\left(Z > \frac{1000-1080}{48.744} = -1.6412\right)\) | ||
| \(= 0.9496\) | A1 | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Given \(\bar{x} = 252.4\), \(s_{n-1} = 24.6\); CI given by \(252.4 \pm 2.576 \times \frac{24.6}{\sqrt{100}}\) | M1 | Correct use of \(252.4\) and \(24.6/\sqrt{100}\). |
| B1 | For 2.576. | |
| \(= 252.4 \pm 6.33(6) = (246.0(63),\ 258.7(36))\) | A1 | c.a.o. Must be expressed as an interval. |
# Question 2:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X < 300) = P\left(Z < \frac{300-260}{24} = 1.6667\right)$ | M1, A1 | For standardising. Award once, here or elsewhere. |
| $= 0.9522$ | A1 | |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $Y \sim N(260 \times 0.6 = 156,\ 24^2 \times 0.6^2 = 207.36)$ | B1, B1 | Mean. Variance. Accept sd $(= 14.4)$. |
| $P(Y > 175) = P\left(Z > \frac{175-156}{14.4} = 1.3194\right)$ | | |
| $= 1 - 0.9063 = 0.0937$ | A1 | c.a.o. |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $Y_1 + Y_2 + Y_3 + Y_4 \sim N(624,\ 829.44)$ | B1, B1 | Mean. Ft mean of (ii). Variance. Accept sd $(= 28.8)$. Ft variance of (ii). |
| $P(\text{this} < 600) = P\left(Z < \frac{600-624}{28.8} = -0.8333\right)$ | | |
| $= 1 - 0.7976 = 0.2024$ | A1 | c.a.o. |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| Require $w$ such that $0.975 = P(\text{above} > w) = P\left(Z > \frac{w-624}{28.8}\right)$ | M1 | Formulation of requirement. |
| $-1.96$ | B1 | |
| $\therefore w - 624 = 28.8 \times -1.96 \Rightarrow w = 567.5(52)$ | A1 | Ft parameters of (iii). |
## Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| $On \sim N(150,\ \sigma = 18)$; $X_1 + X_2 + X_3 + On_1 + On_2 \sim N(1080,\ 2376)$ | B1, B1 | Mean. Variance. Accept sd $(= 48.744)$. |
| $P(\text{this} > 1000) = P\left(Z > \frac{1000-1080}{48.744} = -1.6412\right)$ | | |
| $= 0.9496$ | A1 | c.a.o. |
## Part (vi)
| Answer | Mark | Guidance |
|--------|------|----------|
| Given $\bar{x} = 252.4$, $s_{n-1} = 24.6$; CI given by $252.4 \pm 2.576 \times \frac{24.6}{\sqrt{100}}$ | M1 | Correct use of $252.4$ and $24.6/\sqrt{100}$. |
| | B1 | For 2.576. |
| $= 252.4 \pm 6.33(6) = (246.0(63),\ 258.7(36))$ | A1 | c.a.o. Must be expressed as an interval. |
---2 In the vegetable section of a local supermarket, leeks are on sale either loose (and unprepared) or prepared in packs of 4 .
The weights of unprepared leeks are modelled by the random variable $X$ which has the Normal distribution with mean 260 grams and standard deviation 24 grams. The prepared leeks have had $40 \%$ of their weight removed, so that their weights, $Y$, are modelled by $Y = 0.6 X$.\\
(i) Find the probability that a randomly chosen unprepared leek weighs less than 300 grams.\\
(ii) Find the probability that a randomly chosen prepared leek weighs more than 175 grams.\\
(iii) Find the probability that the total weight of 4 randomly chosen prepared leeks in a pack is less than 600 grams.\\
(iv) What total weight of prepared leeks in a randomly chosen pack of 4 is exceeded with probability 0.975 ?\\
(v) Sandie is making soup. She uses 3 unprepared leeks and 2 onions. The weights of onions are modelled by the Normal distribution with mean 150 grams and standard deviation 18 grams. Find the probability that the total weight of her ingredients is more than 1000 grams.\\
(vi) A large consignment of unprepared leeks is delivered to the supermarket. A random sample of 100 of them is taken. Their weights have sample mean 252.4 grams and sample standard deviation 24.6 grams. Find a $99 \%$ confidence interval for the true mean weight of the leeks in this consignment.
\hfill \mbox{\textit{OCR MEI S3 2008 Q2 [18]}}