| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | June |
| 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 and linear combinations. Parts (i)-(iii) require standard z-score calculations and understanding that the sum of independent normals is normal. Part (iv) is a routine confidence interval calculation. All techniques are standard S3 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04b Linear combinations: of normal distributions5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(C < 9.5) = P\!\left(Z < \dfrac{9.5-10}{0.4} = -1.25\right)\) | M1 A1 | For standardising. Award once here or elsewhere |
| \(= 1 - 0.8944 = 0.1056\) | A1 | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(D - S = D - (C_1+C_2+C_3+C_4) \sim N(-5,\ \sigma^2)\) | B1 | Mean. Accept \(+5\) for \(S-D\) |
| \(\sigma^2 = 3.5^2 + (0.4^2+0.4^2+0.4^2+0.4^2) = 12.89\) | B1 | Variance. Accept sd \((= 3.590\ldots)\) |
| Want \(P(D>S) = P(D-S>0)\) | M1 | Formulation of requirement. Accept \(S-D<0\). This mark could be awarded in (iii) if not earned here |
| \(= 1 - \Phi\!\left(\dfrac{0-(-5)}{3.59} = 1.39(27)\right)\) | ||
| \(= 1 - 0.9182 = 0.0818\) | A1 | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| New \((D-S) = (D \times 1.3)-(C_1+\ldots+C_5) \sim N(-4.5,\ \sigma^2)\) | B1 | Mean. Accept \(+4.5\) for \(S-D\) |
| \(\sigma^2 = (3.5^2 \times 1.3^2)+(0.4^2+\ldots+0.4^2) = 21.5025\) | M1 A1 | Correct use of \(\times 1.3^2\) for variance. c.a.o. Accept sd \((= 4.637\ldots)\) |
| Again want \(P(D>S) = P(D-S>0)\) | Or \(S-D<0\). M1 for formulation in (ii) available here | |
| \(= 1 - \Phi\!\left(\dfrac{0-(-4.5)}{4.637} = 0.9704\right)\) | ||
| \(= 1 - 0.8341 = 0.1659\) | A1 | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| CI given by \(9.73 \pm\) | M1 | |
| \(1.96\) | B1 | 1.96 seen |
| \(\times \dfrac{0.4}{\sqrt{12}}\) | M1 | |
| \(= 9.73 \pm 0.2263 = (9.50(37),\ 9.95(63))\) | A1 | c.a.o. Must be expressed as an interval |
| Since 10 lies above this interval, it seems the cheeses are underweight | E1 | ft c's interval |
| In repeated sampling, 95% of all confidence intervals constructed in this way will contain the true mean | E1 E1 |
# Question 4:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(C < 9.5) = P\!\left(Z < \dfrac{9.5-10}{0.4} = -1.25\right)$ | M1 A1 | For standardising. Award once here or elsewhere |
| $= 1 - 0.8944 = 0.1056$ | A1 | c.a.o. |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $D - S = D - (C_1+C_2+C_3+C_4) \sim N(-5,\ \sigma^2)$ | B1 | Mean. Accept $+5$ for $S-D$ |
| $\sigma^2 = 3.5^2 + (0.4^2+0.4^2+0.4^2+0.4^2) = 12.89$ | B1 | Variance. Accept sd $(= 3.590\ldots)$ |
| Want $P(D>S) = P(D-S>0)$ | M1 | Formulation of requirement. Accept $S-D<0$. This mark could be awarded in (iii) if not earned here |
| $= 1 - \Phi\!\left(\dfrac{0-(-5)}{3.59} = 1.39(27)\right)$ | | |
| $= 1 - 0.9182 = 0.0818$ | A1 | c.a.o. |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| New $(D-S) = (D \times 1.3)-(C_1+\ldots+C_5) \sim N(-4.5,\ \sigma^2)$ | B1 | Mean. Accept $+4.5$ for $S-D$ |
| $\sigma^2 = (3.5^2 \times 1.3^2)+(0.4^2+\ldots+0.4^2) = 21.5025$ | M1 A1 | Correct use of $\times 1.3^2$ for variance. c.a.o. Accept sd $(= 4.637\ldots)$ |
| Again want $P(D>S) = P(D-S>0)$ | | Or $S-D<0$. M1 for formulation in (ii) available here |
| $= 1 - \Phi\!\left(\dfrac{0-(-4.5)}{4.637} = 0.9704\right)$ | | |
| $= 1 - 0.8341 = 0.1659$ | A1 | c.a.o. |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| CI given by $9.73 \pm$ | M1 | |
| $1.96$ | B1 | 1.96 seen |
| $\times \dfrac{0.4}{\sqrt{12}}$ | M1 | |
| $= 9.73 \pm 0.2263 = (9.50(37),\ 9.95(63))$ | A1 | c.a.o. Must be expressed as an interval |
| Since 10 lies above this interval, it seems the cheeses are underweight | E1 | ft c's interval |
| In repeated sampling, 95% of all confidence intervals constructed in this way will contain the true mean | E1 E1 | |4 The weights of Avonley Blue cheeses made by a small producer are found to be Normally distributed with mean 10 kg and standard deviation 0.4 kg .\\
(i) Find the probability that a randomly chosen cheese weighs less than 9.5 kg .
One particular shop orders four Avonley Blue cheeses each week from the producer. From experience, the shopkeeper knows that the weekly demand from customers for Avonley Blue cheese is Normally distributed with mean 35 kg and standard deviation 3.5 kg . In the interests of food hygiene, no cheese is kept by the shopkeeper from one week to the next.\\
(ii) Find the probability that, in a randomly chosen week, demand from customers for Avonley Blue will exceed the supply.
Following a campaign to promote Avonley Blue cheese, the shopkeeper finds that the weekly demand for it has increased by $30 \%$ (i.e. the mean and standard deviation are both increased by $30 \%$ ). Therefore the shopkeeper increases his weekly order by one cheese.\\
(iii) Find the probability that, in a randomly chosen week, demand will now exceed supply.\\
(iv) Following complaints, the cheese producer decides to check the mean weight of the Avonley Blue cheeses. For a random sample of 12 cheeses, she finds that the mean weight is 9.73 kg . Assuming that the population standard deviation of the weights is still 0.4 kg , find a $95 \%$ confidence interval for the true mean weight of the cheeses and comment on the result. Explain what is meant by a 95\% confidence interval.
RECOGNISING ACHIEVEMENT
\hfill \mbox{\textit{OCR MEI S3 2011 Q4 [18]}}