| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2012 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Two or more different variables |
| Difficulty | Standard +0.3 This is a standard S3 question testing routine applications of normal distribution properties and linear combinations. Parts (i)-(iv) require straightforward use of mean/variance formulas for sums and differences of independent normals, while part (v) is a textbook confidence interval calculation. All techniques are direct applications with no novel insight required, making it slightly easier than average overall. |
| 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 | Marks | Guidance |
| \(S \sim N(505,\ 11^2)\), \(L \sim N(1005,\ 17^2)\) | When candidate's answers suggest neglect of difference columns, penalise first occurrence only | |
| \(P(995 < L < 1020) = P\!\left(\dfrac{995-1005}{17} < Z < \dfrac{1020-1005}{17}\right)\) | M1 | For standardising. Award once, here or elsewhere |
| \(= P(-0.5882 < Z < 0.8824)\) | A1 | |
| \(= 0.8113 - (1 - 0.7218) = 0.5331\) | A1 | cao |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_1 - S_2 \sim N(0,\ 11^2 + 11^2 = 242)\) | B1 | Mean and variance. Accept \(\text{sd} = \sqrt{242} = 15.55...\) |
| \(P(-25 < S_1 - S_2 < 25) = P\!\left(\dfrac{-25-0}{\sqrt{242}} < Z < \dfrac{25-0}{\sqrt{242}}\right)\) | M1 | Formulate the problem |
| \(= P(-1.607 < Z < 1.607) = 2\times(0.9459 - 0.5) = 0.8918\) | A1 | cao |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Want \(P(S_1 + S_2 > L)\) i.e. \(P(S_1 + S_2 - L > 0)\) | M1 | Allow \(L-(S_1+S_2)\) provided subsequent work is consistent |
| \(S_1 + S_2 - L \sim N(505 + 505 - 1005,\ 11^2+11^2+17^2) = N(5,\ 531)\) | B1, B1 | Mean; Variance. Accept \(\text{sd} = \sqrt{531} = 23.04...\) |
| \(P(\text{this} > 0) = P\!\left(Z > \dfrac{0-5}{\sqrt{531}}\right) = P(Z > -0.2170) = 0.5859\) | A1 | cao |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Want \(P(S > \tfrac{1}{2}L + 5)\) i.e. \(P(S - \tfrac{1}{2}L > 5)\) | M1 | Allow \(\tfrac{1}{2}L - S\) provided subsequent work is consistent |
| \(S - \tfrac{1}{2}L \sim N\!\left(505 - \tfrac{1005}{2},\ 11^2 + \tfrac{17^2}{2^2}\right) = N(2.5,\ 193.25)\) | B1, B1 | Mean; Variance. Accept \(\text{sd} = \sqrt{193.25} = 13.90...\) |
| \(P(\text{this} > 5) = P\!\left(Z > \dfrac{5-2.5}{\sqrt{193.25}}\right) = P(Z > 0.1798) = 1 - 0.5714 = 0.4286\) | A1 | cao |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| CI is given by \(246 \pm\) | M1 | |
| \(1.96\) | B1 | Must be 1.96. Anything else can get M1B0M1A0 max |
| \(\times \dfrac{14}{\sqrt{20}}\) | M1 | |
| \(= 246 \pm 6.1358 = (239.8(6),\ 252.1(3))\) | A1 | cao. Must be expressed as an interval |
| [4] |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S \sim N(505,\ 11^2)$, $L \sim N(1005,\ 17^2)$ | | When candidate's answers suggest neglect of difference columns, penalise first occurrence only |
| $P(995 < L < 1020) = P\!\left(\dfrac{995-1005}{17} < Z < \dfrac{1020-1005}{17}\right)$ | M1 | For standardising. Award once, here or elsewhere |
| $= P(-0.5882 < Z < 0.8824)$ | A1 | |
| $= 0.8113 - (1 - 0.7218) = 0.5331$ | A1 | cao |
| **[3]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_1 - S_2 \sim N(0,\ 11^2 + 11^2 = 242)$ | B1 | Mean and variance. Accept $\text{sd} = \sqrt{242} = 15.55...$ |
| $P(-25 < S_1 - S_2 < 25) = P\!\left(\dfrac{-25-0}{\sqrt{242}} < Z < \dfrac{25-0}{\sqrt{242}}\right)$ | M1 | Formulate the problem |
| $= P(-1.607 < Z < 1.607) = 2\times(0.9459 - 0.5) = 0.8918$ | A1 | cao |
| **[3]** | | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Want $P(S_1 + S_2 > L)$ i.e. $P(S_1 + S_2 - L > 0)$ | M1 | Allow $L-(S_1+S_2)$ provided subsequent work is consistent |
| $S_1 + S_2 - L \sim N(505 + 505 - 1005,\ 11^2+11^2+17^2) = N(5,\ 531)$ | B1, B1 | Mean; Variance. Accept $\text{sd} = \sqrt{531} = 23.04...$ |
| $P(\text{this} > 0) = P\!\left(Z > \dfrac{0-5}{\sqrt{531}}\right) = P(Z > -0.2170) = 0.5859$ | A1 | cao |
| **[4]** | | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Want $P(S > \tfrac{1}{2}L + 5)$ i.e. $P(S - \tfrac{1}{2}L > 5)$ | M1 | Allow $\tfrac{1}{2}L - S$ provided subsequent work is consistent |
| $S - \tfrac{1}{2}L \sim N\!\left(505 - \tfrac{1005}{2},\ 11^2 + \tfrac{17^2}{2^2}\right) = N(2.5,\ 193.25)$ | B1, B1 | Mean; Variance. Accept $\text{sd} = \sqrt{193.25} = 13.90...$ |
| $P(\text{this} > 5) = P\!\left(Z > \dfrac{5-2.5}{\sqrt{193.25}}\right) = P(Z > 0.1798) = 1 - 0.5714 = 0.4286$ | A1 | cao |
| **[4]** | | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| CI is given by $246 \pm$ | M1 | |
| $1.96$ | B1 | Must be 1.96. Anything else can get M1B0M1A0 max |
| $\times \dfrac{14}{\sqrt{20}}$ | M1 | |
| $= 246 \pm 6.1358 = (239.8(6),\ 252.1(3))$ | A1 | cao. Must be expressed as an interval |
| **[4]** | | |
---2 In a particular chain of supermarkets, one brand of pasta shapes is sold in small packets and large packets. Small packets have a mean weight of 505 g and a standard deviation of 11 g . Large packets have a mean weight of 1005 g and a standard deviation of 17 g . It is assumed that the weights of packets are Normally distributed and are independent of each other.\\
(i) Find the probability that a randomly chosen large packet weighs between 995 g and 1020 g .\\
(ii) Find the probability that the weights of two randomly chosen small packets differ by less than 25 g .\\
(iii) Find the probability that the total weight of two randomly chosen small packets exceeds the weight of a randomly chosen large packet.\\
(iv) Find the probability that the weight of one randomly chosen small packet exceeds half the weight of a randomly chosen large packet by at least 5 g .\\
(v) A different brand of pasta shapes is sold in packets of which the weights are assumed to be Normally distributed with standard deviation 14 g . A random sample of 20 packets of this pasta is found to have a mean weight of 246 g . Find a $95 \%$ confidence interval for the population mean weight of these packets.
\hfill \mbox{\textit{OCR MEI S3 2012 Q2 [18]}}