| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2013 |
| Session | June |
| 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 application of normal distribution properties and linear combinations. Parts (i)-(iii) are routine calculations using standardization and the sum of independent normals. Part (iv) requires forming a linear combination 0.65M - 0.35C, which is slightly more sophisticated but still a textbook technique for S3/Further Stats. The algebraic manipulation to show the inequality is straightforward, and all steps follow standard procedures without requiring novel insight. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(C \sim N(96, 21)\) \(M \sim N(57, 14)\) | When a candidate's answers suggest that (s)he appears to have neglected to use the difference columns of the Normal distribution tables penalise the first occurrence only. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(90 < C < 100)\) | ||
| \(= P\left(\frac{90-96}{\sqrt{21}} < Z < \frac{100-96}{\sqrt{21}}\right)\) | M1 | For standardising. Award once, here or elsewhere. SC – candidates with consistent variances of \(21^2\) and \(14^2\) can be awarded all M and B marks |
| \(= P(-1.3093 < Z < 0.8729)\) | A1 | Either side correct. SC \(-0.2857, 0.1905\) |
| \(= 0.8086 - (1 - 0.9047)\) | A1 | Both table values correct. Or \(0.8086 - 0.0953\) SC \(0.5755-(1-0.6125)\) |
| \(= 0.7133\) | A1 | c.a.o. |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Total weight \(T \sim N(153, 35)\) | B1 | Mean. |
| B1 | Variance. Accept sd \(= 5.916\ldots\) SC \(637\) sd \(= 25.239\) | |
| \(P(T < 145) = P\left(Z < \frac{145-153}{\sqrt{35}} = -1.3522\right)\) | ||
| \(= 1 - 0.9118 = 0.0882\) | A1 | c.a.o. |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T_1 + T_2 + T_3 + T_4 \sim N(612, 140)\) | B1 | Mean. |
| B1 | Variance. Accept sd \(= 11.832\ldots\) SC \(= 2548\) sd \(= 50.478\) | |
| Require \(w\) such that \(P(\text{this} > w) = 0.95\) | M1 | |
| \(\therefore w = 612 - 1.645 \times \sqrt{140}\) | B1 | \(1.645\) seen. |
| \(= 592.5(3)\) | A1 | c.a.o. |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Require \(M \geq 0.35(M + C)\) | M1 | Formulate requirement. |
| \(\therefore 0.65M \geq 0.35C\) | ||
| \(\therefore 0.65M - 0.35C \geq 0\) | A1 | Convincingly shown. Beware printed answer. |
| \(0.65M - 0.35C \sim N((0.65 \times 57) - (0.35 \times 96)) = 3.45,\) | B1 | Mean. |
| \((0.65^2 \times 14) + (0.35^2 \times 21) = 8.4875)\) | M1 | For use of at least one of \(0.65^2 \times \ldots\) or \(0.35^2 \times \ldots\) |
| A1 | Variance. Accept sd \(= 2.913\ldots\) SC variance \(= 136.83\) sd \(= 11.70\) | |
| \(P(\text{This} \geq 0) = P\left(Z \geq \frac{0-3.45}{\sqrt{8.4875}} = -1.1842\right)\) | ||
| \(= 0.8818\) | A1 | c.a.o. |
| [6] |
# Question 4:
| Answer | Mark | Guidance |
|--------|------|----------|
| $C \sim N(96, 21)$ $M \sim N(57, 14)$ | | When a candidate's answers suggest that (s)he appears to have neglected to use the difference columns of the Normal distribution tables penalise the first occurrence only. |
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(90 < C < 100)$ | | |
| $= P\left(\frac{90-96}{\sqrt{21}} < Z < \frac{100-96}{\sqrt{21}}\right)$ | M1 | For standardising. Award once, here or elsewhere. SC – candidates with consistent variances of $21^2$ and $14^2$ can be awarded all M and B marks |
| $= P(-1.3093 < Z < 0.8729)$ | A1 | Either side correct. SC $-0.2857, 0.1905$ |
| $= 0.8086 - (1 - 0.9047)$ | A1 | Both table values correct. Or $0.8086 - 0.0953$ SC $0.5755-(1-0.6125)$ |
| $= 0.7133$ | A1 | c.a.o. |
| **[4]** | | |
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Total weight $T \sim N(153, 35)$ | B1 | Mean. |
| | B1 | Variance. Accept sd $= 5.916\ldots$ SC $637$ sd $= 25.239$ |
| $P(T < 145) = P\left(Z < \frac{145-153}{\sqrt{35}} = -1.3522\right)$ | | |
| $= 1 - 0.9118 = 0.0882$ | A1 | c.a.o. |
| **[3]** | | |
## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $T_1 + T_2 + T_3 + T_4 \sim N(612, 140)$ | B1 | Mean. |
| | B1 | Variance. Accept sd $= 11.832\ldots$ SC $= 2548$ sd $= 50.478$ |
| Require $w$ such that $P(\text{this} > w) = 0.95$ | M1 | |
| $\therefore w = 612 - 1.645 \times \sqrt{140}$ | B1 | $1.645$ seen. |
| $= 592.5(3)$ | A1 | c.a.o. |
| **[5]** | | |
## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| Require $M \geq 0.35(M + C)$ | M1 | Formulate requirement. |
| $\therefore 0.65M \geq 0.35C$ | | |
| $\therefore 0.65M - 0.35C \geq 0$ | A1 | Convincingly shown. Beware printed answer. |
| $0.65M - 0.35C \sim N((0.65 \times 57) - (0.35 \times 96)) = 3.45,$ | B1 | Mean. |
| $(0.65^2 \times 14) + (0.35^2 \times 21) = 8.4875)$ | M1 | For use of at least one of $0.65^2 \times \ldots$ or $0.35^2 \times \ldots$ |
| | A1 | Variance. Accept sd $= 2.913\ldots$ SC variance $= 136.83$ sd $= 11.70$ |
| $P(\text{This} \geq 0) = P\left(Z \geq \frac{0-3.45}{\sqrt{8.4875}} = -1.1842\right)$ | | |
| $= 0.8818$ | A1 | c.a.o. |
| **[6]** | | |4 A company that makes meat pies includes a "small" size in its product range. These pies consist of a pastry case and meat filling, the weights of which are independent of each other. The weight of the pastry case, $C$, is Normally distributed with mean 96 g and variance $21 \mathrm {~g} ^ { 2 }$. The weight of the meat filling, $M$, is Normally distributed with mean 57 g and variance $14 \mathrm {~g} ^ { 2 }$.\\
(i) Find the probability that, in a randomly chosen pie, the weight of the pastry case is between 90 and 100 g .\\
(ii) The wrappers on the pies state that the weight is 145 g . Find the proportion of pies that are underweight.\\
(iii) The pies are sold in packs of 4 . Find the value of $w$ such that, in $95 \%$ of packs, the total weight of the 4 pies in a randomly chosen pack exceeds $w \mathrm {~g}$.\\
(iv) It is required that the weight of the meat filling in a pie should be at least $35 \%$ of the total weight. Show that this means that $0.65 M - 0.35 C \geqslant 0$. Hence find the probability that, in a randomly chosen pie, this requirement is met.
\hfill \mbox{\textit{OCR MEI S3 2013 Q4 [18]}}