| Exam Board | OCR MEI |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2007 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Multiple stage process probability |
| Difficulty | Standard +0.3 This is a straightforward application of linear combinations of normal random variables with standard techniques: single normal probability, sum of two normals, difference of normals, and scaling. All parts follow directly from formulas with no conceptual challenges beyond routine S3 material. The multi-part structure adds length but not difficulty. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Mean | Standard deviation | |
| Mowing | 44 | 4.8 |
| Hoeing | 32 | 2.6 |
| Pruning | 21 | 3.7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(M < 50) = P\!\left(Z < \dfrac{50-44}{4.8} = 1.25\right)\) | M1, A1 | For standardising. Award once, here or elsewhere |
| \(= 0.8944\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H + P \sim N(32+21=53,\ 2.6^2+3.7^2=20.45)\) | B1, B1 | Mean; Variance. Accept sd \(= \sqrt{20.45} = 4.522...\) |
| \(P(H+P < 50) = P\!\left(Z < \dfrac{50-53}{\sqrt{20.45}} = -0.6634\right)\) | ||
| \(= 1 - 0.7465 = 0.2535\) | A1 | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Want \(P(M > H+P)\) i.e. \(P(M-(H+P)>0)\) | M1 | Allow \(H+P-M\) provided subsequent work consistent |
| \(M-(H+P) \sim N(44-(32+21)=-9,\ 4.8^2+2.6^2+3.7^2=43.49)\) | B1, B1 | Mean; Variance. Accept sd \(= \sqrt{43.49} = 6.594...\) |
| \(P(\text{this}>0) = P\!\left(Z > \dfrac{0-(-9)}{\sqrt{43.49}} = 1.365\right)\) | ||
| \(= 1 - 0.9139 = 0.0861\) | A1 | c.a.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mean \(= 44+44+32+32+21+21 = 194\) | B1 | |
| Variance \(= 4.8^2+4.8^2+2.6^2+2.6^2+3.7^2+3.7^2 = 86.98\) | B1 | (sd \(= 9.3263...\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(C \sim N(194 \times 0.15 + 10 = 39.10,\ 86.98 \times 0.15^2 = 1.957)\) | M1, M1, A1 | c's mean in (iv) \(\times 0.15\) \(+10\); ft c's mean in (iv) |
| \(P(C \leq 40) = P\!\left(Z \leq \dfrac{40-39.10}{\sqrt{1.957}} = 0.6433\right)\) | A1 | ft c's variance in (iv) |
| \(= 0.7400\) | A1 | c.a.o. |
| Alternatively: \(P(C \leq 40) = P\!\left(\text{total time} \leq \dfrac{40-10}{0.15} = 200 \text{ min}\right)\) | M1, M1, A1 | \(-10\); \(\div 0.15\); c.a.o. |
| \(= P\!\left(Z \leq \dfrac{200-194}{\sqrt{86.98}} = 0.6433\right) = 0.7400\) | M1, A1 | Correct use of c's variance; ft c's mean and variance in (iv) |
# Question 3:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(M < 50) = P\!\left(Z < \dfrac{50-44}{4.8} = 1.25\right)$ | M1, A1 | For standardising. Award once, here or elsewhere |
| $= 0.8944$ | A1 | | [3 marks] |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H + P \sim N(32+21=53,\ 2.6^2+3.7^2=20.45)$ | B1, B1 | Mean; Variance. Accept sd $= \sqrt{20.45} = 4.522...$ |
| $P(H+P < 50) = P\!\left(Z < \dfrac{50-53}{\sqrt{20.45}} = -0.6634\right)$ | | |
| $= 1 - 0.7465 = 0.2535$ | A1 | c.a.o. | [3 marks] |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Want $P(M > H+P)$ i.e. $P(M-(H+P)>0)$ | M1 | Allow $H+P-M$ provided subsequent work consistent |
| $M-(H+P) \sim N(44-(32+21)=-9,\ 4.8^2+2.6^2+3.7^2=43.49)$ | B1, B1 | Mean; Variance. Accept sd $= \sqrt{43.49} = 6.594...$ |
| $P(\text{this}>0) = P\!\left(Z > \dfrac{0-(-9)}{\sqrt{43.49}} = 1.365\right)$ | | |
| $= 1 - 0.9139 = 0.0861$ | A1 | c.a.o. | [4 marks] |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mean $= 44+44+32+32+21+21 = 194$ | B1 | |
| Variance $= 4.8^2+4.8^2+2.6^2+2.6^2+3.7^2+3.7^2 = 86.98$ | B1 | (sd $= 9.3263...$) | [2 marks] |
## Part (v)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $C \sim N(194 \times 0.15 + 10 = 39.10,\ 86.98 \times 0.15^2 = 1.957)$ | M1, M1, A1 | c's mean in (iv) $\times 0.15$ $+10$; ft c's mean in (iv) |
| $P(C \leq 40) = P\!\left(Z \leq \dfrac{40-39.10}{\sqrt{1.957}} = 0.6433\right)$ | A1 | ft c's variance in (iv) |
| $= 0.7400$ | A1 | c.a.o. | [6 marks] |
| **Alternatively:** $P(C \leq 40) = P\!\left(\text{total time} \leq \dfrac{40-10}{0.15} = 200 \text{ min}\right)$ | M1, M1, A1 | $-10$; $\div 0.15$; c.a.o. |
| $= P\!\left(Z \leq \dfrac{200-194}{\sqrt{86.98}} = 0.6433\right) = 0.7400$ | M1, A1 | Correct use of c's variance; ft c's mean and variance in (iv) |
---3 Bill and Ben run their own gardening company. At regular intervals throughout the summer they come to work on my garden, mowing the lawns, hoeing the flower beds and pruning the bushes. From past experience it is known that the times, in minutes, spent on these tasks can be modelled by independent Normally distributed random variables as follows.
\begin{center}
\begin{tabular}{ | l | c | c | }
\hline
& Mean & Standard deviation \\
\hline
Mowing & 44 & 4.8 \\
\hline
Hoeing & 32 & 2.6 \\
\hline
Pruning & 21 & 3.7 \\
\hline
\end{tabular}
\end{center}
(i) Find the probability that, on a randomly chosen visit, it takes less than 50 minutes to mow the lawns.\\
(ii) Find the probability that, on a randomly chosen visit, the total time for hoeing and pruning is less than 50 minutes.\\
(iii) If Bill mows the lawns while Ben does the hoeing and pruning, find the probability that, on a randomly chosen visit, Ben finishes first.
Bill and Ben do my gardening twice a month and send me an invoice at the end of the month.\\
(iv) Write down the mean and variance of the total time (in minutes) they spend on mowing, hoeing and pruning per month.\\
(v) The company charges for the total time spent at 15 pence per minute. There is also a fixed charge of $\pounds 10$ per month. Find the probability that the total charge for a month does not exceed $\pounds 40$.
\hfill \mbox{\textit{OCR MEI S3 2007 Q3 [18]}}