# Question 6:

## Part 6(a):

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $W = B - 1.1R$ | M1 | May be implied by correct mean or variance |
| $W \sim N(55 - 1.1\times51,\ 1.3^2 + 1.1^2\times1.2^2)$ or $W \sim N(-1.1,\ 3.4324)$ | A1, A1 | ($\pm$)1.1, awrt 3.43 (may be seen in standardisation) |
| $P(W<0) = P\!\left(Z < \frac{0+1.1}{\sqrt{3.4324}}\right)$ | M1 | Standardising with their mean and their sd leading to a probability $> 0.5$ |
| $= P(Z < 0.5937...)$ | | |
| $= 0.7224$ or $0.7237$ | A1 | **awrt 0.72** |

## Part 6(b):

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $X = B_1 - B_2$ | M1 | May be implied by correct mean or variance |
| $X \sim N(55-55,\ 2\times1.3^2)$ or $X \sim N(0,\ 3.38)$ | A1, A1 | 0, 3.38 |
| $P\!\left(Z > \frac{1-0}{\sqrt{3.38}}\right)$ or $P\!\left(Z < \frac{-1-0}{\sqrt{3.38}}\right)$ | dM1 | dep on 1st M1 for standardising with their mean and their sd |
| $P(|X|>1) = 2\times P(X>1)$ | M1 | For $2\times$ seen or implied |
| $= 2\times P(Z>0.5439...) = 2\times(1-0.7054)$ | | $2\times0.2946$ or $2\times0.2932$ (calc) |
| $= 0.5892$ | A1 | **awrt 0.59** |

## Part 6(c):

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $V = B_1+B_2+B_3+B_4+B_5+B_6+B_7+B_8+B_9+B_{10}+S$ and $Y = R_1+R_2+R_3+R_4+R_5+R_6+R_7+R_8+R_9+R_{10}+R_{11}+S$ | M1 | May be implied by either correct distribution |
| $V \sim N(553,\ 16.94)$ and $Y \sim N(564,\ 15.88)$ | A1 | Both correct |
| $D = Y - V$ so $D \sim N(11\times51 - 10\times55,\ 11\times1.2^2 + 10\times1.3^2 + 2\times0.2^2)$ or $D \sim N(11,\ 32.82)$ | M1 | Attempt at their difference for the mean, and their sum for the variance. |
| ($\pm$)11 and awrt 32.8 | A1 | |
| $P(D>0) = P\!\left(Z > \frac{0-11}{\sqrt{32.82}}\right)$ | dM1 | dep on 1st M1 for standardising using their mean and their standard deviation leading to a probability $> 0.5$ |
| $= P(Z > -1.920...)$ | | |
| $= 0.9726$ | A1 | **awrt 0.973** |

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