| Answer/Working | Marks | Guidance |
|---|---|---|
| Never, Sometimes, Regularly, Totals; Males: 30, 132, 78, 240; Females: 26, 143, 91, 260 | M1, A1 | convert % to freq; A1 (26, 91, 30, 132) |
| 56, 275, 169, 500 | B1, B1 | |
| $H_0$: No association (independent) between gender and exercise; $H_1$: association (not independent) between gender and exercise | M1, A1 | |
| Expected Values: Males: 26.88, 132, 81.12, 240; Females: 29.12, 143, 87.88, 260; 56, 275, 169, 500 | A1∨ | |
| $\alpha = 0.05$ $\nu = 2$ ; $CV \chi^2 > 5.991$ | | |
| $\sum \frac{(O-E)^2}{E}$ OR $\sum \frac{O^2}{E} - N = 0.9271$ | | answers in range 0.90 – 0.95 |
| Not in critical region – no evidence of association between gender and exercise | | |

**Total: 12 marks**

## Question 6a

| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim B(3, 1/6)$ | B1 | bino |
| 3, 1/6 | B1 | |

**Total: 2 marks**

## Question 6b

| Answer/Working | Marks | Guidance |
|---|---|---|
| Probability distribution: | | prob – must show working and use B(3,p) or may be implied by correct answer |
| X = 0: $\left(\frac{5}{6}\right)^3$ = 144.68 | M1, M1 | expected |
| X = 1: $3 \times \frac{5}{6} \left(\frac{1}{6}\right)$ = 86.81 | | |
| X = 2: $3 \times \frac{5}{6}\left(\frac{1}{6}\right)^2$ = 17.36 | B2 (-1 ee) | awrt 145.86,8,17.4,1.15/1.16 |
| X = 3: $\left(\frac{1}{6}\right)^3$ = 1.15 (1.16) | | |
| $H_0$: Binomial model is a good fit; $H_1$: Binomial model is not a good fit | B1 | both, no ditto |
| Amalgamate 3 with another group | M1 | |
| $\alpha = 0.01$ $\nu = 2$ ; $CR \chi^2 > 9.210$ | B1; B1∨ | |
| $\sum \frac{(O-E)^2}{E}$ OR $\sum \frac{O^2}{E} - N = 8.6894\ldots$ | M1, A1 | answers in range 8.67 – 8.70 or |
| Evidence that Binomial is a good model. | A1∨ | |

**Total: 11 marks**

## Question 6c

| Answer/Working | Marks | Guidance |
|---|---|---|
| Estimate p; Degrees of freedom reduced by 1 | B1, B1 | |
| Special case: Use of B(3, 0.192) in part (b) | | |
| Expected frequencies: 131.8785, 94.01242, 22.339, 1.769 | M1, M1, B0 | |
| $H_0$: Binomial model is a good fit; $H_1$: Binomial model is not a good fit | B1 | both, no ditto |
| Amalgamate 3 with another group | M1 | |
| $\alpha = 0.01$ $\nu = 1$ ; $CR \chi^2 > 6.635$ | B1; B1∨ | |
| $\sum \frac{(O-E)^2}{E}$ OR $\sum \frac{O^2}{E} - N$ in range 5.45 -5.50 | M1, A1 | |
| Evidence that Binomial is a good model. | A1∨ | |

**Total: 11 marks**

## Question 7a

| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(D) = E(A) - 3E(B) + 4E(C) = 20$ | M1, A1 | |
| $\text{Var}(D) = \text{Var}(A) + 9\text{Var}(B) + 16\text{Var}(C)$ | M1, M1 | Use of a²Var X; Adding 3 Var ie 4 +… |
| $= 341$ | A1 | |
| $P(D < 44) = P\left(z < \frac{44-20}{\sqrt{341}}\right)$ | M1, A1∨ | standardising their mean and sd |
| $= P(z < 1.30)$ | A1 | awrt 1.30 |
| $= 0.9032$ | A1 | |

**Total: 9 marks**

## Question 7b

| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = 20$ | B1 | |
| $\text{Var}(X) = \text{Var}(A) + 3\text{Var}(B) + 16\text{Var}(C)$ | M1, M1 | + and 16; 3 Var (B) |
| $= 287$ | A1 | |
| $P(X > 0) = P\left(z > \frac{-20}{\sqrt{287}}\right)$ | M1 | standardising their mean and sd |
| $= P(z > -1.18)$ | A1 | awrt -1.18 |
| $= 0.8810$ | A1 | |

**Total: 7 marks**