# Question 4:

## Part 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| e.g. The **probability** of an arrow **hitting** the **target** is **constant** | B1 | 1st B1 for **one** suitable comment mentioning "arrows"/"shots"/"hit"/"target", or for **both** comments not in context |
| e.g. The **arrows** are shot **independently** | B1 (2) | 2nd B1 for **both** suitable comments which must mention "arrows"/"shots"/"hit"/"target" at least once |

## Part 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\hat{p} = \dfrac{0\times1 + 1\times10 + 2\times30 + 3\times34 + 4\times17 + 5\times4 + 6\times2 + 7\times0 + 8\times2}{100 \times 8}$ | M1 | For correct attempt at mean or $\hat{p}$ (allow at least 3 correct non-zero products seen) |
| $= \left[\dfrac{2.88}{8}\right] = \mathbf{0.36}$ | A1 (2) | For 0.36 (sight of 2.88 scores M1 and A mark when divided by 8) |

## Part 4(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[\text{Let}\ X \sim B(8,\ \text{awrt "0.36")}]$ $P(X=2) = 0.24936\ldots$ or $P(X=5) = 0.08876\ldots$ | M1 | For sight or use of a correct binomial model to find either prob (ft their 0.36) |
| $(E(2)) = \text{awrt}\ \mathbf{24.94}\ (24.93\text{--}24.95)$ or $(E(5)) = \text{awrt}\ \mathbf{8.88}\ (8.87\text{--}8.89)$ | A1 | For **either** correct expected frequency |
| For using Sum of Expected frequencies $= 100$ (accept awrt 100.01) | B1ft (3) | For two positive values of $r$ and $s$ such that $r + s = 33.82$ or $33.81$ |

## Part 4(d)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Need to pool some values to make all $E_i > 5$ [needn't specify] | B1 | For mention of the need to pool since $E_i < 5$ or to get $E_i > 5$ |
| Two constraints **AND** $p$ was estimated ["2 constraints" $\Leftarrow 5 - 2 = 3$] | B1 (2) | For mention of 2 constraints because $p$ is estimated/calculated from data |

## Part 4(d)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0$: Binomial is a good model for these data; $H_1$: Binomial is not a suitable model for these data | B1 | For both hypotheses correct |
| Pooled expected frequencies: 0 or 1: 15.48; 2: "24.94"; 3: 28.05; 4: 19.73; $\geqslant 5$: "11.80" | M1 | For correct pooling $(0\ \&\ 1)$ and $(\ldots 5)$ (ft their $r$ and $s$); may be implied by awrt 5.18 |
| $\dfrac{(O_i - E_i)^2}{E_i}$: $\dfrac{(11-15.48)^2}{15.48}=1.30$; $\dfrac{(30-\text{"24.94"})^2}{24.94}=1.03$; $\dfrac{(34-28.05)^2}{28.05}=1.26$; $\dfrac{(17-19.73)^2}{19.73}=0.38$; $\dfrac{(8-\text{"11.80"})^2}{11.80}=1.22$ | M1 | For correct use of test statistic (at least one correct calc to at least 2 sf from 3rd or 4th row) |
| Test statistic $= 5.188692\ldots\ \text{awrt}\ \mathbf{5.19}$ (allow awrt 5.18) | A1 | For awrt 5.19 (accept 5.18, 5.20) |
| CV: $\chi^2_3(5\%) = 7.815$ | B1 | For 7.815 (or better); NB $p$-value $= 0.1586\ldots$ and is B0 |
| [Do not reject $H_0$] Data is compatible with Misha's belief/Binomial model is suitable | A1 (6) | dep on both M marks; e.g. $B(8, 0.36)$ is a suitable model is A1 |