# Question 6:

## Part (i):

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $z = 1.25$ | B1 | For 1.25 or better (calculator gives: 1.25027...) |
| $\frac{187.5 - \mu}{\sigma} = 1.25$ | M1 M1 A1 | **1st M1:** for attempting to use a continuity correction i.e. for sight of $188 \pm 0.5$. **2nd M1:** for standardising using $\mu$ and $\sigma$ or $np$ and $\sqrt{np(1-p)}$ (Condone letter $n$ or any integer > 0). **1st A1:** for a correct equation with compatible signs, allow 1.250... If using a value for $n$ it must be 225 |
| $187.5 - \mu = 1.25\sigma$ | | |
| $\mu = 225p$ | M1 | **3rd M1:** for $\mu = 225p$ seen at any stage in the working |
| $\sigma = \sqrt{225p(1-p)}$ | M1 | **4th M1:** for $\sigma = \sqrt{225p(1-p)}$ seen at any stage in the working. Must be for $\sigma$ not $\sigma^2 = 225p(1-p)$ |
| $(187.5 - 225p)^2 = (1.25)^2 \times 225p(1-p)$ or $(150-180p)^2 = 225p(1-p)$ | M1 | **5th M1:** for squaring to get a quadratic equation in $p$ |
| e.g. $900(5-6p)^2 = 225(p - p^2) \Rightarrow 4(25 - 60p + 36p^2) = p - p^2$ | | |
| Leading to $145p^2 - 241p + 100 = 0*$ | A1* | **2nd A1\*:** dep on all previous Ms and use of 1.25 (with correct sign) for at least 1 correct intermediate step from a correct quadratic equation e.g one of those in scheme for 5th M1 |

## Part (ii):

| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| $\left[(29p-25)(5p-4) = 0 \Rightarrow\right] \quad p = 0.8 \quad \text{or} \quad p = \frac{25}{29}$ (accept: 0.862(0689...)) | M1 | For solving the quadratic correctly — leading to $p = \ldots$ or implied by 0.8 or awrt 0.862 |
| $[p =]\ \mathbf{0.8}$ because 0.862 gives a mean greater than 188 (oe) | A1 | For 0.8 **and** a correct reason to eliminate 0.862 |
| | **(10 marks total)** | |