# Question 6:

## Part i(a)
| Working | Mark | Guidance |
|---------|------|----------|
| $n$ is large and $p$ is small | B1 | $n$ is bigger than 10 and $p < 0.25$ |

## Part i(b)
| Working | Mark | Guidance |
|---------|------|----------|
| $[X \sim \text{B}(3000, 0.0025) \rightarrow]\ Y \sim \text{Po}(7.5)$ | B1 | For writing or using Po(7.5) |
| $P(Y > 7) = 1 - P(Y \leq 7) = 1 - 0.5246 = 0.47536...$ awrt **0.475** | M1, A1 | M1 for writing or using $1 - P(Y \leq 7)$ oe |

## Part ii(a)
| Working | Mark | Guidance |
|---------|------|----------|
| A **list/database/register** of **all** employees | B1 | B1 must be **all** employees |

## Part ii(b)
| Working | Mark | Guidance |
|---------|------|----------|
| The **probability** distribution of the **number of employees** that **cycle** to work [from all possible samples of 150] | B1 | Allow **number of employees** that **cycle** to work and their associated **probabilities** |

## Part ii(c)
| Working | Mark | Guidance |
|---------|------|----------|
| $D \sim \text{N}(60, 36)$; $P(C \leq \alpha) = P(Z \leq z) = 0.0668$ | B1 | 1st B1 for Normal approximation with mean $= 60$ and variance $= 36$ / sd $= 6$ |
| $\frac{\alpha - "60"}{" 6"} = -1.5$ or $\frac{"60" - \alpha}{"6"} = 1.5$ | M1, B1 | M1 for standardising with no continuity correction and equating to $z$ using their mean and sd (if not stated clearly they must be correct), with $|z| > 1$; 2nd B1 for $-1.5$ (or 1.5); sign must be compatible with their standardisation; allow if used their 36 instead of 6 |
| $\underline{\alpha = 51}$ | A1cao | |

## Part ii(d)
| Working | Mark | Guidance |
|---------|------|----------|
| $P(C \leq 51) = P(D \leq 51.5)$; $P\!\left(Z \leq \frac{"51" - "60" + 0.5}{"6"}\right)$ | M1 | M1 for standardising with use of continuity correction 'their $51$' $+ 0.5$; allow same mean and sd used in part (c) |
| $= P(Z \leq -1.42) = 1 - 0.9222 = 0.0778/0.0782...$ awrt **0.078** | A1 | |