# Question 5:

## Part (a)
| Working | Mark | Guidance |
|---------|------|---------|
| Width $=4$ (cm) | B1 | Ignore units |
| Area of $14\text{ cm}^2$ represents frequency 28 and area of $4h$ represents 18; or $\frac{4h}{18}=\frac{14}{28}$ | M1 | Clear method using area and frequency, or width $\times$ height $=9$ |
| $h=\mathbf{2.25}$ (cm) | A1 | |

## Part (b)
| Working | Mark | Guidance |
|---------|------|---------|
| $m=(240)+\frac{10}{22}\times 80$ | M1 | Allow use of $(n+1)$ leading to £278.18 or [278, 278.5] |
| $=276.36\ldots$ $\left(\frac{3040}{11}\right)$; $(£)\mathbf{276}\leq m<(£)276.5$ | A1 | Do not award if incorrect end-point seen |

## Part (c)
| Working | Mark | Guidance |
|---------|------|---------|
| $\sum fy=31600$ leading to $\bar{y}=316$ | M1A1 | Attempt $\sum fy$ with at least 3 correct products or answer rounding to 30000 (1 sf) |
| $\sigma_y=\sqrt{\frac{12452800}{100}-(\bar{y})^2}=157.07\ldots$ (awrt $\mathbf{157}$); allow $s=157.86\ldots$ | M1A1 | Correct expression including $\sqrt{\phantom{x}}$; need $\sum fy^2$ correct; condone minor transcription error |

## Part (d)
| Working | Mark | Guidance |
|---------|------|---------|
| Skewness $=0.764\ldots$ (awrt $\mathbf{0.76}$ or $\mathbf{0.75}$) | B1 | For awrt 0.76/0.75 for $m=£276$ or awrt 0.73/0.72 for $m=£278$ |
| Positive skew | B1ft | Correct description based on their measure, or based on mean and median if no measure given |

## Part (e)
| Working | Mark | Guidance |
|---------|------|---------|
| $z=\pm\frac{80}{150}$ | M1 | Attempt to standardise using 320 and 150 and either 240 or 400 |
| $P(240<X<400)=\mathbf{0.40\sim 0.41}$ | A1 | Answer in range $[0.40, 0.41]$; answer only is 2/2 |

## Part (f)
| Working | Mark | Guidance |
|---------|------|---------|
| (e) suggests a reasonable fit for this range BUT (d) since skew it will not be a good fit overall | B2/1/0 | Need 2 comments referencing each of parts (e) and (d); one comment not good since skew, other since matches range in (e); do not use B0B1 |

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