# Question 4:

## Part (a)
| $\sum ft = 4837.5$ (allow 4838 or 4840) | B1 | For 4837.5 or 4838 or 4840 seen |
| $\text{Mean} = \frac{4837.5}{200} = 24.1875$, awrt $\mathbf{24.2}$ or $\frac{387}{16}$ | M1 A1 | For attempt at $\frac{\sum ft}{\sum f}$; allow 1sf so $\sum f =$ awrt 200 and $\sum ft =$ awrt 5000 |
| $\sigma = \sqrt{\frac{134281.25}{200} - \left(\frac{4837.5}{200}\right)^2} = 9.293\ldots$, accept $s = 9.32$, awrt $\mathbf{9.29}$ | M1 A1 | For correct expression including square root; allow transcription error in 134281.25 but not incorrect re-calculation |

## Part (b)
| $Q_2 = [20.5] + \frac{(100/100.5 - 62)}{88} \times 5 = 22.659\ldots$, awrt $\mathbf{22.7}$ | M1 A1 | For a correct fraction $\times 5$; ignore end point but must be $+$; allow use of $(n+1)$ giving $100.5\ldots$ |

## Part (c)
| $Q_1 = 10.5 + \frac{(50/50.25)}{62} \times 10 [= 18.56]$, $(n+1)$ gives $18.604\ldots$ | B1 cso | For fully correct expression including end point; allow use of $(n+1)$ giving $50.25\ldots$ but use of 50.5 scores B0 |

## Part (d)
| $Q_3 = 25.5$ (use of $n+1$ gives $25.734\ldots$) | B1 | For 25.5 (or awrt 25.7 using $n+1$) |
| $IQR = 6.9$ (use of $n+1$ gives 7.1) | B1 ft | For their $Q_3$ - their $Q_1$ (or 18.6) provided $> 0$; accept awrt 2sf |

## Part (e)
| The data is skewed (condone "negative skew") | B1 | Must mention data is skewed or not symmetrical; do not award for "outliers" |

## Part (f)
| Mean decreases and st. dev. remains the same [Must mention mean and st. dev.] | B1 | For one correct comment |
| The median and quartiles would decrease [Must refer to median and at least $Q_1$] | B1 | For two correct comments |
| The IQR would remain unchanged | B1 | For all 3 correct comments |

---