# Question 4:

## Part (a):
Width: 15 minutes is 0.5 cm so 60 mins will be $4 \times 0.5 = \mathbf{2}$ (cm) | B1 | For a width of 2 cm
Height: freq of 25 represented by $6 \times 0.5 = 3\ (\text{cm}^2)$ so freq of 24 is $\frac{24}{25} \times 3$ | M1 | For some calculation linking area and frequency
So height $= \frac{1}{2} \times \frac{24}{25} \times 3 = \mathbf{1.44}$ (cm) | A1 | For 1.44 or exact equivalent e.g. $\frac{36}{25}$
| | (3)

## Part (b):
$[Q_2 =]\ \{30\} + \dfrac{(50 - [25+17])}{28} \times 30$ or e.g. $\dfrac{60-30}{70-42} = \dfrac{m - \{30\}}{50-42}$ | M1 | For $+\frac{8}{28} \times 30$
$= 38.571...\ \ \ \text{awrt}\ \mathbf{38.6}$ | A1 | For awrt 38.6
| | (2)

## Part (c):
Use of midpoints to get $\sum fx = 5070$ (allow 5000 to 1 sf) | M1 |
$[\bar{t}\ \text{or}\ \bar{x}] = \mathbf{50.7}$ | A1 | For 50.7 or exact equivalent e.g. $\frac{5070}{100}$
| | (2)

## Part (d):
$[\sigma] = \sqrt{\dfrac{455512.5}{100} - \text{"50.7"}^2}$ or $\sqrt{1984.635}$ | M1 | For correct expression including square root (ft their mean)
$= 44.5492...\ \ \ \text{awrt}\ \mathbf{44.5}$ | A1 | For awrt 44.5
| | (2)

## Part (e):
$\bar{t}$ or $\bar{x} > Q_2$ [allow "50.7" > "38.6" or formula] so **positive** (skew) | B1 | For positive skew plus a suitable correct reason
| | (1)

## Part (f)(i):
Median: **no change** — since e.g. all 18 values are still below the median | B2/1/0 | All 3 correct: B2; only 2 correct: B1

## Part (f)(ii):
Mean: will be **smaller** — since e.g. changes reduce total of $x$ ($7 \times 7.5$ not $25 \times 7.5$ in 1st class) | B1 |

## Part (f)(iii):
Standard deviation: will be **greater** — since e.g. 18 zeros means data more spread out | | (3) **[13 marks]**

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