## Question 5:

**Part (a):**
18–25 group: area $= 7 \times 5 = 35$; 25–40 group: area $= 15 \times 1 = 15$ | B1, B1 (2 marks) |

**Part (b):**
$(25-20)\times 5 + (40-25)\times 1 = 40$ | M1A1 (2 marks) | $5\times5$ is enough evidence of method for M1. Condone 19.5, 20.5 instead of 20 etc.

**Part (c):**
Midpoints are $7.5, 12, 16, 21.5, 32.5$; $\sum f = 100$ | M1, B1 |
$\frac{\sum ft}{\sum f} = \frac{1891}{100} = 18.91$ | M1A1 (4 marks) | Use of some midpoints, at least 3 correct for M1. $\frac{\sum ft}{\sum f}$ for M1 and anything rounding to 18.9 for A1.

**Part (d):**
$\sigma_t = \sqrt{\frac{41033}{100} - \bar{t}^2}$ or $\sqrt{\frac{n}{n-1}\left(\frac{41033}{100} - \bar{t}^2\right)}$ alternative OK | M1, M1 |
$\sigma_t = \sqrt{52.74...} = 7.26$ | A1 (3 marks) | Clear attempt at $\frac{41033}{100} - \bar{t}^2$ or alternative for first M1. They may use their $\bar{t}$ and gain method mark. Square root for second M1. Anything rounding to 7.3 for A1.

**Part (e):**
$Q_2 = 18$ (or 18.1 if $(n+1)$ used) | B1 |
$Q_1 = 10 + \frac{15}{16} \times 4 = 13.75$ (or 15.25 numerator gives 13.8125) | M1A1 |
$Q_3 = 18 + \frac{25}{35} \times 7 = 23$ (or 25.75 numerator gives 23.15) | A1 (4 marks) | Clear attempt at either quartile for M1. These take the form 'lower limit' + correct fraction $\times$ 'class width'. Anything rounding to 13.8 for lower quartile. 23 or anything rounding to 23.2 dependent on method.

**Part (f):**
$0.376...$; Positive skew | B1, B1 (2 marks) | Anything rounding to 0.38 for B1 or 0.33 for B1 if $(n+1)$ used. Correct answer or correct statement following from their value for second B1.