Higher June 2019 Paper 2 Q13
13 The amounts spent on clothes by 40 boys and 40 girls in one month were recorded.
The table shows information about the amounts spent by the boys.
| Amount, \(x\) (£) | Midpoint | Number of boys | |
|---|---|---|---|
| \(0 \leqslant x < 20\) | 22 | ||
| \(20 \leqslant x < 40\) | 9 | ||
| \(40 \leqslant x < 60\) | 6 | ||
| \(60 \leqslant x < 80\) | 3 | ||
| Total = 40 |
The mean for the girls was £35
Estimate the mean for the girls as a percentage of the mean for the boys. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| Any three of [9.5, 10.5] \(\times\) 22 or [209, 231] and [29.5, 30.5] \(\times\) 9 or [265.5, 274.5] and [49.5, 50.5] \(\times\) 6 or [297, 303] and [69.5, 70.5] \(\times\) 3 or [208.5, 211.5] or 1000 | M1 | |
| (their [209, 231] + their [265.5, 274.5] + their [297, 303] + their [208.5, 211.5]) \(\div\) 40 or \(1000 \div 40\) | M1dep | oe condone bracket error if working seen eg \(220 + 270 + 300 + 210 \div 40\) |
| 25 | A1 | |
| \(\dfrac{35}{\text{their } 25}\) or \(\dfrac{7}{5}\) or 1.4 | M1 | oe eg \(1 + \dfrac{35 - \text{their } 25}{\text{their } 25}\) |
| 140 | A1ft | ft their 25 with 3rd M1 scored |
| Alternative method 2 | ||
| Any three of [9.5, 10.5] \(\times\) 22 or [209, 231] and [29.5, 30.5] \(\times\) 9 or [265.5, 274.5] and [49.5, 50.5] \(\times\) 6 or [297, 303] and [69.5, 70.5] \(\times\) 3 or [208.5, 211.5] or 1000 | M1 | |
| \(35 \times 40\) or 1400 | M1 | |
| 1000 and 1400 | A1 | |
| \(\dfrac{\text{their } 1400}{\text{their } 1000}\) or \(\dfrac{7}{5}\) or 1.4 | M1dep | oe eg \(1 + \dfrac{\text{their } 1400 - \text{their } 1000}{\text{their } 1000}\) dep on M2 |
| 140 | A1ft | ft their 1400 and their 1000 with M3 scored |
Additional guidance
| Alt 1 Correct products seen in the table but a different method not using their products used for the mean shown in the working lines eg \(40 \div 4 = 10\) can score a maximum of M0M0A0M1A1ft | |
| Alt 1 \(1000 \div 4\) (= 250) is not a misread | |
| NB The dependency of the M marks and the requirement for applying A1ft are different for the two alternative methods | |
| Alt 1 3rd M1 Allow any number for their 25 (unless it contradicts their mean) | |
| Alt 1 3rd M1 and A1ft If there is a mean for the boys allow the M mark to be implied by a correct ft answer eg from a mean of 250 allow M1A1ft for 14% | |
| For A1ft allow answers to the nearest whole number or better | |
| Further work after working out the percentage is 3rd M0 eg Mean = 25 \(\dfrac{35}{\text{their } 25} \times 100 = 140\) 140 – 100 = 40 Answer 40 | M1M1A1 M0A0 |