Higher June 2019 Paper 2 Q2
2. Marie works for an optician.
She records the depth of a lens in each of the 100 pairs of glasses on display.
Her results are summarised in the table.
| Depth of lens, \(x\) mm, to the nearest mm | Number of pairs of glasses |
|---|---|
| \(10 \leqslant x \lt 20\) | 5 |
| \(20 \leqslant x \lt 30\) | 20 |
| \(30 \leqslant x \lt 40\) | 23 |
| \(40 \leqslant x \lt 50\) | 52 |
(a)
(i) Calculate an estimate for the mean depth of a lens. [4]
(ii) In which group does the median lie? [1]
(b) In the display of 100 pairs of glasses at Davy’s Opticians, the mean depth of a lens is exactly the same as Marie’s opticians.
Marie says,
“Considering only the mean depth of a lens, our display is certain to be very similar to the display in Davy’s Opticians.”
Explain why Marie is incorrect. [1]
Marie says,
“Considering only the mean depth of a lens, our display is certain to be very similar to the display in Davy’s Opticians.”
Explain why Marie is incorrect. [1]
| Scheme | Marks | Comments |
|---|---|---|
| (a)(i) Mid points: 15, 25, 35, 45 | B1 | |
| \(15\times5 + 25\times20 + 35\times23 + 45\times52\) \((= 75 + 500 + 805 + 2340 = 3720)\) | M1 | FT provided ‘their midpoints’ within the 'bounds' inclusive |
| \(\div 100\) | m1 | |
| 37(.2 mm) | A1 | Unsupported 37.2, award all 4 marks |
| (a)(ii) \(40 \leqslant x \lt 50\) | B1 | Accept any unambiguous indication |
| Scheme | Marks | Comments |
|---|---|---|
| Explanation such as: ‘the mean doesn’t tell you about the spread of the data’, ‘all the depths could be very close to 37.2 mm (whilst Marie’s table shows her display is spread out)’ | E1 | Do not accept indication of ‘Yes’ or an explanation implying ‘Yes’ Do not accept contradictions Allow if range of Marie’s display is used, with lower and upper values considered at 10 mm and 50 mm, or within 1st and last groups Do not accept arguments based on shapes or styles of frames, reasoning and validity of argument must be based only on the depth of lens |
| (6) |