The histogram represents the distribution of the masses, in grams, of individual appre This information is summarised in the table below. Mass (mg): 80 < m ≤ 90, Frequency: 5 Mass (mg): 90 < m ≤ 95, Frequency: 8 Mass (mg): 95 < m ≤ 100, Frequency: p Mass (mg): 100 < m ≤ 102.5, Frequency: q Mass (mg): 102.5 < m ≤ 105, Frequency: 20 Mass (mg): 105 < m ≤ 110, Frequency: 23 Mass (mg): 110 < m ≤ 120, Frequency: 10
- (a)Calculate p and q.[2]
- (b)(i)The mass of each plum in a box is recorded correct to the nearest 5 grams. Mass (to the nearest 5 g): 10-15, Frequency: 6 Mass (to the nearest 5 g): 20-25, Frequency: 18 Mass (to the nearest 5 g): 30-35, Frequency: 25 Mass (to the nearest 5 g): 40-45, Frequency: 10 Mass (to the nearest 5 g): 50-55, Frequency: 1 Calculate an estimate of the mean mass of a plum.[3]
- (b)(ii)Calculate the upper bound for the total mass of plums in the box.[2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2014). The mark or mark type is shown in brackets.
- [2] p = 12, q = 16
- [2] B1 for one correct Or M1 for k× 5 or 1 x 2.5 where k and I are attempts to read from the histogram
- [3] 29.5
- [3] M1 for sum of the midvalues × frequency and M1 for division by 60
- [2] 2070
- [2] M1 for attempt to use upper bounds of individual intervals
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