One day a farmer collected 300 eggs from his chickens. The table below shows the distribution of the masses of the eggs. Mass (m grams) | 42 < m ≤ 46 | 46 < m ≤ 48 | 48 < m ≤ 50 | 50 < m ≤ 54 | 54 < m ≤ 58 | 58 < m ≤ 66 Frequency | 60 | 40 | 48 | 72 | 56 | 24
- (a)(i)An egg is chosen at random. Calculate the probability that the mass of this egg is not greater than 48 grams.[1]
- (a)(ii)An egg is chosen at random from the 300 eggs. Another egg is chosen at random from those that remain. Calculate the probability that the mass of one egg is at most 46 grams, and the mass of the other is more than 58 grams.[2]
- (b)Calculate an estimate of the mean mass of an egg.[3]
- (c)(i)Complete the cumulative frequency table. Mass (m grams) | m ≤ 42 | m ≤ 46 | m ≤ 48 | m ≤ 50 | m ≤ 54 | m ≤ 58 | m ≤ 66 Cumulative Frequency | 0 | 60 | | | | | 300[1]
- (c)(ii)On the grid, draw a smooth cumulative frequency curve to illustrate this information.[2]
- (d)(i)Use your graph to find the median mass of the eggs.[1]
- (d)(ii)Use your graph to find the interquartile range.[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 (May/June 2015). The mark or mark type is shown in brackets.
- [1] 1/3 oe
- [2] 48/1495 oe; M1 for (2x)(60/300 x 24/299) After 0, allow SC1 for 2 x 60/300 x 24/300
- [3] 50.8; M1 for 15240, or 2640+1880+2352+3744+3136+1488, or 44×60+47×40+49×48+52×72+56×56+62×24; B1 for division by 300
- [1] 100 148 220 276
- [2] 7 correct plots and smooth curve; B1 for 5 correct plots
- [1] 50 to 50.5
- [2] 7.25 to 8.00; B1 for 46.5 to 47.0 or 54.25 to 54.50 seen or their reading at 225, or 75 seen
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