﻿ WJEC Maths for AS: Applied sample

# WJEC Maths for AS: Applied sample

2.4 Measures of central tendency (i.e. mean, median and mode) 35 Answer 2 (a) μ = ∑  x n = 6 + 6 + 7 + 8 + 8 + 8 + 9 + 9 + 9 + 9 + 10 + 10 + 11 + 12 14 = 122 14 Mean = 8.7 (1 d.p.) (b) The most frequent size as it appears four times. Modal size = 9 (c) The data is already ordered and as there are an even number of values, there are two middle values which are 9 and 9. Median size = 9 3 A farmer records the number of eggs laid each day by his chickens over a 31-day period and produces the following table: Number of eggs ( x ) 12 13 14 15 16 17 18 Frequency ( f ) 3 7 8 10 2 1 0 Find: (a) the mean (b) the median (c) the mode. Answer 3 (a) μ = ∑  fx ∑  f = (3 × 12) + (7 × 13) + (8 × 14) + (10 × 15) + (2 × 16) + (1 × 17) + (0 × 8) 3 + 7 + 8 + 10 + 2 + 1 + 0 = 14.1 (1 d.p.) (b) The data is arranged in order of size. Remember that it is the x -values that are ordered and as the total frequency is 31 (i.e. an odd number) there will be a single number in the middle. To work out the middle value you can think of the 31 values arranged as 15 1 15 so the 16th value is the middle value (i.e. the median). The 16th value will be 14. Median = 14 (c) The most frequent number of eggs = 15 Mode = 15 You have to think of the 31 items of data arranged like this 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 14, … You now look for the 16th value.

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