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For the frequency distribution below, what is the position of the upper quartile?
Item size: 1, 2, 3, 4, 5, 6, 7
Frequencies: 2, 4, 5, 8, 7, 3, 2
- the (31+1)/4th observation
- the 2((31+1)/4)th observation
- the 3((31+1)/4)th observation
- the 4((31+1)/4)th observation
Correct answer: the 3((31+1)/4)th observation
Solution
Total frequency N = 2+4+5+8+7+3+2 = 31. The upper (third) quartile is at position 3(N+1)/4 = 3(31+1)/4 th observation.
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