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Mean and median

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Mean and median

by vikram4689 » Wed Jun 22, 2011 8:10 am
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If the variables A, B, and C take on the values 1, 2, 3, 4, 5, 6, 7, 8, or 9 with the frequencies indicated by the bars above, for which of the frequency distributions is the mean equal to the median?

(A) A only
(B) B only
(C) C only
(D) A and B only
(E) A and C only

[spoiler]OA:E, but i calculated answer comes out to be B[/spoiler]
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by Frankenstein » Wed Jun 22, 2011 8:28 am
Hi,
In graph problems like that, when the graph is symmetric, you can say mean = median.
Anyway, as you have gone mathematical way I will solve this mathematically

A : I will write the values in a detailed way for this case so that one can understand better:
the values are 1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,4,4,4,4,3,3,3,2,2,1 - total 25 elements
mean = [(1.1 +2.2 + 3.3 +4.4 + 5.5 +6.4 +7.3+ 8.2 +9.1) +5.5]/[2(1+2+3+4)+5] = 125/25 = 5
median is the 13th term as I have already arranged in the increased order. So, median is 5

B: N = 6(4)+4+2(4) = 36
median is average of 18th and 19th terms.
so, median = (3+4)/2 = 3.5
mean = [6(1+2+3+4) + 4.5 + 2(6+7+8+9)]/N = 140/36

C: N = 4+4+6+6+4+6+6+4+4 = 44
median is average of 22nd and 23rd term. i.e. (5+5)/2 =5
mean = [4(1+2+5+8+9)+6(3+4+6+7)]/N = 220/44 =5

Hence, E
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