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Eve Dunn 2022-04-06 Answered
μ = m = q 1 + q 3 2
Let X be a random variable with p.m.f./p.d.f. f X ( x ) that is symmetric about μ ( R ) , i.e., f X ( x + μ ) = f X ( μ x ), x ( , ) .
If q 1 , m and q 3 are respectively the lower quartile, the median and the upper quartile of the distribution of X then show that μ = m = q 1 + q 3 2
How to prove.....
any Hints.
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Answers (2)

Mackenzie Zimmerman
Answered 2022-04-07 Author has 10 answers
Say, q 1 is the lower quartile, i.e. P ( X < q 1 ) = 1 / 4, that is, q 1 f X ( x ) d x = 1 / 4. By symmetry about μ, (by   t 2 μ t), we also have 2 μ q 1 f X ( x ) d x = 1 / 4, that is, P ( X > 2 μ q 1 ) = 1 / 4, so q 3 = 2 μ q 1
Similarly, P ( X < μ ) = 1 / 2 by symmetry.
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Marissa Singh
Answered 2022-04-08 Author has 2 answers
Just want to add another interesting tidbit - if E [ X ] < , then
E [ X ] = μ = m = q 1 + q 3 2
E [ X ] = x f X ( x ) d x = ( 2 μ x ) f X ( x ) d x = 2 μ f X ( x ) d x x f X ( x ) d x = 2 μ E [ X ] E [ X ] = μ
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