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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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.

Mackenzie Zimmerman · Accepted Answer

Say,       q    1   is the lower quartile, i.e.   P  (  X  &amp;lt;      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  &amp;gt;  2  μ  −      q    1    )  =  1      /    4, so       q    3    =  2  μ  −      q    1  Similarly,   P  (  X  &amp;lt;  μ  )  =  1      /    2 by symmetry.

Marissa Singh · Accepted Answer

Just want to add another interesting tidbit - if   E  [  X  ]  &amp;lt;  ∞, 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

μ = m = q 1 </msub>

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