The mean deviation about median of variates 13, 14, 15,...., 99, 100 is 1936 (b) 21.5 (c) 23.5 (d) 22

Explanation

variates are 13, 14, 15,....99,100

here, n=88

so, median= `(44^(th) term + 45^(th) term)/2 = (56+57)/2 = 56.5 `

now, mean deviation = `(sum_(i=1)^88|x_i-56.5|)/88`

= `(43.5+ 42.5 + ..........+ 0.5+0.5+ ..........+ 43.5)/88`

= `(1+3+5+ .....+87)/88` -- eqn (1)

applying Arithmetic progression for the numerator, a=1, d=2

`a_n= a+ (n-1)*d`

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