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Question:

Let X̄ and M.D. be the mean and the mean deviation about X̄ of n observations xi, i=1,2,.....,n. If each of the observations is increased by 5, then the new mean and the mean deviation about the new mean, respectively, are :

X̄+5,M.D.

X̄,M.D.+5

X̄+5m,M.D.+5

X̄,M.D.

Solution:

X̄ = (x1+x2+..+xn)/n
New mean = (x1+5 + x2+5 + .. + xn+5)/n
= (x1+x2+..+xn + 5n)/n
= (x1+x2+..+xn)/n + 5n/n
New mean = Original mean + 5.
M.D. = (|x1 - X̄| + |x2 - X̄| + .. + |xn - X̄|)/n
New M.D. about new mean = |x1+5 - (X̄+5)| + |x2+5 - (X̄+5)| + .. + |xn+5 - (X̄+5)| / n
= |x1 - X̄| + |x2 - X̄| + .. + |xn - X̄| / n
= M.D.