Let $¯ ¯¯¯ ¯ X$ and M.D. be the mean and the mean deviation about $¯ ¯¯¯ ¯ X$ of n observations $x_{i}, i = 1, 2, . . . . ., n .$ If each of the observations is increased by 5, then the new mean and the mean deviation about the new mean, respectively, are :

¯ ¯¯¯ ¯ X, M . D .

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¯ ¯¯¯ ¯ X + 5, M . D .

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¯ ¯¯¯ ¯ X, M . D . + 5

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¯ ¯¯¯ ¯ X + 5 m, M . D . + 5

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Solution

The correct option is D $¯ ¯¯¯ ¯ X + 5, M . D .$
$¯ ¯¯¯ ¯ X = \frac{x_{1} + x_{2} + . . . . . . + x_{n}}{n}$
New mean $= \frac{x_{1} + x_{2} + . . . . . . + x_{n} + 5 n}{n}$
$= \frac{x_{1} + x_{2} + . . . . . . + x_{n}}{n} + \frac{5 n}{n}$
New mean $=$ Original mean $+ 5$ .
M.D. $= (x_{1} - ¯ ¯ ¯ x) + {(x}_{2} - ¯ ¯ ¯ x) + . . . . . . + {(x}_{n} - ¯ ¯ ¯ x)$
New M.D. about new mean $= {x_{1} + 5 - (¯ ¯¯¯ ¯ X + 5)} + {{x}_{2} + 5 - (¯ ¯¯¯ ¯ X + 5)} + . . . . . . + {{x}_{n} + 5 - (¯ ¯¯¯ ¯ X + 5)}$
= M.D.

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Let ¯¯¯¯¯X and M.D. be the mean and the mean deviation about ¯¯¯¯¯X 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 :

Let $¯ ¯¯¯ ¯ X$ and M.D. be the mean and the mean deviation about $¯ ¯¯¯ ¯ X$ of n observations $x_{i}, i = 1, 2, . . . . ., n .$ If each of the observations is increased by 5, then the new mean and the mean deviation about the new mean, respectively, are :