TS ICET 23 May 2018 Shift 2 Paper

Question : 145 of 200

Marks: +1, -0

The standard deviation of n observations

\x_1,\x_2,.....\x_\n

is 15. Then the standard deviation of the observations

5\x_1-11,5\x_2-11,.....5\x_\n-11

Solution:

(5 x_{1}-11,5 x_{2}-11, ... . .5 x_{n}-11)

= 5\;\SD \;(x_1, x_2, …… x_n)

= 5 × 15 = 75.

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