If _ i=1 ¹⁰ (x_i-3 )=7 and _ i=1 ¹⁰ (x_i-3 )^2=27 , then the standard deviation of the 10 items

Step-by-step solution

We have, _ i=1 ¹⁰ (x_i-3 )=7 and _ i=1 ¹⁰ (x_i-3 )^2=27 Standard deviation is remain unchanged, if observation are added or subtracted by a fixed number. Standard deviation aligned & = _ i=1 ¹⁰(x i-3)^2 10 - ( _ i=1 ¹⁰(x i-3) 10 )^2 & = 27 10 - ( 7 10 )^2 & = 270 100 - 49 100 = 221 10 = 14.86 10 =1.486 aligned