If the standard deviation of a variable x is `sigma` , then standard deviation of variable `(a X+b)/c` is (a).`sigma` (b) `a/csigma` (c) `|a/c|sigma` (d) `(asigma+b)/c`

Answer

B

Explanation

Let `y=(alpha x+b)/(c ), i.e., y=(a)/(c )x+(b)/(c )`
i.e., `y=(a)/(c )x+(b)/(c )`
`therefore overline (y)=Aoverline(x)+B`
`therefore y-overline(y)=A(x-overline(x))`
`implies (y-overline(y))^(2)=A^(2)(x-overline(x))^(2)`
`implies sum (y-overline(y))^(2)=A^(2)sum(x-overline(x))^(2)`
`implies n. sigma_(y)^(2)=A^(2). n sigma_(x)^(2)`
`implies sigma_(y)^(2)=A^(2)sigma_(x)^(2)`
`implies sigma_(y)=|A|sigma_(x)`
`implies sigma_(y)=|(a)/(c )| sigma_(x)`
Thus, new SD `=|(a)/(c )|sigma`