Question

If mean of $x_1,x_2.........x_n$ is y then means of $b-a{x}_1,b-a{x}_2........b-a{x}_n$ is

• A. $ay-b$
• B. $b+ay$
• C. $b-ay$
• D. $by+a$

Answer 1

The correct option is C $b-ay$

Given the mean of x_1,x_2,.....,x_n is y

$\frac{x_1+x_2+.....+x_n}{n}=y{x}_1+x_2+.....+x_n=ny$

Means of ${b-ax}_1,{b-ax}_2,.....,{b-ax}_n$ is

$=\frac{b-a{x}_1+b-a{x}_2+.....+b-a{x}_n}{n}=\frac{(b+b+...+b{}^{\prime }{n}^{\prime }times)-a{x}_1-a{x}_2-.....-a{x}_n}{n}=\frac{bn-a{(x}_1+x_2+.....+x_n)}{n}=\frac{bn-a\left(ny\right)}{n}=\frac{bn}{n}-\frac{any}{n}=b-ay$

So Means of ${b-ax}_1,{b-ax}_2,.....,{b-ax}_n$ is $b-ay$