Question

Let a,b,c,d,e be the observations with mean m and standard deviation s. The standard deviation of the observations a+k, b+k, c+k , d+k, e+k is

• A. s
• B. ks
• C. s+k
• D. $\frac{s}{k}$

Answer 1

The correct option is A

s

The given observations are a, b, c, d, e

Mean = m = $\frac{a+b+c+d+e}{5}$

$\Rightarrow \sum x_i=a+b+c+d+e=5m$ ....(i)

Standard deviation , s = $\sqrt{\frac{\sum x_i^2}{5}-m^2}$

Now, consider the observations a+k , b+k , c+k , d+k , e+k

New mean

= $\frac{(a+k)+(b+k)+(c+k)+(d+k)+(e+k)}{5}$

=$\frac{a+b+c+d+e+5k}{5}$

= $\frac{5m+5k}{5}$

= m + k

$\therefore$ New standard deviation

= $\sqrt{\frac{\sum (x_i+k{)}^2}{5}-(m+k{)}^2}$

= $\sqrt{\frac{\sum (x_i^2+k^2+2x_{i}k)}{5}-(m^2+k^2+2mk)}$

$\sqrt{\frac{\sum x_i^2}{5}+\frac{\sum k^2}{5}+\frac{\sum 2x_{i}k}{5}-(m^2+k^2+2mk)}$

$\sqrt{\frac{\sum x_i^2}{5}-m^2+\frac{5k^2}{5}-k^2+\frac{2k\sum x_i}{5}-2mk}$

$=\sqrt{\frac{\sum x_i^2}{5}-m^2+\frac{2k\times 5m}{5}-2mk}$

= $\sqrt{\frac{\sum x_i^2}{5}-m^2}=8$

Hence, the correct answer is option (a).