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

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

$\Rightarrow \sum _{}{x}_i=a+b+c+d+e=5m$ .....(1)

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

$$\begin{gathered} =\frac{a+b+c+d+e+5k}{5} \\ =\frac{5m+5k}{5} \\ =m+k \end{gathered}$$

∴ New standard deviation

$$\begin{gathered} =\sqrt{\frac{\sum _{}{\left(x_i+k\right)}^2}{5}-{\left(m+k\right)}^2} \\ =\sqrt{\frac{\sum _{}\left(x_i^2+k^2+2x_{i}k\right)}{5}-\left(m^2+k^2+2mk\right)} \\ =\sqrt{\frac{\sum _{}{x}_i^2}{5}+\frac{\sum _{}{k}^2}{5}+\frac{\sum _{}2x_{i}k}{5}-\left(m^2+k^2+2mk\right)} \\ =\sqrt{\frac{\sum _{}{x}_i^2}{5}-m^2+\frac{5k^2}{5}-k^2+\frac{2k\sum _{}{x}_i}{5}-2mk} \end{gathered}$$
$$\begin{gathered} =\sqrt{\frac{\sum _{}{x}_i^2}{5}-m^2+\frac{2k\times 5m}{5}-2mk}\left[\mathrm{Using}\left(1\right)\right] \\ =\sqrt{\frac{\sum _{}{x}_i^2}{5}-m^2} \\ =s \end{gathered}$$

Hence, the correct answer is option (a).