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

The given observation are \(a, b, c, d\) and \(e\)
Mean \(=m=\frac{a+b+c+d+e}{5}\)
\(\Rightarrow \quad \boldsymbol{\Sigma} x_i=a+b+c+d+e=5 m\) ....(i)
Now, consider the observations \(a+k, b+k, c+k\), \(d+k, e+k\).
Now, mean
\(=\frac{(a+k)+(b+k)+(c+k)+(d+k)+(e+k)}{5}\)
\(=\frac{a+b+c+d+e+5 k}{5}\)
\(=\frac{5 m+5 k}{5}=m+k\)
\(\therefore\) New standard devaition \(=\sqrt{\frac{\sum\left(x_i+k\right)^2}{5}-(m+k)^2}\)
\(=\sqrt{\frac{\Sigma\left(x_i^2+k^2+2 x_i k\right)}{5}-\left(m^2+k^2+2 m k\right)}\)
\(=\sqrt{\frac{\Sigma x_i^2}{5}-m^2+\frac{5 k^2}{5}-k^2+\frac{2 k \Sigma x_i}{5}-2 m k}\)
\(=\sqrt{\frac{\Sigma x_i^2}{5}-m^2+\frac{2 k \times 5 m}{5}-2 m k}\)
\(=\sqrt{\frac{\Sigma x_i^2}{5}-m^2}=S\)