Question

Find the mean, variance and standard deviation for the following data using short cut method:

$\begin{array}{cccccccccc}{x}_i& 60& 61& 62& 63& 64& 65& 66& 67& 68\\ f_1& 2& 1& 12& 29& 25& 12& 10& 4& 5\end{array}$

Answer 1

Let the assumed mean be A = 64

Then, we may present the table given below:

$\begin{array}{cccccc}{x}_i& f_i& d_i=(x_i-A)& d_i^2& f_i{d}_i& f_i{d}_i^2\\ & & =(x_i-64)& & & \\ 60& 2& -4& 16& -8& 32\\ 61& 1& -3& 9& -3& 9\\ 62& 12& -2& 4& -24& 48\\ 63& 29& -1& 1& -29& 29\\ 64=A& 25& 0& 0& 0& 0\\ 65& 12& 1& 1& 12& 12\\ 66& 10& 2& 4& 20& 40\\ 67& 4& 3& 9& 12& 36\\ 38& 5& 4& 16& 20& 80\\ & 100& & & 0& 286\end{array}$

$\therefore N=\sum f_i=100,\sum f_i{d}_i=0$ and $\sum f_i{d}_i^2$

$\therefore$ mean $=A+\frac{\sum f_i{d}_i}{\sum f_i}=(64+\frac{0}{100})=64$

Variance, ${\sigma }^2=\frac{\sum f_i{d}_i^2}{\sum f_i}-\left(\frac{\sum f_i{d}_i}{\sum f_i}\right)$

$=\frac{286}{100}-{\left(\frac{0}{100}\right)}^2=\frac{286}{100}=2.86$

$\therefore$ standard deviation, $\sigma =\sqrt{2.86}=1.69$