Question

Find the standard deviation for the following distribution :

$\begin{array}{cccccccc}x:& 4.5& 14.5& 24.5& 34.5& 44.5& 54.5& 64.5\\ y:& 1& 5& 12& 22& 17& 9& 4\end{array}$

Answer 1

$\begin{array}{cccccc}{x}_i& f_i& f_i{x}_i& x_i-37.64& (x_i-mean{)}^2& f(x_i-mean{)}^2\\ 4.5& 1& 4.5& -33.14& 1098.45& 1098.45\\ 14.5& 5& 72.5& -23.14& 535.59& 2677.96\\ 24.5& 12& 294& -13.14& 172.73& 2072.82\\ 34.5& 22& 759& -3.14& 9.88& 217.31\\ 44.5& 17& 756.5& 6.86& 47.02& 799.35\\ 54.5& 9& 490.5& 16.86& 284.16& 2557.47\\ 64.5& 4& 258& 26.86& 721.31& 2885.22\\ & N=70& 2635& & & 12308.57\end{array}$

Here , N= 70, $\sum f_i{x}_i=2635$

Mean $\overline{x}=\frac{1}{N}\sum f_i{x}_i$

$\therefore \overline{x}=\frac{1}{N}(\sum f_i{x}_i)=\frac{2635}{70}=37.64$

We have , $\sum f_i(x_i-\overline{x}{)}^2=12308.57$

$\therefore var\left(x\right)=\frac{1}{N}[\sum f_i(x_i-\overline{(}x\left){)}^2\right]=\frac{12308.57}{70}=175.84$

S. D = $\sqrt{var\left(X\right)}=\sqrt{175.84}=13.26$