Question

Find the standard deviation for the following data :

(i)$\begin{array}{cccccc}x:& 3& 8& 13& 18& 23\\ f:& 7& 10& 15& 10& 6\end{array}$

(ii)$\begin{array}{ccccccc}x:& 2& 3& 4& 5& 6& 7\\ f:& 4& 9& 16& 14& 11& 6\end{array}$

Answer 1

(i)$\begin{array}{cccccc}X& f_i& f_i{x}_i& x_i-\overline{X}& (x_i-\overline{X}{)}^2& f_i(x_i-mean{)}^2\\ 3& 7& 21& -9.79& 95.88& 671.13\\ 8& 10& 80& -4.79& 22.96& 229.60\\ 13& 15& 195& 0.21& 0.04& 0.65\\ 18& 10& 180& 5.21& 27.13& 271.26\\ 23& 6& 138& 10.21& 104.21& 625.26\\ & 48& 614& & & 1797.92\end{array}$

Here, N= 48 and $\sum f_i{x}_i=614$

$\therefore \overline{x}=\frac{1}{N}(\sum f_i{x}_i)=\frac{614}{48}=12.79$

$\sum f_i(x-\overline{x}{)}^2=1797.92$

$\therefore var\left(x\right)=\frac{1}{N}[\sum f_i(x_i-\overline{x}{)}^2]=\frac{1797.92}{48}=37.46$

S. D = $\sqrt{varx}=\sqrt{37.46}=6.12$

(ii) $\begin{array}{ccccccc}x:& 2& 3& 4& 5& 6& 7\\ f:& 4& 9& 16& 14& 11& 6\end{array}$

$\begin{array}{ccccc}{x}_i& f_i& d_i=x_i-4& f_i{d}_i& f_i{d}_i^2\\ 2& 4& -2& -8& 16\\ 3& 9& -1& -9& 9\\ 4& 16& 0& 0& 0\\ 5& 14& 1& 14& 14\\ 6& 11& 2& 22& 44\\ 7& 6& 3& 18& 54\\ Total& 60& & \sum f_i{d}_i=37& \sum f_i{d}_i^2=137\end{array}$

$\therefore SD=\sqrt{\frac{\sum f_i{d}^{2}i}{N}-{\left(\frac{\sum f_i{d}_i}{N}\right)}^2}=\sqrt{\frac{137}{60}-{\left(\frac{37}{60}\right)}^2}$

= $\sqrt{2.2833-(0616{)}^2}=\sqrt{2.2833-0.3794}=\sqrt{1.9037}=1.38$