Question

Find the mean , variance and standard deviation for the following data :

(i) 2,4,5,6,8,17

(ii) 6,7,10,12,13,4,8,12

(iii) 227,235,255,269,292,299,312,321,333,348

(iv) 15,22,27,11,9,21,14,9

Answer 1

(i) 2,4,5,6,8,17

Mean = $\overline{X}$ = $\frac{2+4+5+6+8+17}{6}$

= $\frac{42}{6}=7$

$\begin{array}{ccc}{x}_i& x_i-\overline{X}=(x_i-7)& (x_i-7{)}^2\\ 2& -5& 25\\ 4& -3& 9\\ 5& -2& 4\\ 6& -1& 1\\ 8& 1& 1\\ 17& 10& 100\\ & & \\ & & {\sum }_{i=1}^6(x_i-\overline{X}{)}^2=140\end{array}$

n = 6

Variance (X) = ${\sum }_{i=1}^6\frac{(x_i-\overline{X}{)}^2}{n}$

$=\frac{140}{6}=23.33$

Standard deivation = $\sqrt{variance\left(X\right)}=\sqrt{23.33}=4.83$

(ii) 6,7,10,12,13,4,8,12

Mean = $\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}=9$

$\begin{array}{ccc}{x}_i& (x_i-\overline{X})=(x_i-9)& (x_i-\overline{X}{)}^2\\ 6& -3& 9\\ 7& -2& 4\\ 10& 1& 1\\ 12& 3& 9\\ 13& 4& 16\\ 4& -5& 25\\ 8& -1& 1\\ 12& 3& 9\\ & & {\sum }_{i=1}^8(x_i-\overline{X}{)}^2=74\end{array}$

n = 8

$\therefore \left(X\right)={\sum }_{i=1}^8\frac{(x_i-\overline{X}{)}^2}{n}$

$=\frac{74}{8}=9.25$

Standard deviation = $\sqrt{variance\left(X\right)}=\sqrt{9.25}=3.04$

(iii) 227, 235,255,269,292,299,312,321,333,348

Mean = $\frac{227+235+255+269+292+299+312+321+333+348}{10}$

= $\frac{2891}{10}=289.1$

$\begin{array}{ccc}{x}_i& (x_i-\overline{X})=x_i-289.1& (x_i-\overline{X}{)}^2\\ 227& -62.1& 3856.41\\ 235& -54.1& 292.81\\ 255& -34.1& 1162.81\\ 269& -20.1& 404.01\\ 292& 2.9& 8.41\\ 299& 9.9& 98.01\\ 312& 22.8& 524.41\\ 321& 31.9& 1017.61\\ 333& 43.9& 1927.21\\ 348& 58.9& 3469.21\\ & & {\sum }_{i=1}^{10}(x_i-\overline{X}{)}^2=15394.9\end{array}$

n = 10

Variance (X) = ${\sum }_{i=1}^{10}\frac{(x_i-\overline{X}{)}^2}{n}$
$=\frac{15394.9}{10}=1539.49$

Standard deviation = $\sqrt{variance\left(X\right)}=\sqrt{1539.49}=39.24$

(iv) 15,22,27,11,9,21,14,9

Mean $=\frac{15+22+27+11+9+21+14+9}{8}=\frac{128}{8}=16$

$\begin{array}{ccc}{x}_i& (x_i-\overline{X})=(x_i-16)& (x_i-\overline{X}{)}^2\\ 15& -1& 1\\ 22& 6& 36\\ 27& 11& 121\\ 11& 5& 25\\ 9& -7& 49\\ 21& 5& 25\\ 14& -2& 4\\ 9& -7& 49\\ & & {\sum }_{i=1}^8(x_i-\overline{X}{)}^2=310\end{array}$

$\therefore n=8$

Variance (X) = ${\sum }_{i=1}^8\frac{(x_i-\overline{X}{)}^2}{n}$

$=\frac{310}{8}=38.75$

Standard deviation =$\sqrt{variance\left(X\right)}=\sqrt{38.75}=6.22$