Question

. A distribution consists of three components with frequencies 200, 250 and 300 having means 25, 10, and 15 and standard deviations 3, 4, and 5 respectively. Calculate (i) The mean (ii) The standard deviation

Answer 1

Given

Frequencies 200, 250, 300

Means 25, 10, 15

Standard deviations 3, 4, 5

To find

(i) The mean (ii) The standard deviation

Solution

$n_1=200$, $n_2=250$, $n_3=300$

$\bar{x_1}=25$, $\bar{x_2}=10$, $\bar{x_3}=15$

$s_1=3$, $s_2=4$, $s_3=5$

(i) Combined mean is given by

$\bar{x}=\dfrac{n_1\bar{x_1}+n_2\bar{x_2}+n_3\bar{x_3}}{n_1+n_2+n_3}\\\Rightarrow \bar{x}=\dfrac{200\times 25+250\times 10+300\times 15}{200+250+300}\\\Rightarrow \bar{x}=16$

The combined mean is 16.

(ii)

$d_1=\bar{x_1}-\bar{x}=25-16\\\Rightarrow d_1=9$

$d_2=\bar{x_2}-\bar{x}=10-16\\\Rightarrow d_2=-6$

$d_3=\bar{x_3}-\bar{x}=15-16\\\Rightarrow d_3=-1$

Combined standard deviation is given by

$s=\sqrt{\dfrac{n_1(s_1^2+d_1^2)+n_2(s_2^2+d_2^2)+n_3(s_3^2+d_3^2)}{n_1+n_2+n_3}}\\\Rightarrow s=\sqrt{\dfrac{200(3^2+9^2)+250(4^2+(-6)^2)+300(5^2+(-1)^2)}{200+250+300}}\\\Rightarrow s=7.19$

The combined standard deviation is 7.19.