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 the mean and standard deviation.​

Answer 1

The mean and standard deviation of the combined distribution is 16 and 7.192 respectively.

Explanation:

We have given that 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.

And we have to find the mean and standard deviation of the combined distribution.

Firstly let us represent some symbols;

`n_1` = 200
`\bar X_1` = 25
`\sigma_1` = 3

`n_2` = 250
`\bar X_2` = 10
`\sigma_2` = 4

`n_3` = 300
`\bar X_3` = 15
`\sigma_3` = 5

Here,
`\bar X_1, \bar X_2 , \bar X_3` represent the means and
`\sigma_1,\sigma_2,\sigma_3` represent the standard deviations.

Now, as we know that Mean of the combined distribution is given by;

`\bar X = (n_1 * \bar X_1+n_2 * \bar X_2+n_3 * \bar X_3)/(n_1+n_2+n_3)`

Putting the above values in the formula we get;

`\bar X = (200 * 25+250 * 10+300 * 15)/(200+250+300)`

`\bar X = (5000+2500 +4500)/(750)`

`\bar X = (12000)/(750)` = 16

Similarly, the formula for combined standard deviation is given by;

`\sigma = \sqrt{(n_1\sigma_1^(2) + n_1(\bar X_1-\bar X)^(2)+n_2\sigma_2^(2) + n_2(\bar X_2-\bar X)^(2)+n_3\sigma_3^(2) + n_3(\bar X_3-\bar X)^(2) )/(n_1+n_2+n_3) }`

`\sigma = \sqrt{((200 * 3^(2)) + 200 * (25-16)^(2)+(250 * 4^(2)) + 250 * (10-16)^(2)+(300 * 5^(2)) + 300 * (15-16)^(2) )/(200+250+300) }`

`\sigma = \sqrt{(1800 + (200 * 81)+4000 + (250 * 36)+7500 +( 300 * 1) )/(750) }`

`\sigma = \sqrt{(1800 + 16200+4000 + 9000+7500 +300 )/(750) }`

`\sigma = \sqrt{(38800 )/(750) }` = 7.192

Hence, the mean and standard deviation of the combined distribution is 16 and 7.192 respectively.