Question

The arithmetic mean and variance of a set of 10 figures are known to be 17 and 33 respectively. Of the 10 figures one figure 26 was subsequently found inaccurate and was weeded out. What is the new mean and standard deviation

Answer 1

Given: The arithmetic mean and variance of a set of 10 figures are known to be 17 and 33 respectively. Of the 10 figures one figure 26 was subsequently found inaccurate and was weeded out.

To find: New mean and Standard deviation

Solution:

We know that arithmetic mean of certain observations is given by sum of observations divided by number of observations. Therefore,

Mean = Sum of obs. / Number of obs.

=> 17 = (Sum of obs.) / (10)

=> 17 * 10 = Sum of obs.

=> 170 = Sum of obs.

One figure i.e. 26 was removed and number of terms are left 9.

New sum = 170 - 26 => 144

and hence we get,

New mean = 144/9 => 16

$\rule{90mm}{0.9pt}$

Variance is given by:

$\boxed{\sigma^2 = \dfrac{\sum x^2}{n} - (\overline{x})^2}$

${\implies33 = \dfrac{\sum x^2}{10} - (17)^2}$

${\implies33 = \dfrac{\sum x^2}{10} -289}$

${\implies33 + 289 = \dfrac{\sum x^2}{10}}$

${\implies322 = \dfrac{\sum x^2}{10}}$

${\implies3220 = \sum x^2}$

New $\sum{x}^2$ is given by:

${\implies {\sum x^2} =3220 - (26)^2}$

${\implies {\sum x^2} =3220 -276}$

${\implies \sum x^2 =2544}$

New standard deviation is given by:

$\sigma = \sqrt{\dfrac{\sum{x}^2}{n} - (\overline x)^2}$

$\sigma = \sqrt{\dfrac{2544}{9} - (16)^2}$

$\sigma = \sqrt{26.67}$

$\sigma = 5.16$

New standard deviation = 5.16

Note: $\rm Variance = (Standard\,deviation)^2$