Question

The ages, x years, of 18 people attending an evening class are summarized by 

the following totals:

Σx= 745,Σx^2= 33951

 (i) Calculate the mean and standard deviation of the ages of this group of people. 

 (ii) One person leaves the group and the mean age of the remaining 17 people is 

 exactly 41 years. Find the age of the person who left and the standard deviation of the ages of the remaining 17 people.

Answer 1

(i) $E[X]=\frac{1}{18}\sum\limits_{i=1}^{18}x_i=745/18=41.39$

$E[X^2]=\frac{1}{18}\sum\limits_{i=1}^{18}x_i^2=33951/18=1886.17$

$Var[X]=E[X^2]-E[X]^2=1886.17-41.39^2=173.13$

$\sigma(X)=\sqrt{Var[X]}=\sqrt{173.13}=13.16$

(ii) Let X' denotes ages of the remaining 17 people. We may assume that the person who left the group was of age $x_{18}$.

Then $E[X']=\frac{1}{17}\sum\limits_{i=1}^{17}x_i=41$ implies $\sum\limits_{i=1}^{17}x_i=41\cdot 17=697$ and $x_{18}=\sum\limits_{i=1}^{18}x_i-\sum\limits_{i=1}^{17}x_i=745-697=48$

$\sum\limits_{i=1}^{17}x_i^2=\sum\limits_{i=1}^{18}x_i^2-x_{18}^2=33951-48^2=31647$

$E[X'^2]=\frac{1}{17}\sum\limits_{i=1}^{17}x_i^2=31647/17=1861.59$

$Var[X']=E[X'^2]-E[X']^2=1861.59-41^2=180.59$

$\sigma(X')=\sqrt{Var[X']}=\sqrt{180.59}=13.44$

Answer. (i) The mean and standard deviation of the ages of this group of people are $E[X]=41.39$ , $\sigma(X)=13.16$.

(ii) The age of the person who left is 48, the standard deviation of the ages of the remaining 17 people is $\sigma(X')=13.44$.