# There are 60 students in a class .The following is the frequency distribution of the marks obtained by the students in a test : where x is a positive integer . Determine the mean and standard deviation of the marks.

$\begin{array}{1 1}(A)\;1.122\\(B)\;5\\(C)\;7\\(D)\;15\end{array}$

Toolbox:
• The formula used to solve this problem are : Mean $\bar{x} = \large\frac{\sum f_i x_i}{\sum f_i}$
• Standard deviation = $\sigma=\sqrt{\large\frac{\sum f_i x_i^2}{\sum f_i} - \bigg(\large\frac{\sum f_ix_i }{\sum f_i } \bigg)^2}$
• To calculate the mean and SD initially we need to calculate the value of x.
Step 1:
$x-2+x+x^2+(x+1)^2+2x+x+1=60$
$=> 5x-2 +x^2+x^2+1 +2x+1=60$
$=> 2x^2+15x -8x-60 =0$
$=> x(2x+15)-4(2x+15) =0$
$=> (x-4)(2x+15)=0$
=> $x=4 ,x= \large\frac{-15}{2}$
Given x is a positive integer .
So the value of x is 4.
Step 3:
Mean $\bar{x} = \large\frac{\sum f_i x_i}{\sum f_i}$
$\qquad= \large\frac{168}{60}$
$\qquad= 2.8$
Step 4:
Standard deviation = $\sigma=\sqrt{\large\frac{\sum f_i x_i^2}{\sum f_i} - \bigg(\large\frac{\sum f_ix_i }{\sum f_i } \bigg)^2}$
$\qquad= \sqrt { \large\frac{546}{60} -\bigg( \large\frac{168}{60}\bigg)^2}$
$\qquad= \sqrt {\large\frac{546 \times 60 -(168)^2}{60 \times 60}}$
$\qquad= \large\frac{1}{60} $$\sqrt { 32,760 -28,224} \qquad= \large\frac{1}{60 }$$\sqrt {4536}$
$\qquad= \large\frac{67.349}{60}$
$\qquad= 1.122$
Hence A is the correct answer.