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

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- 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.

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