$\begin{array}{1 1}(A)\;11,22\\(B)\;25,1600\\(C)\;5.16,1.54\\(D)\;15,3.45\end{array} $

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- Formulae used to solve this problem are :
- Mean $=\large\frac{\sum x}{n}$
- $SD(\sigma)= \sqrt { \large\frac{ \sum x^2}{n} -\bigg( \large\frac{ \sum x}{n}\bigg)^2}$

Step 1:

Let us calculated the value of $\sum x $ and $ \sum x^2$

Given $ \sum (x-5) =3 \qquad n=18$

$=> \sum x - \sum 5=3$

$=> \sum x - 18 \times 5=3$

$=> \sum x = 3+90$

$=> \sum x=93$

$\sum (x-5)^2=43$

$\sum (x^2 +25-10x) =43$

$\sum x^2+\sum 25 -10 \sum x =43$

$\sum x^2+25 \times 18 -10 \times 93 =43$

$\sum x^2 =43 +930 -450$

$\qquad=523$

Step 2:

Mean $=\large\frac{\sum x}{n}=\frac{93}{18} $$=5.16$

Step 3:

$SD(\sigma)= \sqrt { \large\frac{ \sum x^2}{n} -\bigg( \large\frac{ \sum x}{n}\bigg)^2}$

$\qquad= \sqrt {\large\frac{523}{18} - \bigg( \large\frac{93}{18}\bigg)^2}$

$\qquad= \sqrt { \large\frac{523 \times 18 -93 \times 93}{18 \times 18} }$

$\qquad= \large\frac{1}{18} $$ \sqrt{ 9414-8649}$

$\qquad= \large\frac{27.66}{18} $

$\qquad= 1.54$

Hence C is the correct answer.

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