# If for a distribution $\sum (x-5) =3,\sum (x-5)^2 =43$ and the total number of items is $18$. Find the mean and standard deviation .

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

Toolbox:
• 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.