Question

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

• A. $1.53$
• B. $1.43$
• C. $1.55$
• D. None of these

Answer 1

Answer: (A)

Now, $\Sigma {\left(x-5\right)}^2=43$
$\Rightarrow \Sigma x^2-\Sigma 10x+\Sigma 25=43$
$\Rightarrow \Sigma x^2-10\left\{\Sigma \left(x-5+5\right)\right\}+\left(18\times 25\right)=43$
$\Rightarrow \Sigma x^2-10\Sigma \left(x-5\right)-10\Sigma 5+450=43$
$\Rightarrow \Sigma x^2-10\left(3\right)-\left(10\times 18\times 5\right)+450=43$
$\Rightarrow \Sigma x^2-30-900+450=43$
$\Rightarrow \Sigma x^2=523$ and $\Sigma \left(x-5\right)=3$
$\Rightarrow \Sigma x-\Sigma 5=3$
$\Rightarrow \Sigma x=3+\left(5\times 18\right)=93$
Now, $\sigma =\sqrt{\frac{\Sigma x^2}{n}-{\left(\frac{\Sigma x}{n}\right)}^2}$
$=\sqrt{\frac{523}{18}-{\left(\frac{93}{18}\right)}^2}$
$=\sqrt{29.06-{\left(5.17\right)}^2}$
$=\sqrt{29.06-26.73}$
$=\sqrt{2.33}$
$=1.53$