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} -\Sigma10x +\Sigma25 = 43$
$ \Rightarrow \Sigma x^{2} -10\left\{\Sigma\left(x-5+5\right)\right\}+\left(18\times25\right) = 43$
$\Rightarrow \Sigma x^{2} -10\Sigma\left(x-5\right)-10\Sigma5 +450 =43$
$ \Rightarrow \Sigma x^{2}-10\left(3\right)-\left(10\times18\times5\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-\Sigma5 = 3$
$ \Rightarrow \Sigma x = 3+\left(5\times18\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$