Question

The mean and standard deviation of $50$ observations are $15$ and $2$ respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is $70$ . If the correct mean is $16$ , then the correct variance is equal to :

• A. 10
• B. 36
• C. 43
• D. 60

Answer 1

Answer: (C)

No. of observations: $-50$
$\text{mean}(\overline{x})=15$
Standard deviation $(\sigma )=2$
Let incorrect observation is $x_1\&$ correct observation is $\left(x_1^{\prime }\right)$
Given $x_1+x_1{}_1=70$
$\overline{x}=\frac{x_1+x_2+\dots +x_{50}}{50}=15$ (given)
$\Rightarrow x_1+x_2+\dots \dots x_{50}=750$ ...(i)
Now
Mean of correct observation is $16$
$\frac{x_1^{\prime }+x_2+\dots +x_{50}}{50}=16$
$x_1^{\prime }+x_2+x_3+\dots x_{50}=16\times 50$ ...(ii)
eq. (ii) -eq. (i)
$\Rightarrow x_1^{\prime }-x_1=16\times 50-15\times 50$
$x_1^{\prime }-x_1=50\&x_1+x_1^{\prime }=70$
$x_1^{\prime }=60$
$x_1=10$
$\Rightarrow 4=\frac{x_1^2+x_2^2+\dots .+x_{50}^2}{50}-15^2$ ...(iii)
$\Rightarrow {\sigma }^2=\frac{x_1^{\prime 2}+x_2^2+\dots .x_{50}^2}{50}-16^2$ ...(iv)
From (iii)
$\Rightarrow 4=\frac{(10{)}^2}{50}+\frac{x_2^2+x_3^2+\dots .+x_{50}^2}{50}-225$
$\Rightarrow 4=2-225+\frac{\left(x_2^2+x_3^2+\dots .+x_{50}^2\right)}{50}$
$\Rightarrow 227=\frac{\left(x_2^2+x_3^2+\dots .x_{50}^2\right)}{50}$
From (iv)
${\sigma }^2=\frac{(60{)}^2}{50}+\left(\frac{x_2^2+x_3^2+\dots .+x_{50}^2}{50}\right)-(16{)}^2$
${\sigma }^2=\frac{60\times 60}{50}+227-256$
${\sigma }^2=72+227-256$
${\sigma }^2=43$