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 $x _{1}+ x _{1}{ }_{1}=70$
$\overline{ x }=\frac{ x _{1}+ x _{2}+\ldots+ x _{50}}{50}=15$ (given)
$\Rightarrow x _{1}+ x _{2}+\ldots \ldots x _{50}=750$ ...(i)
Now
Mean of correct observation is $16$
$\frac{ x _{1}^{\prime}+ x _{2}+\ldots+ x _{50}}{50}=16$
$x _{1}^{\prime}+ x _{2}+ x _{3}+\ldots x _{50}=16 \times 50$ ...(ii)
eq. (ii) -eq. (i)
$\Rightarrow x _{1}^{\prime}- x _{1}=16 \times 50-15 \times 50$
$x _{1}^{\prime}- x _{1}=50\, \&\, x _{1}+ x _{1}^{\prime}=70$
$x _{1}^{\prime}=60$
$x _{1}=10$
$\Rightarrow 4=\frac{x_{1}^{2}+x_{2}^{2}+\ldots .+x_{50}^{2}}{50}-15^{2}$ ...(iii)
$\Rightarrow \sigma^{2}=\frac{x_{1}^{\prime 2}+x_{2}^{2}+\ldots . x_{50}^{2}}{50}-16^{2}$ ...(iv)
From (iii)
$\Rightarrow 4=\frac{(10)^{2}}{50}+\frac{x_{2}^{2}+x_{3}^{2}+\ldots .+x_{50}^{2}}{50}-225$
$\Rightarrow 4=2-225+\frac{\left(x_{2}^{2}+x_{3}^{2}+\ldots .+x_{50}^{2}\right)}{50}$
$\Rightarrow 227=\frac{\left(x_{2}^{2}+x_{3}^{2}+\ldots . x_{50}^{2}\right)}{50}$
From (iv)
$\sigma^{2}=\frac{(60)^{2}}{50}+\left(\frac{x_{2}^{2}+x_{3}^{2}+\ldots .+x_{50}^{2}}{50}\right)-(16)^{2}$
$\sigma^{2}=\frac{60 \times 60}{50}+227-256$
$\sigma^{2}=72+227-256$
$\sigma^{2}=43$