Question

The mean and variance of $7$ observations are $8$ and $16$ respectively. If two observations are $6$ and $8,$ then the variance of the remaining $5$ observations is

• A. $\frac{536}{25}$
• B. $\frac{134}{5}$
• C. $\frac{112}{5}$
• D. $\frac{92}{5}$

Answer 1

The correct option is A $\frac{536}{25}$

Let $\overline{x}$ be the mean of $7$ observations
$\therefore \overline{x}=8\Rightarrow \sum _{i=1}^7{x}_i=56$ and
$\frac{\sum _{i=1}^7{x}^2}{7}-(\overline{x}{)}^2=16\Rightarrow \sum _{i=1}^7{x}^2=560$
Sum of remaining $5$ observations
$=\sum _{i=1}^{5}x=56-(8+6)=42$
Sum of squares of $5$ observation $=\sum _{i=1}^5{x}^2=560-6^2-8^2=460$
Variance of remaining $5$ observations
$=\frac{\sum _{i=1}^5{x}^2}{5}-{\left(\frac{\sum _{i=1}^{5}x}{5}\right)}^2=\frac{460}{5}-{\left(\frac{42}{5}\right)}^2=\frac{536}{25}$