Question

The mean and variance of the marks obtained by the students in a test are $10$ and $4$ respectively. Later, the marks of one of the students is increased from $8$ to $12$ . If the new mean of the marks is $10.2$, then their new variance is equal to :

• A. $3.92$
• B. $3.96$
• C. $4.04$
• D. $4.08$

Answer 1

Answer: (B)

$\displaystyle \sum_{ i =1}^{ n } x _{ i }=10 n $
$ \displaystyle\sum_{ i =1}^{ n } x _{ i }-8+12=(10.2) n$
$ \therefore n =20$
Now $ \frac{\displaystyle\sum_{ i =1}^{20} x _{ i }^2}{20}-(10)^2=4 \Rightarrow \displaystyle\sum_{ i =1}^{20} x _{ i }^2=2080 $
$ \frac{\displaystyle\sum_{ i =1}^{20} x _{ i }^2-8^2+12^2}{20}-(10.2)^2 $
$ =108-104.04=3.96$