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 :

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

Tardigrade Admin · Accepted Answer

displaystyle ∑ i =1 n x i =10 n displaystyle∑ i =1 n x i -8+12=(10.2) n ∴ n =20 Now ( displaystyle∑ i =120 x i 2/20)-(10)2=4 ⇒ displaystyle∑ i =120 x i 2=2080 ( displaystyle∑ i =120 x i 2-82+122/20)-(10.2)2 =108-104.04=3.96

Q. 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 :

$3.92$

$3.96$

$4.04$

$4.08$

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

Q. 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 :

3.92

3.96

4.04

4.08