Mean and variance of marks obtained by students in a test are 10 and 4 respectively. Later, marks of

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13.Statistics

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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

$4.04$

B

$4.08$

C

$3.96$

D

$3.92$

(JEE MAIN-2023)

Solution

$\sum \limits_{ i =1}^{ n } x _{ i }=10\,n$

$\text { Now } \frac{\sum \limits_{ i =1}^{ n } x _{ i }^2}{20} x _{ i }-8+12=(10.2) n \quad \therefore n =20$

$\frac{\sum \limits_{ i =1}^{20} x _{ i }2-8^2+12^2}{20}-4 \Rightarrow \sum \limits_{ i =1}^{20} x _{ i }^2=2080$

$=108-104.04=3.96$

Standard 11

Mathematics

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