JEE Mains · Maths · STD 11 - 13. statistics
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 \(3.92\)
Answer & Solution
Correct Answer
(C) \(3.96\)
Step-by-step Solution
Detailed explanation
\(\sum \limits_{ i =1}^{ n } x _{ i }=10\,n\) \(\text { Now } \frac{\sum \limits_{ i =1}^{ n } x _{ i }^2}{20} x _{ i }-8+12=(10.2) n \quad \therefore n =20\) \(\frac{\sum \limits_{ i =1}^{20} x _{ i }2-8^2+12^2}{20}-4 \Rightarrow \sum \limits_{ i =1}^{20} x _{ i }^2=2080\)…
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