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.3.96
- B.3.92
- C.4.04
- D.4.08
Show answer and solution
Correct option: A
Solution
Let n be the number of students. New mean = (Σx + 4)/n = 10.2, and old mean = Σx/n = 10, so 4/n = 0.2, giving n = 20.
Old variance = 4, so Σx² /n - (mean)² = 4 → Σx² = 20(4 + 100) = 2080.
New Σx² = 2080 - 8² + 12² = 2080 - 64 + 144 = 2160.
New variance = 2160/20 - (10.2)² = 108 - 104.04 = 3.96.
Old variance = 4, so Σx² /n - (mean)² = 4 → Σx² = 20(4 + 100) = 2080.
New Σx² = 2080 - 8² + 12² = 2080 - 64 + 144 = 2160.
New variance = 2160/20 - (10.2)² = 108 - 104.04 = 3.96.
[JEE Main 2023 · 25 Jan · Shift 1]