NTA Abhyas JEE Main2020MathematicsStatisticsPractice
The mean of five observations is 4 and their variance is 2.8 . If three of these observations are 2 ,   2 and 5 , then the other two are
Options
- A2 and 9
- B3 and 8
- C4 and 7
- D5 and 6
Correct answer
D. 5 and 6
Step-by-step solution
x ¯ = 4 , N = 5 and Σ x - x ¯ 2 N = 2.8 ⇒ Σ x - x ¯ 2 = 2.8 5 ∴ Σ x - x ¯ 2 = 1 4 ∴ 2 - 4 2 + 2 - 4 2 + 5 - 4 2 + α - 4 2 + β - 4 2 = 1 4 Where α , β are the other two observations. ∴ 4 + 4 + 1 + α - 4 2 + β - 4 2 = 1 4 ∴ α - 4 2 + β - 4 2 = 5 Also, 2 + 2 + 5 + α + β 5 = 4 ∴ α + β = 20 - 9 = 1 1 Clearly 5 , 6 only satisfy the above equations in α , β . Hence, the required numbers are 5,6 .