NTA Abhyas JEE Main2020MathematicsStatisticsPractice
If both the mean and the standard deviation of 50 observations x 1 , x 2 , … , x 50 are equal to 16 , then the mean of x 1 - 4 2 , x 2 - 4 2 , … , x 50 - 4 2 is
Options
- A525
- B480
- C400
- D380
Correct answer
C. 400
Step-by-step solution
For observations x 1 ,   x 2 ,   . . . . . . . . . . . x 50 Mean, x - = ∑ x i   50 = 16 . . . . . (i) Variance, σ 2 = ∑ x i 2 50 - ( x - ) 2 = 16 2 ⇒ ∑ x i 2 50 = 16 2 + ( x - ) 2 = 16 2 + 16 2 = 512 . . . . . (ii) So, the mean value of ( x 1 - 4 ) 2 ,   ( x 2 - 4 ) 2 ,   . . . . . . . ( x 50 - 4 ) 2 will be = ∑ ( x i - 4 ) 2 50 = ∑ x i 2 - 8 ∑ x i + 16 × 50 50 = ∑ x i 2 50 - 8 ∑ x i 50 + 16 = 512 - 8 × 16 + 16 = 400 (