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^(1-42), x^(2-42), …,x^(50-42) is

Question

MARKS Content Team · Accepted Answer

Correct answer: (C) 400. For observations x_1, x_2, ...........x_50 Mean, x-=(∑x)/(i 50)=16 . . . . . (i) Variance, σ^2=(∑x)/i250-(x^(-)2)=16^2 ⇒(∑x)/i250=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 =(∑(xi-4)^2)/50=(∑xi2-8∑x_i+16×50)/50 =(∑xi2)/50-8(∑x_i)/50+16=512-8×16+16=400 (using (i) and (ii))

If both the mean and the standard deviation of $50$ observations $x_{1}, x_{2}, \dots, x_{50}$ are equal to $16$ , then the mean of ${(x_{1} - 4)}^{2}, {(x_{2} - 4)}^{2}, \dots, {(x_{50} - 4)}^{2}$ is

Solution

More Statistics Questions

If both the mean and the standard deviation of 50 observations x1, x2, …, x50 are equal to 16, then the mean of x1-42, x2-42, …,x50-42 is

Solution

More Statistics Questions

If both the mean and the standard deviation of $50$ observations $x_{1}, x_{2}, \dots, x_{50}$ are equal to $16$ , then the mean of ${(x_{1} - 4)}^{2}, {(x_{2} - 4)}^{2}, \dots, {(x_{50} - 4)}^{2}$ is