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Let x1,""x2,"". . . . . . ,""xn be n obs...

Let `x_1,""x_2,"". . . . . . ,""x_n` be n observations, and let ` bar x` be their arithematic mean and `sigma^2` be their variance. Statement 1: Variance of `2x_1,""2x_2,"". . . . . . ,""2x_n""i s""4""sigma^2` . Statement 2: Arithmetic mean of `2x_1,""2x_2,"". . . . . . ,""2x_n""i s""4x` . (1) Statement 1 is false, statement 2 is true (2) Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1 (3) Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1 (4) Statement 1 is true, statement 2 is false

A

Statement 1 is false, statement 2 is true.

B

Statement 1 is true, statement 2 is true, statement 2 is a correct explanation for statement 1.

C

Statement 1 is true, statement 2 is true , statement 2 is not a correct explanation for statement 1.

D

Statement 1 is true, statement 2 is false.

Text Solution

Verified by Experts

The correct Answer is:
D

A.M. of `2x_(1), 2x_(2)..2x_(n) " is" (2x_(1)+2x_(2)+..+2x_(n))/(n)`
`=2((x_(1)+x_(2)+..+x_(n))/(n))=2overline(x)`
So statement 2 is false.
variance `(2x_(i)=2^(2)` variance `(x_(i)=4sigma^(2)`
So statement 1 is true.
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