Question

Test grades on the last statistics exam had a mean = 78 and standard deviation = .14. Suppose the teacher decides to curve by subtraction 31 from all scores then doubling the values. If Y represents the new test scores, what is the mean and standard deviation of Y?

Answer 1

Answer:

The mean and standard deviation of Y are, 94 and 0.28 respectively.

Step-by-step explanation:

Let the random variable , 'Test grades on the last statistics exam' be X.

Then according to the question,

E(X) = 78 ------------(1)

and

$\sigma_{X}$ = 0.14------------(2)

Now, according to the question,

Y = 2(X - 31)

⇒E(Y) = 2(E(X) - 31)

          = $2 \times (78 - 31)$

          = 94  ----------------(4)

and

V(Y) = 4V(X)

⇒$\sigma_{Y} = 2 \times \sigma_{X}$

⇒$\sigma_{Y} = 2 \times 0.14$ = 0.28

So, the mean and standard deviation of Y are, 94 and 0.28 respectively.