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

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

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 * (78 - 31)`

= 94 ----------------(4)

and

V(Y) = 4V(X)

⇒
`\sigma_(Y) = 2 * \sigma_(X)`

⇒
`\sigma_(Y) = 2 * 0.14` = 0.28

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