Question

Test grades on the last statistics exam had a mean = 77 and standard deviation = 24 Suppose the teacher decides to curve by subtraction 32 from all scores then doubling the values. If Y represents the new test scores, what is the mean and standard deviation of Y? a) EM-90; ơ--59.2 b) EM = 154;Oy=9.6 c) EM-122; σ-48 EM : 45; σ--27.2 e) None of the above

Answer 1

Test grades on the last stats exam had aa mean=77 and standard deviation =2.4.

Let,these scores are denoted by the random variable X.

The teacher subtracts 32 from all and doubles them then.

So,the new set of values are

Y=2(X-32).

E(X)=77 and var(X)=2.4²=5.76

So,

E(Y)

=2(E(X)-32)

=2(77-32)

=2*45

=90

var(Y)

=4var(X-32)

=4var(X)

=4*5.76

=23.04

So,

standard deviation of Y

=sqrt(23.04)

=4.8

Thus,

(d)E(Y)=90 and sd(Y)=4.8 is the correct answer.