Test grades on the last statistics exam had a mean 79 and s

Test grades on the last statistics exam had a mean = 79 and standard deviation = .12. Suppose the teacher decides to curve by subtraction 29 from all scores then doubling the values. If Y represents the new test scores, what is the mean and standard deviation of Y? a) E[Y] = 100; Y = -57.76 b) E[Y] = 100; Y = .24 c) E[Y] = 50; Y = -28.76 d) E[Y] = 158; Y = .48 e) E[Y] = 129; Y = .24 f) None of the above

Solution

E(Y)= (79-29)*2=100

V(Y)= 4*0.24^2

Std dev = 2×0.12=0.24