Question

The mean and standard deviation in midterm tests for a probability course are 64 and 16, respectively. These quantities for the final exam are 59 and 20. What grade on the final exam is comparable to Velma's 76 on the
midterm?

Answer 1

Let X be the final grade that is comparable to Velma's in the midterm. Then, we have

μ_f = 59, and σ_f = 20

For the midterm, we have X_m = 76 , μ_m = 64, σ_m = 16

To find the final grade that is comparable to Velma's 76 in the midterm, we need to find standardized variable for X and 76.

If X is a random variable with the mean μ and the standard deviation σ , then the standardized random variable of X is X_m^* = X - μ / σ

Since we have X_m = 76 , μ_m = 64 and σ_m = 16 for the midterm, we get

X_m^* = 76-64 / 16 = 12/16 = 3/4

Since we have X_f = X , μ_f = 59, and σ_f = 20 for the midterm, we get

X_f^* = X - 59 / 20

Since the final grade is comparable with the midterm test, we get

X_f^* = X_m^*

On substituting X_f^* = X - 59 / 20 and X_m^* =3/4 , we get

X - 59 / 20 = 3/4

4X -236 = 60

4X = 296

X = 296/4

X = 74