Statistics class

Question

The scores on a college entrance exam have an approximate normal distribution with a mean of 64, and a standard deviation of 12. Which scores do 95% of the student scores lie between?

Answer 1

We will need to compute P(z-score 1 < z < z-score 2) = 0.95

The 0.95 will be the middle of the normal distribution curve, with 0.025 tails on both sides.

First find z-score corresponding to 0.975 (right tail) = 1.96

Then find value x corresponding to that z-score: (x - 64 ) / 12 = 1.96. That is the upper value (87.52)

Since we are dealing with a symmetric distribution, the lower value will be equidistant from the mean = 64 - (87.52 - 64) = 40.48.