Question

find the z-scores that seperate the middle 16 % of the distribution from the area in the tails of the standard normal distribution. the z-scores are?

Answer 1

solution

middle 16% of score is

P(-z < Z < z) = 0.16

P(Z < z) - P(Z < -z) = 0.16

2 P(Z < z) - 1 = 0.16

2 P(Z < z) = 1 + 0.16 = 1.16

P(Z < z) = 1.16 / 2 = 0.58

P(Z < 0.20) = 0.58

z  ± 0.20 using z table