Question

FIND THE Z SCORE FOR WHICH 86% OF THE DISTRIBUTION'S AREA LIES BETWEEN -Z AND Z

Answer 1

We want to find, the z-score such that, P(-z < Z < z) = 0.86

P(-z < Z < z) = 0.86

=> 2 * P(Z < z) - 1 = 0.86

=> 2 * P(Z < z) = 1 + 0.86

=> 2 * P(Z < z) = 1.86

=> P(Z < z) = 1.86/2

=> P(Z < z) = 0.93

Using standard normal z-table we get, the z-score corresponding probability of 0.93 is z = 1.48

=> P(-1.48 < Z < 1.48) = 0.86

=> z-score = 1.48