Question

Let Z be the standard normal variable. Find the values of z if z satisfies the given probabilities. (Round your answers to two decimal places.) (a) P(Z > z) = 0.9678 z =

(b) P(−z < Z < z) = 0.7888 z =

Answer 1

a) We want to find, the z-score such that, P(Z > z) = 0.9678

P(Z > z) = 0.9678

=> 1 - P(Z < z) = 0.9678

=> P(Z < z) = 1 - 0.9678

=> P(Z < z) = 0.0322

Using standard normal z-table we get, z-score corresponding probability of 0.0322 is, z = -1.85

=> P(Z > -1.85) = 0.9678

=> Z = -1.85

b) We want to find, the z-score such that, P(-z < Z < z) = 0.7888

P(-z < Z < z) = 0.7888

=> 2 * P(Z < z) - 1 = 0.7888

=> 2 * P(Z < z) = 1.7888

=> P(Z < z) = 1.7888/2

=> P(Z < z) = 0.8944

Using standard normal z-table we get z-score corresponding probability of 0.8944 is, z = 1.25

=> P(-1.25 < Z < 1.25) = 0.7888

=> Z = 1.25