Question

If Z is a standard normal random variable, find the value z₀ for the following probabilities. (Round your answers to two decimal places.)

(a) P(Z > z₀) = 0.5

z₀ =

(b) P(Z < z₀) = 0.9279

z₀ =

(c) P(−z₀ < Z < z₀) = 0.90

z₀ =

(d) P(−z₀ < Z < z₀) = 0.99

z₀ =

Answer 1

Using standard normal table,

P(Z > z0) = 0.5

= 1 - P(Z < z0) = 0.5

= P(Z < z0) = 1 - 0.5.

= P(Z < z0 ) = 0.5

= P(Z < 0 ) = 0.5

z0 =0 (using standard normal (Z) table )

(B)

Using standard normal table,

P(Z < z0) = 0.9279

= P(Z < z0) = 0.9279

= P(Z < 1.46) = 0.9279

z0 = 1.46 Using standard normal table,

(C)

middle 90% of score is

P(-z < Z < z) = 0.90

P(Z < z) - P(Z < -z) = 0.90
2 P(Z < z) - 1 = 0.90

2 P(Z < z) = 1 + 0.90 =1.90

P(Z < z) = 1.90 / 2 = 0.95

P(Z < 1.65) = 0.95

z0 ±1.65 using z table

(D)

middle 99% of score is

P(-z < Z < z) = 0.99

P(Z < z) - P(Z < -z) = 0.99
2 P(Z < z) - 1 = 0.99

2 P(Z < z) = 1 + 0.90 =1.99

P(Z < z) = 1.99/ 2 = 0.995

P(Z < 2.58) = 0.995

z0 ±2.58 using z table