Question

Find the following z values for the standard normal variable Z. (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.)

a. P(Z ≤ z) = 0.8133

b. P(Z > z) = 0.8125

c. P(−z ≤ Z ≤ z) = 0.96

d. P(0 ≤ Z ≤ z) = 0.2991

Answer 1

Solution

Using standard normal table,

(a)

P(Z z) = 0.8133

P(Z 0.89) = 0.8133

z value = 0.89

(b)

P(Z > z) = 0.8125

1 - P(Z < z) = 0.8125

P(Z < z) = 1 - 0.8125 = 0.1875

P(Z < -0.89) = 0.1875

z value = -0.89

(c)

P(-z Z z) = 0.96

P(Z z) - P(Z -z) = 0.96

2P(Z z) - 1 = 0.96

P(Z z) = 1 + 0.96 = 1.96

P(Z z) = 1.96 / 2 = 0.98

P(Z 2.05) = 0.98

z value = 2.05

(d) .

P(0 Z z) = 0.2991

P(Z z) - P(Z 0) = 0.2991

P(Z z) = P(Z 0) + 0.2991

P(Z z) = 0.5 + 0.2991 = 0.7991

P(Z 0.84) =0.7991

z value = 0.84