Question

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

a. P(Z ≤ z) = 0.9478

b. P(Z > z) = 0.6321

c. P(-z ≤ Z ≤ z ) = 0.88

d. P( 0 ≤ Z ≤ z) = 0.3232

Please show work, thank you.

Answer 1

a)

P( Z <= z) = 0.9478

From the Z table, z-score for the probability of 0.9478 is 1.624

z = 1.62 ( Rounded to two decimal places)

b)

P( Z > z) = 0.6321

So,

P( Z <= z) = 1 - 0.6321

P( Z <= z) = 0.3679

From Z table, z-score for the probability of 0.3679 is -0.34

z = -0.34

c)

P( -z <= Z <= z) = 0.88

That is

P( Z <= z) - P( Z <= -z) = 0.88

P( Z <= z) - ( 1 - P( Z <= z) ) = 0.88

P( Z <= z) - 1 + P( Z <= z) = 0.88

2 * P( Z <= z) = 1.88

P( Z <= z) = 0.94

From the Z table, z-score for the probability of 0.94 is 1.555

z = 1.56 (Rounded to 2 decimal places)

d)

P( 0 <= Z <= z) = 0.3232

P( Z <= z) - P( Z <= 0) = 0.3232

P( Z <= z) - 0.5 = 0.3232

P( Z <= z) = 0.8282

From the Z table, z-score for the probability of 0.8232 is 0.93

z = 0.93 (Rounded to nearest integer)