Question

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

a.	P(Z ≤ z) = 0.8485
b.	P(Z > z) = 0.7329
c.	P(−z ≤ Z ≤ z) = 0.96
d.	P(0 ≤ Z ≤ z) = 0.2517

Answer 1

a)

P(Z <= z) = 0.8485

Using Excel,

= NORM.S.INV ( probability)

= NORM.S.INV ( 0.8485 )

= 1.03

b)

P(Z > z) = 0.7329

P(Z < z) = 1 - 0.7329

P(Z < z) = 0.2671

NORM.S.INV ( probability)

= NORM.S.INV ( 0.2671 )

= -0.62

c)

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

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

P(Z < z) - ( 1 - P(Z < z) ) = 0.96

P(Z < z) - 1 + P(Z < z) = 0.96

2 P(Z < z) = 1.96

P(Z < z) = 0.98

= NORM.S.INV ( probability)

= NORM.S.INV ( 0.98 )

z = 2.05

d)

P(0 <= Z <= z) = 0.2517

P(Z <= z) - P(Z <= 0) = 0.2517

P(Z <= z) - 0.5 = 0.2517

P(Z <= z) = 0.7517

= NORM.S.INV ( probability)

= NORM.S.INV ( 0.7517 )

z = 0.68