Question

Given that Z is a standard normal random variable, find z for each situation using the Norm.S.Inv function in excel.

a. The area to the right of z is .01

b. The area between -z and z is .9030

c. The area between -z and z is .9948

e. The area to the right of z is .025

f. The area to the right of z is .3300

Answer 1

a. The area to the right of z is .01 = The area to the left of z is 1 - 0.01 = 0.99

P[ Z > z ] = 1 - P[ Z < z ]

=NORM.S.INV(0.99)  

z = 2.33

b. The area between -z and z is .9030

P[ -z < Z < z ] = 0.9030

P[ Z < z ] - P[ Z < -z ] = 0.9030

P[ Z < z ] - P[ Z > z ] = 0.9030

P[ Z < z ] - ( 1 - P[ Z < z ] )= 0.9030

2*P[ Z < z ] - 1 = 0.9030

2*P[ Z < z ] = 1.9030

P[ Z < z ] = 1.9030/2

P[ Z < z ] = 0.9515

=NORM.S.INV(0.9515)

z = 1.66

c. The area between -z and z is .9948

P[ -z < Z < z ] = 0.9948

P[ Z < z ] - P[ Z < -z ] = 0.9948

P[ Z < z ] - P[ Z > z ] = 0.9948

P[ Z < z ] - ( 1 - P[ Z < z ] )= 0.9948

2*P[ Z < z ] - 1 = 0.9948

2*P[ Z < z ] = 1.9948

P[ Z < z ] = 1.9030/2

P[ Z < z ] = 0.9974

=NORM.S.INV(0.9974)

z = 2.79

e. The area to the right of z is .025 = The area to the left of z is 1 - 0.025 = 0.975

P[ Z > z ] = 1 - P[ Z < z ]

=NORM.S.INV(0.975)  

z = 1.96

f. The area to the right of z is .3300 = The area to the left of z is 1 - 0.33 = 0.67

P[ Z > z ] = 1 - P[ Z < z ]

=NORM.S.INV(0.67)  

z = 0.44