Question

Problem 3-23 (Algorithmic) For the standard normal random variable z, find z for each situation. If required, round your answers to two decimal places. For those boxes in which you must enter subtractive or negative numbers use a minus sign. (Example: -300)

A)The area to the left of z is 0.2175. z =

B)The area between −z and z is 0.9458. z =

C)The area between −z and z is 0.2036. z =

D)The area to the left of z is 0.9904. z =

E)The area to the right of z is 0.6941. z =

Answer 1

Solution:

Given that,

A ) P ( Z < z ) = 0.2175

Using standard normal table

z = - 0.78

B ) P(-z < Z < z) = 0.9458.

P(Z < z) - P(Z < z) = 0.9458.

2P(Z < z) - 1 = 0.9458.

2P(Z < z) = 1 + 0.9458.

2P(Z < z) = 1.9458.

P(Z < z) = 1.9458. / 2 = 0.82

P(Z < z) = 0.9729

Using standard normal table

z = 1.93

C ) P(-z < Z < z) =0.2036

P(Z < z) - P(Z < z) = 0.2036

2P(Z < z) - 1 =0.2036

2P(Z < z) = 1 + 0.2036

2P(Z < z) = 1.2036

P(Z < z) = 1.2036. / 2 = 0.6018

P(Z < z) = 0.6018

Using standard normal table

z = 0.26

D ) P ( Z < z ) = 0.9904

Using standard normal table

z = 2.34

E ) P ( Z > z ) = 0.6941

1 - P ( Z < z ) = 0.6941

Using standard normal table

= 1 -  0.6941

= 0.3059

Using standard normal table

z = - 0.51