Question

Find a value of the standard normal random variable z ​, call it z 0​, such that the following probabilities are satisfied.

a. ​P(z less than or equals z 0​) equals 0.3027

b. ​P(minus z 0less than or equals z less than z 0​) equals 0.1518

c. ​P(z less than or equals z 0​) equals0.7659

d. ​P(z 0 less than or equals z less than or equals​ 0) equals 0.2706

e. ​P( minus z 0 less than or equals z less than z 0​) equals 0.8146

f. ​P(minus1less than z less than z 0​) equals 0.5757

Answer 1

Solution:

Given that,

a ) P ( -z z0 ) = 0.3027

Using standard normal table

z0 = -0.52

b) P(-z0 Z < z0) = 0.1518

P(Z z) - P(Z z) = 0.1518

2P(Z z0) - 1 = 0.1518

2P(Z z0) = 1 + 0.1518

2P(Z -z0) = 1.1518

P(Z z0) = 1.1518 / 2 = 0.5759

P(Z -z0) = 0.5759

Using standard normal table

z0 = 0.19

z0 = - 0.19

c ) P ( -z z0 ) = 0.7659

Using standard normal table

z0 = 0.73

d ) P ( -z z0 ) = 0.2706

Using standard normal table

z0 = -0.61

e) P(-z0 Z < z0) = 0.8146

P(Z z) - P(Z z) =   0.8146

2P(Z z0) - 1 =   0.8146

2P(Z z0) = 1 +  0.8146

2P(Z -z0) = 1. 8146

P(Z z0) = 1.8146 / 2 = 0.5759

P(Z -z0) = 0.9073

Using standard normal table

z0 = 1.32

z0 = - 1.32

f ) P ( -1 Z z0 ) =  0.5757

P ( Z z0 ) - P ( Z -1 ) =  0.5757

P ( Z z0 ) = 0.5757 - 0.1587

P ( Z z0 ) = 0.4170

Using standard normal table

z0 = - 0.21