Question

 

Given that z is a standard normal random variable, find z for each situation. (Round your answers to two decimal places.)

(a)

The area to the left of z is 0.1841.

(b)

The area between −z and z is 0.9398.

(c)

The area between −z and z is 0.2052.

(d)

The area to the left of z is 0.9948.

(e)

The area to the right of z is 0.6915.

Answer 1

Solution:-

Given that,

Using standard normal table,

a) P(Z < z) = 0.1841

.= P(Z < -0.90 ) = 0.1841  

z = -0.90

b) P( -z < Z < z) = 0.9398

= P(Z < z) - P(Z <-z ) = 0.9398

= 2P(Z < z) - 1 = 0.9398

= 2P(Z < z) = 1 + 0.9398

= P(Z < z) = 1.9398 / 2

= P(Z < z) = 0.9699

= P(Z < 1.88) = 0.9699

= z  ± 1.88

c) P( -z < Z < z) = 0.2052

= P(Z < z) - P(Z <-z ) = 0.2052

= 2P(Z < z) - 1 = 0.2052

= 2P(Z < z) = 1 + 0.2052

= P(Z < z) = 1.2052 / 2

= P(Z < z) = 0.6026

= P(Z < 0.26) = 0.6026

= z  ± 0.26

d) P(Z < z) = 0.9948

.= P(Z < 2.56) = 0.9948

z = 2.56

e) P(Z > z) = 0.6915

= 1 - P(Z < z) = 0.6915  

= P(Z < z) = 1 - 0.6915

= P(Z < z ) = 0.3085

= P(Z < -0.50 ) = 0.3085  

z = -0.50