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 0.9534.

(c)

The area between −z and z is 0.2206.

(d)

The area to the left of z is 0.9948.

(e)

The area to the right of z is 0.6915.

Answer 1

(a)

The area to the left of z is 0.1841.

P(Z<z) = 0.1841

From standard normal tables ,

z = -0.9

(b)

The area between −z and z is 0.9534.

P(-z<Z<z) = 0.9534

Because of symmetry;

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

P(-z<Z<z) = 1 - [P(Z>z)+P(Z<-z)]

i.e

P(Z>z) + P(Z<-z) = 1-P(-z<Z<z)=1 - 0.9534=0.0466

2P(Z<-z) = 0.0466

P(Z<-z) =0.0466/2 = 0.0233

From standard normal tables :

-z = -1.99

z=1.99

(c)The area between −z and z is 0.2206.

P(-z<Z<z) = 0.2206

Because of symmetry;

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

P(-z<Z<z) = 1 - [P(Z>z)+P(Z<-z)]

i.e

P(Z>z) + P(Z<-z) = 1-P(-z<Z<z)=1 - 0.2206 =0.7794

2P(Z<-z) = 0.7794

p(Z<-z) = 0.7794/2 = 0.3897

From standard normal tables,

-z = -0.28

z= 0.28

(d)

The area to the left of z is 0.9948.

P(Z<z) = 0.9948

From standard normal tables,

z = 2.56

(e)

The area to the right of z is 0.6915.

P(Z>z) = 0.6915

i.e P(Z>z) = 1-P(Z<z) = 0.6915

P(Z<z) = 1- 0.6915 = 0.3085

From standard normal tables

z = -0.50