Question

A normal distribution has a mean of 21.51 and standard deviation of 2.1. Determine the value below which 18.41 percent of the observations will occur.

Use the following probability distribution function for the next two questions.

x P(x)

0 0.20

3 0.41

4 0.39

What is the expected value of x?

What is the variance of x?

Answer 1

Solution:

a ) Given that,

mean = = 21.51

standard deviation = = 2.1

Using standard normal table,

P(Z < z) = 18.41%

P(Z < z) = 0.1841

P(Z < - 0.90 ) = 0.1841

z = - 0.90

Using z-score formula,

x = z * +

x = - 0.90 * 2.1 + 21.51

= 19.62

x = 19.62

b ) The expected value of x is E(X)

E(X). = X*P (x )

= (0*0.20 + (3*0.41 ) +( 4*0.39 )

= ( 0 + 1.23 + 1.56 )

= 2.79

The expected value of x = 2.79

The variance of x is V(X)

V(X) = x² P (x ) -  ²

= (0²0.20 + 3²0.41 + 4²0.39 ) - 2.79²

= (0 + 3.69 + 6.24 - 7.28 )

= 9.93 - 7.28

= 2.65

The variance of x is V(X) = 2.65