Question

The amounts of money requested on home loan applications at Down River Federal Savings follow the normal distribution, with a mean of $72,000 and a standard deviation of $18,000. A loan application is received this morning. What is the probability that: (Round z-score computation to 2 decimal places and the final answer to 4 decimal places.) a. The amount requested is $86,000 or more? Probability b. The amount requested is between $61,000 and $86,000? Probability c. The amount requested is $61,000 or more? Probability

Answer 1

Solution :

Given that,

mean = = 72,000

standard deviation = = 18,000

a ) P (x > 86,000 )

= 1 - P (x < 86,000 )

= 1 - P ( x -  / ) < ( 86,000 - 72,000 / 18,000)

= 1 - P ( z < 14000 / 18,000 )

= 1 - P ( z < 0.78 )

Using z table

= 1 - 0.7823

= 0.2177

Probability = 0.2177

b ) P ( 61,000 < x < 86,000 )

P ( 61,000 - 72,000 / 18,000) < ( x -  / ) < ( 86,000 - 72,000 / 18,000)

P ( - 11000 / 18,000 < z < 14000 / 18,000)

P (-0.61 < z < 0.78)

P ( z <0.78 ) - P ( z < -0.61)

Using z table

= 0.7823 - 0.2709

= 0.5114

Probability = 0.5114

c ) P (x > 6,1000 )

= 1 - P (x < 61,000 )

= 1 - P ( x -  / ) < ( 61,000 - 72,000 / 18,000)

= 1 - P ( z < -11000 / 18,000 )

= 1 - P ( z < -0.61 )

Using z table

= 1 - 0.2709

= 0.7291

Probability = 0.7291