Question

In: Statistics and Probability

The amounts of money requested on home loan applications at Down River Federal Savings follow the...

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

Solutions

Expert Solution

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


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