Question

***PLEASE SHOW HOW TO SOLVE IN EXCEL*** NOT HANDWRITTEN

7) For borrowers with good credit scores, the mean debt for revolving and installment accounts is $15,015. Assume the standard deviation is $3,540 and that debt amounts are normally distributed.

a. What is the probability that the debt for a borrower with good credit is more than $18,000?

b. What is the probability that the debt for a borrower with good credit is less than $10,000?

c. What is the probability that the debt for a borrower with good credit is between $12,000 and $18,000?

d. What is the probability that the debt for a borrower with good credit is no more than $14,000?

Answer 1

Given that mean = 15015 and standard deviation = 3540

(A) To find P(x>18000)

we can P(X>18000) =1 - P(x<18000)

using excel function NORMDIST(x,mean,sd, cumulative)

put x = 18000, mean = 15015 and standard deviation = 3540 and cumulative = TRUE

we get

P(X>18000) =1 - NORMDIST(18000,15015,3540,TRUE)

= 0.1996

(B)

To find P(x<10000)

using excel function NORMDIST(x,mean,sd, cumulative)

put x = 10000, mean = 15015 and standard deviation = 3540 and cumulative = TRUE

we get

P(X<10000) =NORMDIST(10000,15015,3540,TRUE)

= 0.0783

(C)

To find P(12000<x<18000)

using excel function NORMDIST(x,mean,sd, cumulative)

put x = 12000 and y = 18000, mean = 15015 and standard deviation = 3540 and cumulative = TRUE

we get

P(x<X<y) =NORMDIST(y,mean,sd,TRUE)-NORMDIST(x,mean,sd,TRUE)

this implies

P(12000<x<18000) = NORMDIST(18000,15015,3540,TRUE) - NORMDIST(12000,15015,3540,TRUE)

= 0.6033

(D)

To find P(x<14000)

using excel function NORMDIST(x,mean,sd, cumulative)

put x = 14000, mean = 15015 and standard deviation = 3540 and cumulative = TRUE

we get

P(X<14000) =NORMDIST(14000,15015,3540,TRUE)

= 0.3872