Question

The loan department of Southern Bank has found that car loans they have issued over the past year are normally distributed with a mean of $25,000 and a standard deviation of $2,000. One car loan account issued by the bank during the past year is randomly selected from the files of the loan department. Find the probability the amount of the loan was: (a) no more than $24,000 (b) Between $22,000 and $24,000

Answer 1

a)0.3084

b)0.15

Step-by-step explanation:

the key to answer this question is to see that is assumed thar car loans are normally distributed, and as we are giving the mean and variance, we can use the standard normal distribution. let´s first remmember how to do that:

`P[x\leq a]=P[(x-mu)/(sigma) \leq (a-mu)/(sigma)]`

there x is a random variable which is normal but when substrating the mean and dividing by standard deviation is a normal standard distribution, so applying to the problem we have:

a)

`P[(x-25,000)/(2,000) \leq (24,000-25,000)/(2,000)]`

so solving this we have that we are looking for the prbability less than -0.5 we can use any table for normal standard distribution which you can find in any statistics book, and the probability will be 0.30854

b) for the b part as the normal distribution is simetric and has a cumulative probability function we can calucalte the probability for 24,000 and substract it from 22,000 probability, we already calculated te 24,000 so lets calculate the remaining:

`P[(x-25,000)/(2,000) \leq (22,000-25,000)/(2,000)]`

and this probability is 0.1586. so the answer will be 0.30854-0.1586=0.15