1. ## Probability Problem

I am stumped on this one a bit.

In 2001, the average conventional first mortgage for new single-family homes was \$245,000. Assuming a normal distribution and a standard deviation of \$30,000, what first-mortgage amount would have been exceeded by only 5% of the mortgage customers?

I approached this problem with the idea that I need to find the Z value corresponding .45 in the right side of the normal distribution. Given that they want to know which amount is only exceeded by 5% of first time buyers than I assumed that we need to know the probability in the right tail which is .05. So then I used .05 with the equation z=x-mean/std dev

After all, I am looking for an X value. Well Here is the math that I did:

.0199 (Prob of right tail) = x - 245,000/30000
x=245,597 which is not correct according to the answer in the back of our textbook

Any ideas? Thanks much.

2. In addition, I tried calculating the X value using .45 and finding the corresponding Z value:

.1736=x-245,000/30000
x=\$250,208 which is still not the correct answer according to our textbook.

Any ideas?

Thanks much.

3. Originally Posted by jkeelsnc
I am stumped on this one a bit.

In 2001, the average conventional first mortgage for new single-family homes was \$245,000. Assuming a normal distribution and a standard deviation of \$30,000, what first-mortgage amount would have been exceeded by only 5% of the mortgage customers?

I approached this problem with the idea that I need to find the Z value corresponding .45 in the right side of the normal distribution. Given that they want to know which amount is only exceeded by 5% of first time buyers than I assumed that we need to know the probability in the right tail which is .05. So then I used .05 with the equation z=x-mean/std dev

After all, I am looking for an X value. Well Here is the math that I did:

.0199 (Prob of right tail) = x - 245,000/30000
x=245,597 which is not correct according to the answer in the back of our textbook

Any ideas? Thanks much.

How's about using the corresponding Z-Score of 1.6445 where to the right of this point only 5% of the unit normal distribution lies.

Thus, X = (1.6445)*30,000 + 245,000 = 294,335.

4. ## Probability

Thanks for the help. I got it right now.

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