[cowgirl3]

I am looking for some help on probabilites of Normal Distributions:

Suppose it is know that the distibution of purchase amounts by customers entering a popular retail store is approximately normal with a mean of $25 and standard deviation of $8.

a)Find the dollar amount such that 75% of all customers spend no more than this amount
b) Find the dollar amount such that 80% of all customers spend at least this amount
c) Find two dollar amounts, equidistant from the mean of $25 such that 90% of all customers purchases are between these values.

I am using the NORMINV function in excel to answer these questions where the first input is probability (x).

My question is for part a, can someone clarify if the probability should be 0.75 or 0.25 (1-.75). I think it should be 0.75

For part b, can someone clarify if the probability should be 0.80 or 0.20 (1-.80). I think it should be 0.20

I am confused between the wording of "Spend no more than" and "spend at least" for part a and b.

For part c, I am not quite sure what I should be doing.

Any help would be appreciated.

Thanks!

[jcaron2]

Cowgirl, you're correct about parts (a) and (b). For part (c), they're asking you to find the middle 90%. In other words, what are the values corresponding to 5% and 95%? 90% of the customers spend amounts between those two values.

[cowgirl3]

So for part C, I found the z score for 90% confidence interval which is 1.65. To find the upper and lower values, I used the equation:
mean (+/- ) z score * standard deviation. So the answers I came up with are $11.80 ($25 - (1.65* $8)) and $38.20 ($25 + (1.65 *$8). Can anyone confirm if I am on the right track with this answer.

Thanks so much!

[Unknown008]

Well, I don't know how to use the confidence interval method, but when I do it using the normal table, I get something very close to your answer. I guess yours is good then :) (My z score is 1.645 while yours is 1.65)

Anyway, the mean is at 50% of the people.

Since the question asks for the amounts equidistant from the mean, such that you get 90% of the people, it means that there is 90/2 = 45% to the right of the mean and 45% to its left.

Now, my normal table gives me the probabilities to the left of the amount. 45% right of the mean is at the 45+50 = 95 percent amount.

That is from a z-score of 1.645.

Converting this to amounts in dollar, I get:





And the other lower value is