Probability distribution; what is probability that i wont spend more than 75% of the customers

Question

Very simple question but I am kinda stuck.

Suppose that evidence in sales data suggests that average amount spent by a customer in a store averages 35.60 with a standard deviation of 16.4. A sample of 64 customers is selected at random for study.

I cant figure out how to find answer to part D.

For part A,B,C I used

• mu=35.6

  n=64

  sigma=16.4

  Standard Error= 16.4/sqrt 64 = 2.05

  z=(x-u)/(q/sqrt of n)

  so x minus mu over Standard error (sigma over sqrt of n).

If you guys could help me check that my answers are right?

A. What is the probability that the sample mean of the amount spent exceeds 38.00?

Answer:z= 38-35.6/2.05 = 1.171; P(0.1210=12.10%)

B. What is the probability that the sample mean of the amount spent is between 35.00$(x1) and 38.00$(x2) ?

z of x1 = -0.293; P = 0.3859

z of x2 = 0.195; P = 0.4247

1-(0.3859+0.4237)= 0.18936

C. What is the probability that the sample mean of the amount spent is no more than 34.00?

z= -0.78; P(0.2236)

D. what is the maximum you are willing to spend if you dont want to spend more than 75% of the customers? I dont know how to do this one.

I am prepping for exam and this question is taken directly out of past ones. If anyone has sometime to help with some questions it would really mean a lot to me.

Answer 1

Part A is correct.

Part B has the wrong z value for $x_2$. It should be the same you had before, 1.17. But the method is correct.

Part C has the wrong p value for $z=-0.78$. Check the tables.

For part D you are considering the distribution of X (a typical shopper) and not the sample mean, so use $SD=16.4$. First you need the z value corresponding to the upper quartile, approx $0.675$

Then, $$0.675=\frac{X-35.6}{16.4}$$

Find $X$

I hope this is OK! Good luck with the exam.