12. Refer to the data in the previous problem. expect to see them get total tips
ID: 3362289 • Letter: 1
Question
12. Refer to the data in the previous problem. expect to see them get total tips between $100 and $200? What proportion of Saturdays do you a. 9044 b. .9522 c. 0478 d. 0956 13. Expenditures per shopper at Upscale Clothing have a mean of $120 and a standard deviation of $40 and are normally distributed. What proportion of shoppers do you expect to have expenditures exceeding $200? a. .0401 b. .0228 c. 4773 d. 9772 14. Using the data in the previous problem (# 17 above) if you want to give the top 20% of your shoppers special treatment, approximately what expenditure level should be exceeded so that a shopper receives special treatment? a. $154 b. $120 c. $200 d. $141 15. Insurance claim amounts follow a normal distribution with a mean of $800 and a standard deviation of $300. If a random sample of 16 insurance claims is selected, what is the probability for the sample mean to exceed $900? a. .0912 b. 3694 c. .6306 d. .9088 16. Refer to the data in the previous problem. If a random sample of 9 insurance claims is selected, what is the probability for the sample mean to be less than $700? a. .0912 b. .1587 c. .3694 d. .8413Explanation / Answer
13)
mean = 120
std. dev. = 40
P(X > 200) = P(z > (200 - 120)/40) = P( z > 2) = 0.0228
Option (B) is correct.
15)
mean = 800
std. dev. = 300
n = 16
P(X > 900) = P(z > (900 - 800)/(300/sqrt(16))) = P( z > 1.333) = 0.0912
Option (A) is correct.
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