A grocery store's receipts show that Sunday customer purchases have a skewed distribution with a mean of $30 and a standard deviation of $18. Suppose the store had 316 customers this Sunday. a) Estimate the probability that the store's revenues were at least $10 comma 100. b) If, on a typical Sunday, the store serves 316 customers, how much does the store take in on the worst 1% of such days? Also if you know which program to use on a Ti-83 for these questions please let me know!
a)
expected value of 316 customers =316*30 =9480
and standard deviation =18*sqrt(316)=319.975
for normal distribution z score =(X-μ)/σx |
probability that the store's revenues were at least 10000:
probability =P(X>10000)=P(Z>(10000-9480)/319.975)=P(Z>1.63)=1-P(Z<1.63)=1-0.9479=0.0521 |
(
if using ti-84 press 2nd -vars -use command :normalcdf(10000,1000000,9480,319.975)) |
b)
for 1th percentile critical value of z= | -2.33 | ||
therefore corresponding value=mean+z*std deviation= | 8735.63 |
(for ti-84 : 2nd -vars -command : invnorm(0.01,9480,319.975)
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