Question

# A grocery​ store's receipts show that Sunday customer purchases have a skewed distribution with a mean...

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)

#### Earn Coins

Coins can be redeemed for fabulous gifts.