Please check my work:

Q; A random sample of 50 check out times at Corner Convenience store was obtained. The checkout time was recorded for each customer in the sample. Suppose the mean of the sample was 65 seconds and the standard deviation was 8.5 seconds.(26 points)

a.Compute the standard score (z - score) for a customer that takes 83 seconds to checkout. Round your answer to the nearest hundredth and interpret the meaning of your answer as it pertains to this problem.(6 points)

83-65= 18 18/8.5= 2.168 A Zscore of of 83 seconds is atleats 2.17 stand devations above the mean

8.5

b.What is checkout time for a customer in the sample who had a standard score (Z - score) of -1.45?Round your answer to the nearest tenth. (5 points)

per Z table of -1.45 = 0.0735 7.4%

c.At least what percent of the checkout times in this sample should we expect to find within2.7 standard deviations of the mean? Give your answer to the nearest percent.(5points)

99.7%

d.At most what percent of the checkout times in this sample should we expect to find less than 48 seconds and more than 82 seconds? (5points)

48-65= -17/8.5= -2 z score 2 is 0.9772 97.7%???????

82-65=17/8.5= 2

e.At least what percent of the checkout times in this sample should we expect to find between 45 seconds and 85 seconds? Give your answer to the nearest percent.(5points)

45-65=20/8.5= -2.35 ???????? do use the Z table??

85-25=20/8.5= 2.35