Keith is the manager of an auto-care center. Based on service records from the past year, the time (in hours) that a technician requires to complete a standard oil change and inspection follows a right-skewed distribution with p. = 30 minutes and o = 20 minutes. For a promotion, Keith randomly selects 40 current customers and offers them a free oil change and inspection if they redeem the offer during the next month. Keith budgets an average of 35 minutes per customer for a technician to complete the work. If he wants to be 99% certain that he doesn't go "over budget"? To convert 99% into a z-score, we will use ＿ a..normalcdf(35, 99999999, 30, 20)
b..invNorm(0.99, 0, 1)
c...invNorm(0.01, 0, 1)
d...normalcdf(35, 99999999, 30, 3.16)
e.normal cdf(35, 99999999, 0, 1)

plz need help number 1 2 3 4 5 are pic