Inactive Tutor answered 03/05/22
(646-613)/47 = 33/47 = about 0.7021= the z score.
Look it up on z tables or on an online z score calculator or a scientific calculator with statistics functions.
P(>646) = 0.2413. about 24.13%
Sally Y.
asked 03/05/22The amount of money spent weekly on cleaning, maintenance, and repairs at a large restaurant was observed over a long period of time to be approximately normally distributed, with mean 𝜇 = $613 and standard deviation 𝜎 = $47.
(a) If $646 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount? (Round your answer to four decimal places.)
(b) How much should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only 0.11? (Round your answer to the nearest dollar.)
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Inactive Tutor answered 03/05/22
(646-613)/47 = 33/47 = about 0.7021= the z score.
Look it up on z tables or on an online z score calculator or a scientific calculator with statistics functions.
P(>646) = 0.2413. about 24.13%
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