Question

A waiter believes the distribution of his tips has a model that is slightly skewed to the right, with a mean of $8.20 and a standard deviation of $5.80. He usually waits on about 40 parties over a weekend of work.

​a) Estimate the probability that he will earn at least $450.

​b) How much does he earn on the best 5% of such​ weekends?

Show work on TI-84

Answer 1

a) average =450/40 = 11.25

=>  P(X-bar >= 11.25) = P ( Z > ((11.25-8.2) / (5.80/sqrt(40)) )

= P(Z > 3.325)

= 1 − P(Z < 3.325)

= 1 − 0.9996

= 0.0004

b)P ((x -mu)/(s/sqrtn)) = z

= ((X - 8.2) / (5.80/sqrt(40) = 2.33

= (X - 8.2) = 2.33*0.9171

= X - 8.2 = 2.1368

= X = 8.2+  2.2841

= X = 10.3368

=) tips for 40 parties = 10.3368* 40 = 413.472 best 5% of the weekends

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