Question

A water believes the distribution of his tips has a model that is slightly skewed to the left, with a mean of 58.90 and a standard deviation of $4.40 He usually waits on about 30 parties over a wecked of work a) Estimate the probability that he will eam at least 5350 b) How much does he car on the best 5% of such weekends? a) Pltips from 30 parties > $350) - (Round to four decimal places as needed)

Answer 1

Answou Given that maan : 4.90 standard deviation o 4:40 (a) PC Hips torom 30 parries > $350) : The average value for one tip 350 = 11.666 30 Plavorage lip > $11.666): P(2> 11-666 - 8.90 4.40 130 - P(Z > 3.443) :1-PCZC3:443) = 1-0.9997 - 0.0003 L (b) P(x > 2 ) = 0.05 X-M - 9.90 = 1-0.05 ou 4.40/530 5-6.90 2 < = 0.95 from 2-value corresponds 2C-8.90 = 1.645 P-value 0.95 is 1.645 4.40/330 I $10.2214 For 30 Partie the tip amount would be 30 X$10.2214 = $ 306.64217 -642