Question

A waiter believes the distribution of his tips has a model that is slightly skewed to the left, with a mean of $8.40 and a standard deviation of $4.20. He usually waits on about 60 parties over a weekend of work. a) Estimate the probability that he will earn at least $600. b) How much does he earn on the best 5% of such weekends?

Answer 1

a) The average amount of each2 party = 600/60=10

This is a normal distribution question with

z = 2.9509

This implies that

P(x > 10.0) = P(z > 2.9509) = 1 - 0.9984

b) Given in the question

P(X < x) = 0.95

This implies that

P(Z < 1.6448536269514722) = 0.95

With the help of formula for z, we can say that

x = 9.2918

For 60 parties total will be 9.2918*60 = $ 555.708

PS: you have to refer z score table to find the final probabilities.