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 ​$6.50. He usually waits on about 60 parties over a weekend of work. ​

a) Estimate the probability that he will earn at least ​$550.

​b) How much does he earn on the best 10​% of such​ weekends?

Answer 1

Given μ = 8.40, σ = 6.50, n = 60

a)

We have to estimate the probability that he will earn at least​ $550, which means P(X ≥ 550/60) (from each party)

P(X ≥ 9.17) = P(Z≥​(9.17−8.40)/(6.50/√60)) = Pr(Z≥0.9176) = 1 - Pr(Z < 0.9176) = 1−0.8206 = 0.1794

Where Z = (X - μ)/(σ/√n)

the probability that he will earn at least ​$550 is 17.94%

b)

earn on the best 10​% of such​weekends

P(Z = (X - μ)/(σ/√n)) = 0.1

(X - μ)/(σ/√n) = 1.2815

(X−8.40)/(6.50/√60) = 1.2815

X- 8.40 = 1.0748

X = $9.4748

For 60 parties, X = 9.4748*60 = 568.488 = $569 (rounded)