Question

2) A waiter believes the distribution of his tips has a model that is slightly skewed to the right, with a mean of $9.60 and a standard deviation of $5.40.

a) Explain why you cannot estimate the probability that the next 4 parties will tip an average of at least $15.

b) On an average weekend, he serves 40 tables. Estimate the probability that he will earn at least $12.50 per table on such weekends.

c) How much does he earn on the best 10% of such weekends (i.e. 40 tables per weekend)?

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	9.6
std deviation   =σ=	5.400
sample size       =n=	4
std error=σx̅=σ/√n=	2.70

probability that the next 4 parties will tip an average of at least $15:

probability =P(X>15)=P(Z>(15-9.6)/2.7)=P(Z>2)=1-P(Z<2)=1-0.9772=0.0228

b)

sample size       =n=	40
std error=σx̅=σ/√n=	0.8538

probability =P(X>12.5)=P(Z>(12.5-9.6)/0.854)=P(Z>3.4)=1-P(Z<3.4)=1-0.9997=0.0003

c)

for top 10% or at 90th percentile critical value of z=		1.28
therefore corresponding value=mean+z*std deviation=9.6+1.28*0.8538 =			$10.69