Question

A waiter believes the distribution of his tips has a model that is slightly skewed to the left?, with a mean of ?$10.50 and a standard deviation of ?$6.60. He usually waits on about 50 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?

PLEASE MAKE SURE IT IS CORRECT BECAUSE I ALREADY POSTED IT , THE ANSWERS WAS WRONG

Answer 1

a)

Given,

sample size n = 50

mean  = 10.50

mean for 50 parties = 50* 10.50 = 525

standard deviation  = 6.60

standard deviation for 50 parties = * 6.60 = 46.67

probability that he will earn at least 600 P(X 600) = P( X > 600) //continuous distribution

x =

= (600 - 525) / 46.67

= 1.60

P(X 600) = 0.0548 //from standard normal table

b)

P(X > ) = 5 / 100 = 0.05

P( X < ) = 1 - 0.05 =  0.95

= inverse Norm (0.95)

- = (1.645) * 46.67 //from excel's "NORMS INV" function

= 601.77215