Question

A waiter believes the distribution of his tips has a model that is slightly skewed to the left​, with a mean of ​$8.50 and a standard deviation of ​$4.10. He usually waits on about 30 parties over a weekend of work. ​

Round to 4 decimal places

a) Estimate the probability that he will earn at least ​$300. ​

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

Answer 1

The sample size is sufficiently large (at least 30)According to central limit theorem, the distribution of mean tip amount will be normally distributed with mean, = and =

= $8.50

= $4.1

= $8.50

= = 0.75

P( < A) = P(Z < (A - )/)

a) P(he will earn at least ​$300) = P( 300/30)

= P( 10)

= 1 - P( < 10)

= 1 - P(Z < (10 - 8.5)/0.75)

= 1 - P(Z < 2)

= 1 - 0.9772

= 0.0228

b) Let his average earning on the best 10​% of such​ weekends be B

P( > B) = 0.10

P( < B) = 1 - 0.10 = 0.90

P(Z < (B - 8.5)/0.75) = 0.90

Take z value corresponding to probability of 0.9000 from standard normal distribution table

(B - 8.5)/0.75 = 1.28

B = 9.46

Amount earned in the best 10% of such weekends = 9.46 x 30 = $283.80