Question

A waitress believes the distribution of her tips has a model that is slightly skewed to the right​, with a mean of ​$9.80 and a standard deviation of ​$6.80. She usually waits on about 40 parties over a weekend of work. a) Estimate the probability that she will earn at least ​$500. ​b) How much does she earn on the best 1​% of such​ weekends?

Answer 1

Given,

= 9.80 , = 6.80

For n = 40 parties,

X = N ( 40 * 9.80 , 40 * 6.80²) = N ( 392 , 1849.6)

We convert this to standard normal as

P( X < x) = P( Z < x - / )

a)

P( X >= 500) = P( Z >= 500 - 392 / sqrt(1849.6) )

= P( Z >= 2.5112)

= 0.006

b)

We have to calculate X such that

P( X > x) = 0.01

That is

P( X < x) = 0.99

P( Z < x - / ) = 0.99

From the Z table, z-score for the probability of 0.99 is 2.3263

x - / = 2.3263

x - 392 / sqrt( 1849.6) = 2.3263

Solve for x .

x = $492.05