A waiter believes the distribution of his tips has a model that is slightly skewed to the left, with a mean of $9.40 and a standard deviation of $4.40. He usually waits on about 40 parties over a weekend of work. a) Estimate the probability what he will earn at least $400. b) how how much does he earn on the best 1% of such weekends? a) P(tips from 40 parties>$400=

a) 0.194

b) $440.8

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $9.40

Standard Deviation, σ = $4.40

Formula:

Standard error due to sampling =

a) P(tips from 40 parties > $400)

Thus, the average pay is

P(x > 10)

Calculation the value from standard normal z table, we have,  

b) We have to find the value of x such that the probability is 0.01

Calculation the value from standard normal z table, we have,  

Thus, the mean tip is $11.02

Tip for 40 parties =

Thus, he earns $440.8 on the best 1% of such weekends.

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts