A waitress believes the distribution of her tips has a model that is slightly skewed to the left, with a mean of $8.30 and a standard deviation of $4.70. She usually waits on about 60 parties over a weekend of work. a) Estimate the probability that she will earn at least $550.

probability = 0.9286

Step-by-step explanation:

given data

mean = $8.30

standard deviation = $4.70

waits parties over a weekend of work =  60

solution

when we earn at least $550 in 60 parties

so average per party will be x =

average x = 9.1667

so  now we get here z value that is

    ...............1

put here value and we get

z =    

z = 0.184404

so  

P(x>9.1667) = P(z>0.184404)

use standard normal table

probability = 1 - 0.0714 = 0.9286

answer: howdy

step-by-step explanation: