Question

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.

Answer 1

Answer:

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 = $\frac{550}{60}$

average x = 9.1667

so  now we get here z value that is

$z = \frac{x-\mu }{\sigma }$     ...............1

put here value and we get

z =  $\frac{9.1667-8.30}{4.70}$  

z = 0.184404

so  

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

use standard normal table

probability = 1 - 0.0714 = 0.9286