The average hourly wage of workers at a fast food restaurant is $6.50/hr with a standard deviation of $0.45. assume that the distribution is normally distributed. if a worker at this fast food restaurant is selected at random, what is the probability that the worker earns more than $6.75? place your answer, rounded to 4 decimal places, in the blank. for example, 0.1776 would be a legitimate entry.

The probability that the worker earns more than $ 6.75 is 0.7108

Step-by-step explanation:

A normal random variable with mean Mu = 6.5 and standard deviation sd = 0.45 is standardized with the transformation:

Z = (X - Mu) / sd = (X - 6.5) / 0.45

For a value of $ 6.75, Z = (6.75 - 6.5) / 0.45 = 0.5556.

P (X > 6.75) = P (Z > 0.5556) = 0.7108.

The probability that the worker earns more than $ 6.75 is 0.7108

the answer is 11.2 (i know i’m late)