The average hourly wage of workers at a fast food restaurant is $6.50/hr. assume the wages are normally distributed with a standard deviation of $0.45. if a worker at this fast food restaurant is selected at random, what is the probability that the worker earns more than $6.75

Salaries are normally distributed

Average salary per hour: $6.50 ⇒ μ=$6.50

σ=$0.45 (Standard deviation)

Probability that the selected worker will earn $6.75:

P (x>6.75)

Now you must standardize the normal random variables:

If x~N (μσ²) = > Z = (x-μ) / σ

Z = (6.75-6.50) / 0.45

Z=0.556 = > P (Z>0.556) = ? (To find this probability, you must use the tables for Standard normal distribution).

Then:

P (Z>0.556) = 0.2877

The probability of selecting a worker who earns more than $ 6.75 is 28.7%