Question

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

Answer 1

Answer:

The probability of selecting a worker who earns more than $ 6.75 is 0.2877 or 28.77%

Step-by-step explanation:

We are given;

The wages are normally distributed;

Average wage per hour; μ = $6.50

Standard deviation; σ = $0.45

Now, we want to find the probability that the worker earns more than $6.75

So, we'll find the z-value of P(x > 6.75) using the formula Z = (x-μ)/σ

Thus,

Z = (6.75-6.50)/0.45

Z = 0.556

So, looking at the z-distribution table P(Z>0.556) = 1 - P(Z < 0.556) = 1 - 0.7123 = 0.2877

Thus, The probability of selecting a worker who earns more than $ 6.75 is 0.2877 or 28.7%