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?
A) 5.2%
B) 28.8%
C) 42.1%
D) 27.6%

Question

Quizplus · Accepted Answer

To find the probability that a worker earns more than $6.75, we use the Z-score formula: $Z = \frac{X - \mu}{\sigma}$, where $X$ is the value of interest ($6.75), $\mu$ is the mean ($6.50), and $\sigma$ is the standard deviation ($0.45). Plugging in the values, we get $Z = \frac{6.75 - 6.50}{0.45} = \frac{0.25}{0.45} \approx 0.56$. Looking up 0.56 on the standard normal distribution table gives us the area to the left of 0.56, which is approximately 0.7123. To find the probability of earning more than $6.75, we subtract this value from 1 (since the total area under the curve is 1). Thus, $1 - 0.7123 = 0.2877$, or 28.8%.

The Average Hourly Wage of Workers at a Fast Food