The average hourly wage of employees of a certain company is $14.24. Assume the variable is normally distributed. If the standard deviation is $3.27, find the probability that a randomly selected employee earns more than $18.98.
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To find the probability that a randomly selected employee earns more than $18.98, we need to calculate the area under the normal distribution curve to the right of $18.98.
First, we need to standardize the value $18.98 to convert it into a Z-score. The formula for calculating the Z-score is:
Z = (X - μ) / σ
Where:
- X is the value we want to standardize ($18.98 in this case)
- μ is the mean (average) of the distribution ($14.24 in this case)
- σ is the standard deviation of the distribution ($3.27 in this case)
Using the provided values, the Z-score can be calculated as:
Z = (18.98 - 14.24) / 3.27
Z = 1.448
Once we have the Z-score, we can use a standard normal distribution table or calculator to find the probability associated with it. Since we're interested in the area to the right of the Z-score, we need to subtract the cumulative probability from 1.
P(X > 18.98) = 1 - P(Z < 1.448)
Using a standard normal distribution table or calculator, we find that the cumulative probability for Z = 1.448 is approximately 0.9265.
P(X > 18.98) = 1 - 0.9265
P(X > 18.98) ≈ 0.0735
Therefore, the probability that a randomly selected employee earns more than $18.98 is approximately 0.0735 or 7.35%.