The average hourly wage of employees of a certain company is $14.58. Assume the variable is normally distributed. If the standard deviation is $3.99, find the probability that a randomly selected employee earns less than $9.79.
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To find the probability that a randomly selected employee earns less than $9.79, we need to use the standard normal distribution.
First, we need to standardize the $9.79 value using the formula:
Z = (X - μ) / σ
where:
Z is the standard score,
X is the value,
μ is the mean, and
σ is the standard deviation.
In this case, X = $9.79, μ = $14.58, and σ = $3.99.
Z = ($9.79 - $14.58) / $3.99
Now we can calculate the Z-score:
Z = (-$4.79) / $3.99
Z ≈ -1.20
Next, we need to find the probability associated with the Z-score of -1.20. This can be done using a standard normal distribution table or a calculator. We'll use the standard normal distribution table.
Looking up the Z-score of -1.20 in the table, we find that the corresponding probability is approximately 0.1151.
Therefore, the probability that a randomly selected employee earns less than $9.79 is approximately 0.1151, or 11.51%.