Questions: Suppose that, on average, electricians earn approximately μ=74,000 per year in the United States. Assume that the distribution for electricians' yearly earnings is normally distributed and that the standard deviation is σ=12,000. What is the probability that the average salary of a randomly selected electrician is less than 60,000?

Solution Steps

We need to find the probability that the average salary of a randomly selected electrician is less than \$60,000, given that the average salary \( \mu \) is \$74,000 and the standard deviation \( \sigma \) is \$12,000. The distribution of electricians' yearly earnings is assumed to be normally distributed.

To find the probability, we first calculate the Z-scores for the given salary threshold. The Z-score is calculated using the formula:

\[ Z = \frac{X - \mu}{\sigma} \]

where:

• \( X \) is the value we are interested in (in this case, \$60,000),
• \( \mu = 74000 \),
• \( \sigma = 12000 \).

Calculating the Z-score for \$60,000:

\[ Z_{end} = \frac{60000 - 74000}{12000} = \frac{-14000}{12000} = -1.1667 \]

The Z-score for the lower bound (negative infinity) is:

\[ Z_{start} = -\infty \]

Using the Z-scores, we can find the probability that the average salary is less than \$60,000. This is given by:

\[ P(X < 60000) = \Phi(Z_{end}) - \Phi(Z_{start}) \]

Where \( \Phi \) is the cumulative distribution function (CDF) of the standard normal distribution. Since \( \Phi(Z_{start}) = \Phi(-\infty) = 0 \), we have:

\[ P(X < 60000) = \Phi(-1.1667) - 0 = \Phi(-1.1667) \]

From the calculations, we find:

\[ P(X < 60000) \approx 0.1217 \]

Final Answer