Suppose that, on average, electricians earn approximately [tex]$\mu=$54,000[/tex] per year in the United States. Assume that the distribution for electricians' yearly earnings is normally distributed and that the standard deviation is [tex]$\sigma=$12,000[/tex] dollars. What is the probability that the average salary of four randomly selected electricians is more than [tex]$50,000[/tex] but less than [tex]$60,000[/tex]?
A. 0.8413
B. 0.7486
C. 0.5889
D. 0.9048
Answer (1)
Calculate the standard error of the mean: σ x ˉ = 4 12000 = 6000 .
Calculate the z-scores for $50,000 and 60 , 000 : z_1 = -\frac{2}{3} an d z_2 = 1$.
Find the probabilities: P ( z < 1 ) ≈ 0.8413 and P ( z < − 3 2 ) ≈ 0.2525 .
Calculate the probability: P ( − 0.6667 < z < 1 ) = 0.8413 − 0.2525 = 0.5888 .
Explanation
Understand the problem First, we need to understand the problem. We are given that the average salary of electricians is $54,000 with a standard deviation of $12,000. We want to find the probability that the average salary of a sample of 4 electricians is between $50,000 and $60,000.
Calculate the standard error of the mean To solve this, we need to calculate the standard error of the mean, which is the standard deviation divided by the square root of the sample size. In this case, the standard error is: σ x ˉ = n σ = 4 12000 = 2 12000 = 6000
Calculate the z-scores Next, we need to calculate the z-scores for $50,000 and 60 , 000. T h ez − score i sc a l c u l a t e d a s : z = σ x ˉ x − μ $For 50 , 000 : z 1 = 6000 50000 − 54000 = 6000 − 4000 = − 3 2 ≈ − 0.6667 $For 60 , 000 : z 2 = 6000 60000 − 54000 = 6000 6000 = 1 $
Find the probability Now, we need to find the probability that the z-score is between -0.6667 and 1. This is the same as finding the area under the standard normal curve between these two z-scores. We can find this by subtracting the probability that z is less than -0.6667 from the probability that z is less than 1: P ( − 0.6667 < z < 1 ) = P ( z < 1 ) − P ( z < − 0.6667 ) Using a standard normal table or calculator, we find that: P ( z < 1 ) ≈ 0.8413 P ( z < − 0.6667 ) ≈ 0.2525 Therefore, P ( − 0.6667 < z < 1 ) = 0.8413 − 0.2525 = 0.5888
Final Answer Therefore, the probability that the average salary of four randomly selected electricians is more than $50,000 but less than $60,000 is approximately 0.5888.
Examples
Understanding the distribution of salaries can help in various real-world scenarios. For instance, a company might want to estimate the probability of needing to offer a certain salary range to attract qualified electricians. Similarly, labor economists can use this type of analysis to study wage inequality and the impact of various policies on income distribution. By understanding the distribution and probabilities, informed decisions can be made regarding compensation and economic planning.
