Question

Suppose that, on average, electricians earn approximately μ= $54,000 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 sample mean is greater than $66,000?

Answer 1

Answer:

0.15866.

Step-by-step explanation:

We have been given that on average, electricians earn approximately μ= $54,000 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. We are asked to find the probability that the sample mean is greater than $66,000.

First of all, we will find the z-score corresponding to 66,000 using z-score formula.

$z=\frac{x-\mu}{\sigma}$

$z=\frac{66000-54000}{12000}$

$z=\frac{12000}{12000}$

$z=1$

Now, we need to find the probability that z-score is greater than 1 that is $P(z>1)$.

Upon using formula $P(z>A)=1-P(z$, we will get:

$P(z>1)=1-P(z$

Upon using normal distribution table, we will get:

$P(z>1)=1-0.84134$

$P(z>1)=0.15866$

Therefore, the probability that the sample mean is greater than $66,000 would be 0.15866 or approximately $15.87\%$.