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 dollars. What is the probability that the average salary of four randomly selected electricians is more than $50,000 but less than $60,000?

Answer 1

Answer:

$P(50000$

And we can find this probability with this difference:

$P(-0.33$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

$P(-0.33$

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(54000,12000)$  

Where $\mu=54000$ and $\sigma=12000$

We are interested on this probability

$P(50000$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

$P(50000$

And we can find this probability with this difference:

$P(-0.33$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.  

$P(-0.33$