Given the mean = $12

Standard deviation = $2.50

Suppose that a large state university graduate research assistants are paid by the hour. Data from the personnel office show the distribution of hourly wages paid to graduate students across the campus is approximately normal with a mean of $12, and standard deviation of $2.50

Determine

G) probability of selecting at random from the personnel files a graduate assistant whose hourly wage is extreme in either direction---either $10 or below or $14 or above

I am stuck on this would I do (14-12)/2.5 and (10-12)/2.5 to find the z-score? I'm confused about the proper steps

Continue calculations to find Z.

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the least proportion/probability for each Z score.

Either-or probabilities are found by adding the individual probabilities.

Well, let me explain it to you in my own way, using a little bit of humor!

To find the probability of selecting a graduate assistant with an hourly wage of $10 or below or $14 or above, we need to calculate the z-scores for these values. So, you're on the right track!

For $10 or below, we have (10 - 12) / 2.50, which gives us a z-score of -0.80. Now, for $14 or above, we have (14 - 12) / 2.50, which gives us a z-score of 0.80.

To find the probability of selecting a graduate assistant with an hourly wage at these extreme values, we need to look up the z-scores in the standard normal distribution table.

However, here's where my humor comes in! Picture this: You're at a comedy club, and the comedian says, "Hey, did you know that the standard normal distribution table has all the answers to life's questions?" Well, it might not have ALL the answers, but it sure has the answer we're looking for!

So, let's consult this magical table. When we look up the z-scores of -0.80 and 0.80, we find that the corresponding probabilities are approximately 0.2119 and 0.7881.

Now, to find the probability of selecting a graduate assistant with an hourly wage that is either $10 or below OR $14 or above, we simply add these two probabilities together.

Therefore, the probability of selecting such a graduate assistant is approximately 0.2119 + 0.7881 = 1.

So, there you have it, my friend! The probability of selecting at random from the personnel files a graduate assistant whose hourly wage is extreme in either direction is approximately 1, which means it's highly likely to find such an extreme value.

Hope that brought a smile to your face while understanding the concept!

To find the probability of selecting a graduate assistant whose hourly wage is extreme in either direction, we can calculate the z-scores for the given values and then use the z-table.

Let's first find the z-score for $10:
z = (x - mean) / standard deviation
z = (10 - 12) / 2.50
z = -0.80

Next, let's find the z-score for $14:
z = (x - mean) / standard deviation
z = (14 - 12) / 2.50
z = 0.80

Now, we will use the z-table to find the probability associated with these z-scores.

From the z-table, we can find that the probability associated with a z-score of -0.80 is 0.2119, and the probability associated with a z-score of 0.80 is also 0.2119.

Since we are interested in extreme values (either $10 or below, or $14 or above), we need to find the probability of selecting a graduate assistant whose hourly wage is less than $10 or greater than $14.

To do this, we can subtract the probability associated with a z-score of -0.80 from the total probability of a normal distribution, which is 1. Additionally, we need to add the probability associated with a z-score of 0.80.

Probability of selecting at random a graduate assistant whose hourly wage is extreme in either direction = 1 - 0.2119 + 0.2119 = 1 - 0.2119 + 0.2119 = 1 - 0 = 1.

Therefore, the probability of selecting at random from the personnel files a graduate assistant whose hourly wage is extreme in either direction (either $10 or below or $14 or above) is 1.

To determine the probability of selecting a graduate assistant whose hourly wage is extreme in either direction ($10 or below or $14 or above), you need to find the area under the normal distribution curve beyond these two thresholds.

You are correct in using the formula for calculating the z-score, which is (value - mean) / standard deviation. Let's calculate the z-scores for $10 and $14 using the given mean and standard deviation:

For $10:
z-score = ($10 - $12) / $2.50 = -0.8

For $14:
z-score = ($14 - $12) / $2.50 = 0.8

Now, we need to find the probability associated with each of these z-scores. Since the normal distribution is symmetric, the area in the tail beyond either threshold is the same. Thus, we can calculate the probability for one of the tails and then double it to get the probability for both tails.

To find the probability associated with a z-score, we can use a standard normal distribution table or a calculator. Since both -0.8 and 0.8 have positive z-scores, we need to find the area in the upper tail of the standard normal distribution.

Using a standard normal distribution table or calculator, you can find that the area beyond 0.8 is approximately 0.2119. This means the probability of selecting a graduate assistant with an hourly wage of $14 or above is 0.2119.

Doubling this probability to account for the lower tail as well, the probability of selecting a graduate assistant whose hourly wage is extreme in either direction is approximately 0.2119 * 2 = 0.4238, or 42.38%.

Therefore, there is a 42.38% chance of randomly selecting a graduate assistant whose hourly wage is either $10 or below, or $14 or above.