Join our free STEM summer bootcamps taught by experts. Space is limited.Register Here 🏕

# The average teacher's salary in Connecticut (ranked first among states) is $\$ 57,337$. Suppose that the distribution of salaries is normal with a standard deviation of$\$7500 .$a. What is the probability that a randomly selected teacher makes less than $\$ 52,000$per year?b. If we sample 100 teachers' salaries, what is the probability that the sample mean is less than$\$56,000 ?$

### Discussion

You must be signed in to discuss.

### Video Transcript

right in this problem. We have two parts, even a and A B part, and both parts are centered around the fact that we have an average teacher salary. So we're talking about the population of teachers in Connecticut having an average salary of 57,000 $337 and the standard deviation is$7500. So we're going to use that in both huts. So let's focus on part A and in part A. We want the probability that a teacher one teacher has an A salary that's less than $52,000 a year. So with that, we're going to have to draw a bell curve. And with the bell curve, we always put the average in the center. So we're gonna have 57,000 337 in the center, and we're looking for on average or s aureus salary less than 52,000. So the next thing we're gonna have to do is we're going to have to calculate the Z score associated with 52,000, So we're going to 52,000 minus 57,337 divided by the standard deviation of 7500 and the Z score associated with the 52,000 is approximately negative 0.71 So what we can do is we can go back to our picture, and we could say that 52,000 is equivalent to a Z score of negative 0.71 So we can also rewrite our problem to say the probability that the Z score is less than negative 0.71 sear, then going to use your standard normal distribution chart in the back of your textbook. And when we look up negative 0.71 we're going to get a value of point 2389 So for part a, the probability that a teacher selected has a salary of less than$52,000 would be 0.2389 Now let's go to Part B and in Part B. We're going to use the same information about the population, and we know that the average salary of the teacher in Connecticut was 57,337 and the standard deviation was 7500. But in part B, we're now going to select a sample of teachers and in this instance were selecting 100 teachers. So our sample size is 100 and we now want to find out what the probability is that the sample mean of those 100 is less than \$56,000. So since we have a sample of 100 we're talking sample means the central limit theorem is going to go into play. And according to the Central Limit Theorem, the average of the distribution of means will be the same as the average of the population, which in this case is 57,300 37. And the standard error of the means is going to be the standard deviation of the population divided by the square root of the sample size. And again, in this case, it's going to be 7500 divided by the square root of when, So we're going to want to draw another bell curve, and again we're gonna put the average in the center, and 56,000 is gonna be over here. And we're looking for, on average to be less than 56. That was so we need to find busy score, associate it with 56,000. So our Z score is going to be 56,000 minus 57,300 37 divided by the standard deviation of those means with standard error of the mean, which is our Sigma, or the 7500 divided by the square root of 100. And that turns out to be approximately negative 1.78 So we can now go back to our picture. And we could say that the Z score associated with 56,000 is negative 1.78 so we can rewrite our problem instead of us talking about the probability that the average is less than 56,000. We can talk about the probability that the Z score is less than negative 1.78 So we're then going to go to our standard normal distribution chart in the back of the book, and we're going to find the probability to be 0.375 So in summary, Part B was asking us what's the probability that when we select 100 teachers, their average salary is less than 56,000 and the answer would be 0.3 75

WAHS