Question

Consider a normally distributed population with a mean ? = 199 and standard deviation ? = 81. Suppose random samples of size n = 9 are selected from this population. 1. a) What is the mean of the distribution of the sample mean? ?_x? = b) What is the standard error of the mean? (2 decimal places) ?_x? = In the following questions, round the standard error and z-scores to exactly 2 decimal places before determining probabilities. Report probabilities accurate to at least 3 decimal places. 2. What is the probability that a randomly selected sample mean is: a) greater than 182.8? b) less than 207.1? c) greater than 182.8 and less than 207.1? d) greater than 182.8 given less than 207.1? e) greater than 182.8 or greater than 207.1? 3. Between what values would you expect to find the middle 56% of the sample means? Round to the nearest integer. Place the smaller value in the first box and the larger value in the second box. Between and . 4. Why are we able to use the normal distribution in the calculations above? Because the original population is normal Because the sample mean is large enough Because the sample size is large enough

          Consider a normally distributed population with a mean ? = 199 and standard deviation ? = 81. Suppose random samples of size n = 9 are selected from this population.

1. a) What is the mean of the distribution of the sample mean?
?_x? =
b) What is the standard error of the mean? (2 decimal places)
?_x? =

In the following questions, round the standard error and z-scores to exactly 2 decimal places before determining probabilities. Report probabilities accurate to at least 3 decimal places.

2. What is the probability that a randomly selected sample mean is:
a) greater than 182.8?
b) less than 207.1?
c) greater than 182.8 and less than 207.1?
d) greater than 182.8 given less than 207.1?
e) greater than 182.8 or greater than 207.1?

3. Between what values would you expect to find the middle 56% of the sample means?
Round to the nearest integer.
Place the smaller value in the first box and the larger value in the second box.
Between and .

4. Why are we able to use the normal distribution in the calculations above?
Because the original population is normal
Because the sample mean is large enough
Because the sample size is large enough

Show more…

Consider a normally distributed population with a mean ? = 199 and standard deviation ? = 81. Suppose random samples of size n = 9 are selected from this population.

1. a) What is the mean of the distribution of the sample mean?
?x? =
b) What is the standard error of the mean? (2 decimal places)
?x? =

In the following questions, round the standard error and z-scores to exactly 2 decimal places before determining probabilities. Report probabilities accurate to at least 3 decimal places.

2. What is the probability that a randomly selected sample mean is:
a) greater than 182.8?
b) less than 207.1?
c) greater than 182.8 and less than 207.1?
d) greater than 182.8 given less than 207.1?
e) greater than 182.8 or greater than 207.1?

3. Between what values would you expect to find the middle 56% of the sample means?
Round to the nearest integer.
Place the smaller value in the first box and the larger value in the second box.
Between and .

4. Why are we able to use the normal distribution in the calculations above?
Because the original population is normal
Because the sample mean is large enough
Because the sample size is large enough

Added by Richard M.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Jeff Vermeire

07/20/2022

Step 1

Step 1: Calculate the mean of the distribution of the sample mean: The mean of the population is 199. Show more…

Show all steps

lock

Thanks for your feedback!

Consider a normally distributed population with a mean μ = 199 and standard deviation σ = 81. Suppose random samples of size n = 9 are selected from this population. 1. a) What is the mean of the distribution of the sample mean? μ_x̄ = b) What is the standard error of the mean? (2 decimal places) σ_x̄ = In the following questions, round the standard error and z-scores to exactly 2 decimal places before determining probabilities. Report probabilities accurate to at least 3 decimal places. 2. What is the probability that a randomly selected sample mean is: a) greater than 182.8? b) less than 207.1? c) greater than 182.8 and less than 207.1? d) greater than 182.8 given less than 207.1? e) greater than 182.8 or greater than 207.1? 3. Between what values would you expect to find the middle 56% of the sample means? Round to the nearest integer. Place the smaller value in the first box and the larger value in the second box. Between and . 4. Why are we able to use the normal distribution in the calculations above? Because the original population is normal Because the sample mean is large enough Because the sample size is large enough

Play audio

Feedback

Powered by NumerAI

Jeff Vermeire and 59 other subject Intro Stats / AP Statistics educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Recommended Videos

Recommended Textbooks

Transcript

00:01 We have a normally distributed population with a mean of 199, a standard deviation of 81, and a sample size of 9.

00:11 And we want to calculate a couple things.

00:14 The first thing at access is what is the sample mean.

00:19 And here that would just be the mean of the population, so 199.

00:25 And the second part of this question is what is the standard error of the mean, which is the same as the same standard deviation, which is just our population standard deviation divided by the square root of the number of people in our trial.

00:45 So 81 divided by the square root of 9, which is 81 divided by 3 or 27.

00:56 Next, we're asked, what is the probability that our sample mean is greater than 182 .8.

01:07 And for this one, you use the z score, or z is equal to our statistic minus the sample mean divided by the standard error of the mean.

01:22 So that's going to equal 182 .8 minus 199 divided by 27, which gives us a value of negative 0 .60.

01:36 And if we look at a z table, that'll give us an area under the curve of 0 .2742.

01:49 But that's the area to the left.

01:53 That's the less than value.

01:55 And since we're looking for the probability that it's greater than 182 .8, we have to take the complement of this.

02:03 So the probability that the sample mean is greater than 182 .8.

02:08 82 .8 is equal to 1 minus 0 .2742 or 0 .7258.

02:27 The next thing we're asked is the probability that the sample mean is less than 207 .1.

02:35 And we do the same thing we just did with this previous problem, except since we're looking for the value that is less than.

02:42 Then we don't need to take the complement.

02:45 So starting with the z score, z is equal to 207 .1 minus 199 by 27, which is equal to 0 .3.

03:00 And that gives us an area of, let me just write this x bar, less than 207 .1.

03:15 That gives us an area of zero.

03:18 0 .6179.

03:22 It's kind of ugly, so i'll rewrite it.

03:24 0 .6179.

03:32 Next, we're asked to calculate the probability that the sample mean is between 182 .8 and 207 .1.

03:51 Well, we can already use the information that we've calculated to figure this out.

03:57 We've already figured out that the probability that's less than 207 .1 is 0 .6179.

04:07 And the probability that it's greater than 182 .8 would be this area, the .272...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later