Olivia F.

asked • 03/27/23

with a mean of 19.6 inches, and standard deviation of 3.2 inches. If 47 items are chosen at random, what is the probability that their mean length is greater than 20.8 inches?

James B.

is this data normally distributed?
Report

03/28/23

Olivia F.

Yes it is.
Report

03/28/23

Marla G.

tutor
The hypothesis you want to test is: x > 20.8, where x is the mean value for the length. If your data is normally distributed, then use the information they give you in the problem, to convert the mean to a std normal value (i.e. x-mean(19.6)/std. dev(3.2)=Z. Now that you have a Z score, or value, you can go to the tables for a std. normal distribution to find out the probability.
Report

04/29/23

Marla G.

tutor
My comment above is actually the answer to the problem! Having technical issues, so I just put it in a comment!
Report

04/29/23

Marla G.

tutor
The hypothesis you want to test is: x > 20.8, where x is the mean value for the length. If your data is normally distributed, then use the information they give you in the problem, to convert the mean to a std normal value (i.e. [x-(mean=19.6) /std. dev(=3.2)]=Z. Now that you have a Z score, or value, you can go to the tables for a std. normal distribution to find out the probability.
Report

04/29/23

Marla G.

tutor
The hypothesis you want to test is: x > 20.8, where x is the mean value for the length. If your data is normally distributed, then use the information they give you in the problem, to convert the mean to a std normal value (i.e. x-mean(19.6)/std. dev(3.2)=Z. Now that you have a Z score, or value, you can go to the tables for a std. normal distribution to find out the probability. I've tried several times to add this as an ANSWER to your question, but I've not had any success trouble shooting my technical issue. I can add an answer to other questions, but not this one!! I hope you see my solution!!
Report

04/29/23

1 Expert Answer

By:

David B. answered • 07/23/23

Tutor
5.0 (257)

Math and Statistics need not be scary

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.