What Kyla Polee said below. I was nearly in tears after hours of pounding my head against the wall on this. You explained perfectly. Thank you so much.
As I was about to cry and give up, I found this!! Praise you, you math genius! I appreciate how well you are able to explain this. I listened to my professor for hours and could not grasp it until you explained it.
This is brilliant. I just watched my lecture and as always, came immediately to your page so I could understand what I just watched. What I DON't understand is why the professor didn't show us how to do these using the calculator - even though we are supposed to have one. Is there a good reason for this? I can understand the need to know the formulas but I'm more likely to make a mistake when doing everything by hand/without a calculator.
This was so helpful! thank you so much! I do have a question/example of a problem on my homework I was wondering if you could make a video about with this type of question with a TI-84... the question is "The mean lifetime of a tire is 42 months with a variance of 49. If 145 tires are sampled, what is the probability that the mean of the sample would differ from the population mean by less than 0.8 months?" just not sure what to do with the 0.8..... I've gone through your other videos and I can't find anything that relates to this, thanks!!!
Allison, you would do the same process as I did in the video: normalcdf(lower, upper, mu, sigma/sqrt(n). For the lower do the mean of 42-0.8 and for the upper 42+0.8, since you want the mean of the sample to be within 0.8 of the population mean. Since they give variance, you would sqrt to get standard deviation, sqrt (49)=7. Hope that helps.
The video is so helpful but how do I solve for probability that a sample mean is greater than another sample mean when there is no x given. Population mean are: 30 and 25. Standard deviation are 2 and 3 respectively And sample mean are 40 and 50 respectively. Please how do I solve the above
You’re saving my life with these videos. My professor explains nothing has lectures from 1970 with no calculator videos.
I'm happy to hear my videos are helping! Spread the word to your classmates!
@@MathandStatsHelp these videos are more helpful then the book or actual lectures from the curriculum, I will spread the word!
What Kyla Polee said below. I was nearly in tears after hours of pounding my head against the wall on this. You explained perfectly. Thank you so much.
Glad it helped
Omg! You saved my life! I was super stressed over this and was actually on the brink of crying. Thank you a thousand times! ❤
Glad it helped
As I was about to cry and give up, I found this!! Praise you, you math genius! I appreciate how well you are able to explain this. I listened to my professor for hours and could not grasp it until you explained it.
Was feeling doomed about the central limit theorem, your videos turned everything around, and now feeling confident in the subject thank you!
Super helpful. Why has no one explained it so well?! Thank you!
Glad you found my explanation helpful.
This channel saved my day again! I am so grateful!
I am so happy I found you. Thank you.
This is brilliant. I just watched my lecture and as always, came immediately to your page so I could understand what I just watched. What I DON't understand is why the professor didn't show us how to do these using the calculator - even though we are supposed to have one. Is there a good reason for this? I can understand the need to know the formulas but I'm more likely to make a mistake when doing everything by hand/without a calculator.
Some professors just don't show it.
You explained in 10 minutes what it took my professor an hour to explain.
Thank you so much, this saved me so much time.
This was so helpful! thank you so much! I do have a question/example of a problem on my homework I was wondering if you could make a video about with this type of question with a TI-84... the question is "The mean lifetime of a tire is 42 months with a variance of 49.
If 145 tires are sampled, what is the probability that the mean of the sample would differ from the population mean by less than 0.8 months?" just not sure what to do with the 0.8..... I've gone through your other videos and I can't find anything that relates to this, thanks!!!
Allison, you would do the same process as I did in the video: normalcdf(lower, upper, mu, sigma/sqrt(n). For the lower do the mean of 42-0.8 and for the upper 42+0.8, since you want the mean of the sample to be within 0.8 of the population mean. Since they give variance, you would sqrt to get standard deviation, sqrt (49)=7. Hope that helps.
@@MathandStatsHelp Thank you so much for the help! It's my first time taking a stats class and your videos have helped tremendously!
If the question says “at least” or “less than or equal too” is it the same as finding “less than” on the calculator ?
Less than doesn't include the value, where the other two do.
This helped a lot!!! Thank You
3:35 Yes, for the calculator, but E(99) is still infinitely far away form infinity.
The video is so helpful but how do I solve for probability that a sample mean is greater than another sample mean when there is no x given.
Population mean are: 30 and 25.
Standard deviation are 2 and 3 respectively
And sample mean are 40 and 50 respectively.
Please how do I solve the above
hi just wanted to say thank you and i love you!!@!@%#@!!!˜˜ so helpful thx!!!!!!!!
I'm glad you found it helpful!
Thank you so much :)
I wanted a looney toons clip
Thank you!
I think my calculator is messed up it didn't give me 0.0015 it gave me 0.4705
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