i might be a bit late but when you are calculating the percentiles with normalcdf, you dont enter the mean and stdev of the statistics. can you explain why? thanks
@@hurboglan1003 @Hurb Oglan using both his method of finding and using the z-score and then using mean of 0 and standard devation of 1 and the method of directly using the statistic given and using the mean and standard deviation given will result in the same answer. You can try this yourself on the problem at 3:47. He finds a z-score of .755, converting the normal distribution with a mean of 64 and SD (standard deviation) of 2.65 into a standard normal distribution, with a mean of 0 and SD of 1. The amount of standard devations away from the mean the statistic is has not changed, only the unit measuring the standard deviation. It's like saying that I am 100% in front of you in a race (I have ran double the amount of space you have). This means I am 50 feet in front of you. If we change the feet to meters, I am still 100% in front of you. Using the z-score he found, he use the calculator function normalcdf to find how the area above his z-score because the problem is asking what percent of girls are ABOVE 66 inches. He does normalcdf(lower=.755(his z-score), upper=99999 (just a big number), and mean=0 and SD=1 (because he converted into a standard normal distribution)) and gets .2251. However, just putting the given mean and standard deviation also provides the same number. Doing normalcdf(lower=66 (because we are trying to find the percent of girls above 66 inches), upper =1000 (just a big number), and mean =64 and SD=2.65 (given in the problem)) gives the answer of .2252, proving that it does not matter if you use z-score with a standard normal distribution or a normal distribution with the given mean and SD. The slight difference in answers is caused by him rounding the z-score from .75471... up to .755, but AP testers will not take off points for this.
For the standard deviation of a sampling distribution for sample means, yes! The problem I was doing towards the end was not asking about a mean but a total time.
at 14:22, you say that normalcdf only deals with z-scores, I'm pretty sure you can enter the true sample mean and standard deviation and enter the true value you want instead of converting to a z-score and assuming a normal curve with mean 0 and std. dev 1.
Yes you are correct. But without technology the only way to find probabilities is with z-scores and a z-table. I personally like to use z-score so my students understand what they are doing instead of having calculator do everything.
Why not explain the normal distribution placement calculation in terms of a definite integral. Just save the integral on the calculator haha. This would work for all these problems. So for area you write an integral either to or from infinite and set equal to the proportion etc.
Few AP stats teachers will teach the normal model using calculus. Many of our students have not taken calculus yet and it is not in the AP stats course of study. Thanks for comment though.
Thank you, really great crash course before the exam!
Thank you! My AP test is this Thursday .
Best of luck to you!!!!
ayyyy same. anyways why doesnt he use 1E99 instead of 99999
@@rahulshah7611 1E99 represents such a high number that it's practically infinity. Typing in 99999 is faster than 1E99
This is soooooooo useful for reviewing! Thank you so much!
you are my savior!!! i’m gonna ace the ap test because of u
Awesome, so glad to hear! Good luck!
i might be a bit late but when you are calculating the percentiles with normalcdf, you dont enter the mean and stdev of the statistics. can you explain why? thanks
He said in another comment that it's because he wants his students to understand what they are doing instead of having the calculator do everything
@@prao2tapper yeah but the calculation would be wrong wouldnt it
@@hurboglan1003 @Hurb Oglan using both his method of finding and using the z-score and then using mean of 0 and standard devation of 1 and the method of directly using the statistic given and using the mean and standard deviation given will result in the same answer.
You can try this yourself on the problem at 3:47. He finds a z-score of .755, converting the normal distribution with a mean of 64 and SD (standard deviation) of 2.65 into a standard normal distribution, with a mean of 0 and SD of 1. The amount of standard devations away from the mean the statistic is has not changed, only the unit measuring the standard deviation. It's like saying that I am 100% in front of you in a race (I have ran double the amount of space you have). This means I am 50 feet in front of you. If we change the feet to meters, I am still 100% in front of you.
Using the z-score he found, he use the calculator function normalcdf to find how the area above his z-score because the problem is asking what percent of girls are ABOVE 66 inches. He does normalcdf(lower=.755(his z-score), upper=99999 (just a big number), and mean=0 and SD=1 (because he converted into a standard normal distribution)) and gets .2251. However, just putting the given mean and standard deviation also provides the same number. Doing normalcdf(lower=66 (because we are trying to find the percent of girls above 66 inches), upper =1000 (just a big number), and mean =64 and SD=2.65 (given in the problem)) gives the answer of .2252, proving that it does not matter if you use z-score with a standard normal distribution or a normal distribution with the given mean and SD.
The slight difference in answers is caused by him rounding the z-score from .75471... up to .755, but AP testers will not take off points for this.
🙏🏻 😁
shouldn't be the standard deviation formula: SD/squareroot n???
For the standard deviation of a sampling distribution for sample means, yes! The problem I was doing towards the end was not asking about a mean but a total time.
Not too many songs over 5.8. Laughs in power metal. Though in all seriousness these reviews are very good prep for my AP test next week.
awesome glad I good help! Good luck!!
at 14:22, you say that normalcdf only deals with z-scores, I'm pretty sure you can enter the true sample mean and standard deviation and enter the true value you want instead of converting to a z-score and assuming a normal curve with mean 0 and std. dev 1.
Yes you are correct. But without technology the only way to find probabilities is with z-scores and a z-table. I personally like to use z-score so my students understand what they are doing instead of having calculator do everything.
Why not explain the normal distribution placement calculation in terms of a definite integral. Just save the integral on the calculator haha. This would work for all these problems. So for area you write an integral either to or from infinite and set equal to the proportion etc.
Few AP stats teachers will teach the normal model using calculus. Many of our students have not taken calculus yet and it is not in the AP stats course of study. Thanks for comment though.