Applying the central limit theorem to find probability example 1

[@esfasdasasdasdas3888]

Thank you! By far this was the clearest explanation I could find.

[@dmcguckian]

Glad it was helpful!

[@christian_uu]

you have no idea how helpful this was, thank you so much!

[@dmcguckian]

I’m glad it was helpful!

[@prettyone9883]

Wowwwwwwww😳😳😳😳a real professor indeed, my Lecturer only lecture us and rap at the same time so I couldn't get the concept well but per your explanation I've been able to figure it out well, Please Prof. Do more videos and make more examples for us pls 🥺🥺🥺 and if You've your private page that you do tutorials on all statistics topic and also all economics topic please kindly add me up 🥺🥺🥺🥺 am pleading 🙏🏻🙏🏻🙏🏻

[@dmcguckian]

I have a website www.STATSprofessor.com. It’s free, and it makes finding the videos you need easier.

[@davidmoretz5389]

One of the best explained examples--thank you!

[@dmcguckian]

Thanks so much!

[@jocelyncamacho482]

DUDE YOU JUST SAVED MY LIFE. THANK YOU!!!!!!!!!

[@dmcguckian]

You’re welcome

[@rababshahzad]

THANKYOU SOOO MUCH, I COULDNT FIND ANY THING USEFUL BEFORE THIS!

[@dmcguckian]

That’s good to hear. I’m happy the video was helpful!

[@AndyIsHereBoi]

this was the most helpful one that i have ever found

[@dipankarbanerjee1130]

Thanks Prof.🙏 lots of love from India ❤❤

[@coopermau4277]

I wanna be this guys friend. Saving my stats grade on video at a time lets go!

[@DaPower17]

God I wish literally any of my professors could teach this well

[@sajimathew7798]

thank you, sir !! with love from India

[@dmcguckian]

You’re most welcome!!

[@stekichung]

great explanation, thank you so much for making this!

[@dmcguckian]

Thank you! I’m glad it was helpful.

[@sanchitakanta1018]

@@dmcguckian Can you please say when to use the formula z= X-mu/sigma And when to use z= X-mu/sigma÷sqrt(n)

[@dmcguckian]

@@sanchitakanta1018 if the problem asks for the probability that the mean for a sample is … vs the probability of an individual measurement being somewhere. Go to my website site and watch the STATS1, 6.2 videos: www.statsprofessor.com/chapters.php?id=5#ptop

[@MrKingoverall]

I LOVE YOU MAN !!

[@dmcguckian]

Thanks :)

[@epitomeapex]

Well done Teacher💯

[@dmcguckian]

Thank you 🙏

[@bobmcbobface1602]

So we’re basically integrating a Gaussian?

[@dmcguckian]

Correct, we are always finding an area under the curve when calculating these probabilities. The z table I'm using provides the areas obtained by integrating between 0 and the given z score.

[@shazmeenshakeel751]

thank you so much, this was very helpful

[@zakarias86]

Thank you so much for the enlightening explanation! However, I still can not understand why you directly applied the CLT on the 32 women, while this theorem assumes the calculation of the mean of the sampling distribution. Here is an example: Maybe we need to observe the mean of n=5(or more) of the 32 women(with replacement), and we do it repeatedly 100 or 1000 times. Then we can apply the CLT. Otherwise, the sample age at first marriage of the 32 women might not be normally distributed. Plz, correct me if I am wrong🙏.

[@vainbow4632]

The sample mean, the one calculated value, is already a random variable that has a probability distribution. It's that distribution (or rather the standardized version of it) that the CLT is concerned with. You do not need very many sample means. That is often done just to show the CLT in action, but it's not necessary.

[@aliteralnormalguy9046]

Thank you so much sir! I thought I was a goner then I saw this video. Thanks again!

[@dmcguckian]

You’re welcome

[@ksiddarthadshetty3334]

Great explanation , thanks 👍

[@dmcguckian]

You’re welcome

[@randym1788]

Thanks for the wonderful explanation. It would be very helpful if you could elaborate how did you derive the standard deviation of the normal curve (for the average) to be 4/sqrt(n)

[@vainbow4632]

Calculate Var(sum(Xi)/n) using the properties of variance and assuming independence of the Xi's. The formula rolls out of that calculation.

[@therice8024]

Can you explain why standard deviation divide by square root of n?

[@baguls8084]

amazing video

[@dmcguckian]

Thanks

[@aitothechamp7263]

If we are trying to create our own problem, what situations would this apply to? This is only when we have the population mean and population st dev, correct? Like, what if I have all of the sample data, and want to find the probability that the sample data meets a certain criteria (xbar). So in this case I have the literal sample mean and literal sample st dev, but don't have the population mean or population st dev.

[@dmcguckian]

This is preparing you to test hypotheses and to create confidence intervals when you get into inference. For example, we might want to know the population mean, but we can’t often gather all of the data to calculate a population mean. A sample mean could be helpful in that case. We know the mean of the sample means is the population mean, so any sample mean is part of a distribution with a center equal to the population mean. This means the sample mean is likely near the true mean on the number line. One simple thing to do is to reach out and scoop up the values close to your sample mean. This set of values (an interval on the number line) is likely to contain the true mean, since the sample mean is usually nearby the population mean. How do we know this though? Well the Central Limit Theorem says that if n is large enough, samples of size n will have sample means that fall on a bell shaped curve. We don’t need to know the true mean, but knowing the true standard deviation is really helpful. Unfortunately, we usually can’t know that. However, we can estimate it using the sample data. If we assume s is close to the true sigma, we can find the standard deviation for the sample means. S/root(n).

[@ahaha731]

It is very useful🎉

[@emoney716]

also what if I didn't have a z table? how can I figure out the z score?

[@dmcguckian]

The z score is determined from the formula. You don't need a table for the z score. You need a table to find the corresponding probability. Without a table, you either need to use Calculus to find the integral or you need to use software.

[@audryk.7825]

Where does CLT come into play?

[@dmcguckian]

In the problem, they ask for the probability that the average of a sample is between two values. We need to know the distribution of the mean. That’s where the central limit theorem comes in. You cannot assume the bell curve applies without it. Without that, we could not use our normal probability table.

[@sanchitakanta1018]

The average for a normal distribution lies at the middle with a z value of 0. Isn't it? How can it lie between 26 and 27.

[@dmcguckian]

The average of a standard normal distribution is zero. There are an infinite number of normal distributions. The standard normal curve has mean zero and standard deviation 1, but heights for males in the USA could have a normal distribution with a mean of 69 inches and sd of 2.8 inches. There is no limit to the number of normal curves possible.

[@sanchitakanta1018]

@@dmcguckian yes got it. Thank you so much!

[@sutofana]

THANK YOU❤❤❤❤❤❤

[@garrettrains8997]

How do you know how many rows to go over in the z table once you have the z scores?

[@dmcguckian]

If your looking up z=1.36, you use the left column in the table to find 1.3. Then in that row, you go over until you find the 0.06 position

[@oumaimaelberdai05]

Thank u so much for that 🙏

[@dmcguckian]

You’re welcome

[@pijararity7207]

hello why most z score tables i googled state that 1.41 is 0.9207

[@dmcguckian]

There are many z tables not just one. A z table provides areas between two limits under the standard normal curve. You have to make sure the table your using is for the same region. The one you’re talking about is providing the area from 1.41 to negative infinity. It actually doesn’t matter what table you’re using. The task is the same. You must compare the area provided by the table to the area you are looking for. If they are not the same, you have to figure out how to find what you need. This will involve either subtracting or adding the value from the table to 1 or 1/2 or another value from the table. The table I am using shows the area it provides at the top. All tables have a drawing at the top that shows the area it provides. The table you are finding is very common now because a couple of best selling textbooks have switched to that design. They use two tables actually (a neg and a positive one) to do the work I do with just one table.

[@100nitishyadav8]

@@dmcguckian thank you

[@tanjonzou9840]

ok so i calculated: 26-25 which would be the numerator and (4/sq.rt.32) i got 2 but you got 1.4. i don't understand

[@quakewakey]

(26-25) = 1 is the numerator, and the denominator should be (4 / sqrt32) just like you said. sqrt32 = roughly 5.66, and dividing 4 by 5.66 gives you roughly 0.71. That would leave you with 1 / 0.71 = his 1.41. I'm not sure where exactly you went wrong, but perhaps you simply miss-clicked the numbers on your calculator or otherwise you must've accidentally done some mistake in the order of which you calculated it...

[@luisrodrigueziii7316]

Hello my friend. Why did you divide the given standard deviation by the square root of 32?

[@dmcguckian]

In this problem, we are looking at the probability of x-bar being inside a given range of values for a sample of 32. X-bar, the sample mean for a random sample of 32 measurements, has a standard deviation of sigma divided by the square root of n (in this example, the square root of 32). If the problem asked about the probability of an individual measurement being in the given range, we would use the provided standard deviation. We would not divide by the root of n in that case.

[@luisrodrigueziii7316]

@@dmcguckian that makes perfect sense. Thank you for taking the time to reply sir.

[@tinodaisheganyani3935]

Excellent

[@dmcguckian]

Thank you

[@jamalelmazouni2738]

Thanks It's clear

[@emoney716]

Im a little confused on how you knew you could use the bell curve?

[@dmcguckian]

The CLT states that when the sample size is large enough, the sampling distribution of the sample mean is approximately normal, so the fact that the problem has a sample size larger than 30 allows me to assume the bell curve applies.

[@2025yos]

Very clear

[@sthefanyguzman8209]

thank you so much

[@dmcguckian]

You’re welcome

[@ilovejingle]

That's so clear

[@inthemidwest3514]

how every guy calculates his odds at the club.

[@sruthirajan1454]

Thanks.

[@dmcguckian]

You're welcome

[@BolandianBolandian]

excellent!

[@dmcguckian]

Thank you!

[@Diego_Rangel]

Thank you

[@dmcguckian]

You’re welcome

[@furkankocak3142]

Thanks

[@dmcguckian]

You’re welcome

[@Dawgmawm_to2]

I’d like to just say you have given me hope that I can keep my 4.0. I’m studying for my statistics cumulative final and could not for the life of me remember CLT. I was getting so frustrated trying to work it out on my own and here this video popped up and by the end of the video I can confidently say I remember the application! Thank you so much for making videos like this and explaining so clearly without being condescending 🤍 p.s the way you write your fives had me entranced 😂

[@dmcguckian]

😂 I never realized I wrote my 5s differently than everyone until pretty recently. Thanks for watching!

[@besochiqvanaia9668]

blessed!

[@clairemarie-i3r]

it should be 0.4207-0.4977 isn't it? the answer would be - 0.077

[@vainbow4632]

If you get a negative probability as an answer, some alarm bells should go off.

[@shanicenbarnes4339]

Your z scores are incorrect

[@dmcguckian]

I assure you they are correct. This problem is one of my university students’ homework examples. Literally thousands of students have solved this problem over the years, but I’m happy to help you understand if you have a question.

[@Occult_with_Sidhhy]

It's coming 77% not 7.7

[@dmcguckian]

0.0770 is 7.7%. Also, the whole upper half of the curve can have only 50% of the area, so the answer can't be more than 50% (since we want only a part of the upper half of the curve).

[@temwanitamil_]

thank you

[@dmcguckian]

You’re welcome

[@Snaggletooth44]

Thank you