Standardizing Normally Distributed Random Variables
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- čas přidán 21. 08. 2024
- I discuss standardizing normally distributed random variables (turning variables with a normal distribution into something that has a standard normal distribution). I work through an example of a probability calculation, and an example of finding a percentile of the distribution. It is assumed that you can find values from the standard normal distribution, using either a table or a computer.
The mean and variance of adult female heights in the US is estimated from data found in a National Health Statistics Report:
McDowell MA, Fryar CD, Ogden CL, Flegal KM. Anthropometric reference data for children and adults: United States, 2003-2006. National health statistics re- ports; no 10. Hyattsville, MD: National Center for Health Statistics. 2008.
For those using R, here is the R code for the examples used in this video:
American female heights example (approximately normally distributed with a mean of 162.2 and a standard deviation of 6.8).
Finding the probability that a randomly selected female is taller than 170.5 cm.
Easiest way:
1-pnorm(170.5,162.2,6.8)
[1] 0.111121
Standardizing route:
1-pnorm((170.5-162.2)/6.8,0,1)
[1] 0.111121
The default in R's pnorm is the standard normal distribution (mean=0, SD=1), so the mean and SD can be left out when dealing with the standard normal.
1-pnorm((170.5-162.2)/6.8)
[1] 0.111121
Finding the probability that a randomly selected female has a height between 150.5 and 170.5.
Easiest way:
pnorm(170.5,162.2,6.8)-pnorm(150.5,162.2,6.8)
[1] 0.8462162
Standardizing route:
pnorm((170.5-162.2)/6.8)-pnorm((150.5-162.2)/6.8)
[1] 0.8462162
10th percentile of heights of adult American females.
Easiest:
qnorm(.1,162.2,6.8)
[1] 153.4854
Alternatively, via the standard normal distribution:
qnorm(.1)
[1] -1.281552
That's the 10th percentile of the standard normal distribution. Converting to the distribution of heights,
-1.281552*6.8+162.2
[1] 153.4854
Videos like these.. 4 years old and still saving the lives of students around the world. Good shit.
Thanks for the compliment! I tried to make them stand the test of time (e.g. examples that play well through time, no Justin Bieber or Rebecca Black references :).
@@jbstatistics thanks to uuu
And up until 6 yeaaars
Up until 7 years now
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I was stupid forgot the password. I would appreciate any assistance you can offer me!
Just want to thank you for this even after 7 years
In fewer than 3 days of taking notes from your videos, I finally understood what I couldn't figure out in more than 10 days of classes. I got a 96% on my first exam thanks to your videos, and the more I watch the more I understand. Thank you, sir.
For me, it was not a 10 days but a whole semester 🙃
Thank you for providing such helpful videos. Keep it ip @jbstatistics 👍
Hey I just stopped by to tell you how good professor you are!
Thanks for making those videos. Just like other comments told you,you are able to explain things that others try and fail and even create more confusion.
Thanks again! Cheers!
+Marlom Oliveira Thanks Marlom! I'm very glad I can be of help. All the best.
even 10 years later, you're still saving people from failing their exams last-minute :D
I'm glad you find my videos helpful.
The random variable X has a mean of mu (E(X) = mu). (X-mu) is itself a random variable, with a mean of 0. E(X-mu) = E(X) - E(mu) = mu - mu = 0. As a simple example, suppose we have a distribution with a mean of 3. If we create a new distribution by subtracting 3 from every possible value, then the mean of the new distribution would be 0.
that's what i needed...thanks man
@@cogitateandabet Man I didn't get it. Can you please explain
I usually never comment, but god damn after watching 1 minute of your video I actually understood more than 1 hour of trying to understand my textbook.... you are amazing!
you actually saved me. Thank you for explaining this in a digestable manner. I wish I had found you earlier!
video of 7 years old and it is saving my life for the statistics exam. Definitely awesome
Man you are awesome! You just made my life so much easier. I tried so many channels and two different teachers and so far you can explain in 10 minutes what others take hours! Thank you!
+Jorge Hurtado You're very welcome Jorge!
I am having a really hard time understanding my statistics lectures and these videos are amazing. Better than Khan Academy and everything else I have found.
I have an online exam today so it is nice that these videos are very short. I have to come back when I am doing the homework.
of all the videos ive watched today, yours are the best and makes things easier. You're so far the best in explaining these statistics topics😊🤓
Best Statistics videos on CZcams.
Your video rescued us from the hell of our quiz in stats! Thank you so much!!
You are very welcome Amy!
I wish I could have found this channel earlier in the semester. THIS helped so much !!!!!!!!!! Thanks!!!
+Valeria Santoyo I'm glad I could help!
I just watched the fast version in 2X and now watching this makes me feel I am watching it in -2X :D
So glad I found this channel. Thanks so much for the step by step explanation. Your video was very helpful.
you can't imagine how helpful your videos have been ! thank you
You are very welcome!
Thank you so much! I now can pass my class because of you!
When I look at 1.22 in the table, I find 0.8888 and NOT 0.111. What is wrong?
Your table is giving you the area to the left of the value of z. In my example, I needed to find the area to the right. Since the area under the entire curve is 1, the area to the right of 1.22 is 1-0.8888 = 0.1112.
ok, thank you!
you then subtract 1
You make it so simple....thank you so much 🤩👍
You are very welcome!
Yes, videos from 2006 and up are still helping.
One of the best videos out there. Thanks man !
We can find the appropriate areas using software or a standard normal table. I have a playlist "Using a Standard Normal Table" which contains videos for the two main table types. Cheers.
Best stats videos on CZcams
Thanks!
EACH LECTURE IS SHORT AND VERY VERY EXCELLENT -LANGUAGE IS VERY CLEAR - EXCELLENT THANK U AMARJIT ADVOCATE DELHI HIGH COURT INDIA
+Amarjeet Singh You are very welcome!
1. I watched your videos on how to read standard tables but still can not understand how the area got to be 0.111.
I think we should look at the table with positive numbers and that gives an area of 0.8888. The table with negative numbers gives an area of 0.111, as you said in the video, but I do not see why we should look at that table in the first place.
2. In the percentiles section, you have marked 0.1 on the left side of our graph. Left side is for negative values, right? Why should 0.1 not come to the right side?
I would be glad if you could help me.
Great video. Thank you.
we have to subtract the .8888 from 1 in order to have the area greater than z=1.22 and that becomes equal to .111
Your videos are helping me to finally understand econometrics. Thank you so much.
You are very welcome Adam. All the best!
Thank you so much for helping me shine light on the previously dark world that is statistics :D If I would get an A its honestly 99% because of you!!
You are very welcome. I'm glad to be of help!
Thanks for these videos. They're some of the more sensible ones out there ...
You are welcome Brolnox.
Still relevant. Love this!
How do you find the area between -1.72 and 1.22 without using any software computations? or is there a formula for it?:-o
There isn't a formula -- the area is obtained by numerical integration. If you don't have access to software that will find the area (e.g. R, SAS, Excel), then you would need to look up the relevant values in the standard normal table and make the appropriate calculation.
Is this by using the (+ and -) zscore table? And thank you for responding! Your videos helped me a lot!
Yes, you'd use areas from the Z table to find the appropriate area.
Great stuff! I never thought I would ever be interested in probibility and statistics (having had a subpar education in the field), but apparently I was wrong!
That's good to hear! And trust me, there are more interesting things in probability and statistics than standardizing normally distributed random variables!
I'm hooked!
amazing, thank you so much. so easy with this explanation. you deserve an award.
Thanks for the very kind words!
You're a lifesaver. Thanks!
This is a great explaination!
thank you for all of your videos
You are very welcome!
Perfect pace, good stuff!
Thanks!
Very well explained and clear, well done.
Thanks!
You are Incredible.Thanks For the Videos.
Good Video
Going through almost all your videos, haha.
Superb video...!!! Very useful!! Thank you.
You are very welcome!
Can someone explain why the 10th percentile is a value that has an area 10% to the left. Why not the right?
+Jono Hermes in this example it means, that you are taller, than 10% of people.
The equivalent situation on the right side would be the 90th percentile, ie a height that is taller than 90% of the population. 100% is the area under the entire curve.
VERY WELL EXPLAINED, THANK YOU
Really amazing video.Thanks a lot!
Thanks! nicely explained and really helpful .
Great explanation
What i was looking for. Thanks
You are very welcome!
Hello JB statistics. I really like your videos they are so simple to follow and understand with appropriate examples. Please, could you do a video for a transformation of random variables and also Convergence of Random variables.. i tried to find something to learn on this but i couldn't. Or else if its already available please give the name or link so that i can check it out. Stay blessed, your videos are life saver trust me.
+Alan Lwanga Hi Alan. Thanks for the compliment, and I'm glad to be of help. I doubt that I'll have time in the near future to make videos on those topics. I'm sure there are some good resources out there, but I don't have any suggestions. All the best.
you're the best bro
Thank you so much !
god bless u for making these
waTCHING IT IN 2022
Still good :)
Very useful ❤thanks for this😂 helping us in 2018 too! Phew
Standardizing is something I'm running into a lot lately, and I'm curious about something. If I have a variable x and I want to identify outliers, I could standardize and remove anything above +/- 1.96, or whichever cutoff I choose. I wonder, if x is not normally distributed (e.g. are quite skewed), are z-scores no longer valid? I think that because z-scores are calculated using the mean, but since x is not normal, the mean is no longer a reasonable value for this. Is this correct?
How did you get get .846?
+MadMax and how do u get 0.111 ?
+Killan TRAPSA I have videos outlining how to find areas under the standard normal curve. You can use software or a standard normal table to find that area.
see how it is calculated: See, P(Z1-1.72)=0.9573, what you actually want is the Z1 and Z2. Because the left side of Z2 = 1-Z2 = 0.0427, and we would know that the "AND" part will be Z1 - (1-Z2) = 0.8888-0.0427=0.8461
+Killan TRAPSA1, because in the previous example, it is to ask P(Z>1.22), while, actually from the Normal Distribution value table, you can get 1.22 as 0.8888, which means value smaller than 1.22, as you already knew that the total shadowing is 1, then the P(Z>1.22)= 1-0.8888=0.1112, hope this helps
you subtract 0.047 from 0.8888 and if u r asking how i got this numbers its the area of -1.72 and 1.22
Thank you for this!
As I opened my Chinese Version book of probability in Gausion's distribution chapter.
I thought...w*f
but after your explaination...
Thx
Great videos, thank you for taking the time upload these lessons!
How did you find the Z value for the 10th percentile on this distribution curve?
I am familiar finding the value of the 10th percentile (using norminv fx in excel) but not sure how you derived Z score of 10th percentile
I look forward to your response,
+promar In the video I needed to find the 10th percentile of the standard normal distribution. This can be found using software, or a standard normal table if software is not available. I used the R command qnorm(.1,0,1) to find it. I know little of Excel, but the NORMINV command would perform the same function. NORMINV(0.1,0,1) would result in the correct value of -1.281552.
I still don't get the point of standardizing it? Why not just find the probability without stadardizing
There are a number of reasons. It is often helpful to have a baseline distribution from which to work. In later videos, I discuss statistical inference and Z tests. It is helpful to standardize in these spots, so that various different scenarios end up having test statistics with similar properties. We can then discuss "large" and "small" z values, with these remaining consistent in the various situations. Also, when coding software to actually find the probabilities, this can be done just for the standard normal distribution then generalized to the rest with just a line or two.
Man you are amazing. you should teach in my University !!!
Thanks! You never know what the future might bring :)
Thanks for useful video and if possible please make video for Random Process. Thanks again.
simply superb
Thanks!
why is 0.111??????
Your videos have been fucking helpful for my short summer class. THANK YOU!
You are very welcome!
Thank you
Sorry, why do you say that by subtracting the mean from X you necessarily get a new mean of 0? I lost you in that reasoning. Thanks! (Very helpful videos)
Good presentation
Thanks!
why is youtube better than my university which I pay a lot of money?
Could you explain why in statistical inference one should never standardize a variable according to a sample distribution of the same data and why that would be circular reasoning?
In regard to the probability of a randomly selected adult American female taller than 170.5cm, the answer I received is 0.1112
The answer you provided was .111
Did you just shave off the 2? or did I receive the incorrect answer?
I rounded to 3 decimal places. To 6 decimal places, the z value is 1.220588. The area to the right of this is 0.111121. The area to the right of 1.22 is 0.1112324. A table will give the area to the right of 1.22 as 0.1112.
I notice many instructors seem to round to the nearest hundredth. Is this something that is mandatory?
Most standard normal tables list z values to 2 decimal places, so if you're using a table to find areas, you typically end up rounding the z value to 2 d.p. There's nothing mandatory about rounding to 2 decimal places -- if you're using software, then don't round until you get the answer and then round to a reasonable number of decimal places.
Simply Great
Thanks!
Thanks, this really helps!
You are very welcome.
Could you or someone explain me with more details how to find the probability when there are two values like (a
thanks this helps me a lot.
hi, you are great at explaining sir. I have question . I look at the value 1.22 on the table it have the probability 0.8888 and -1.22 had probability of 0.1112 . if the standard deviation is 1.22 how come the prob is 0.1111 ? thanks
+Seerat Raseen Yes apparently the negative value has been mistaken for the positive one! had the same doubt.
+Bhagirath Tallapragada The question is asking for the probability that a randomly selected female is *taller* than 170.5 cm. So we need the area to the *right* of 170.5 under the appropriate normal curve. This is equal to the area to the *right* of the calculated z-score (1.22) under the standard normal curve. When you run off to the z table and look up 1.22, the table gives the area to the *left*, which is not what we need. The standard normal distribution is symmetric about 0, so the area to the *right* of 1.22 is equal to the area to the *left* of -1.22. This is why when you look up -1.22 in the table, you find the final answer. You can also find the final answer by subtracting the area to the left of 1.22 from 1.
Why convert to a standard normal if you can already find the probability from the normal distribution, using integration?
The formula to b integrated is complex so we try to standardize and use pre calculated values. If u hev the energy of integrating it's no big deal u get the same thing
i cant thank you enough.
i have a better understanding of the z value now
thank you so much
can you please show all the steps because its very frustrating seeing answers and not knowing where they come from
I'll assume you're referring to finding the area under the curve. Those areas can be found using software or a standard normal table. There are many different software packages that find those areas (e.g. R, SAS, SPSS, STATA, Excel), and so you may very well also find it frustrating if, for example, I show you how to find it in R but your class uses Excel. If you're relying on standard normal tables, I have videos on how to use those tables.
As I say in the description, "It is assumed that you can find values from the standard normal distribution, using either a table or a computer."
Thank you.
hi can u please explain the difference between central and non central chi square distribution..I want to know the mean and variance for each of them..
Hi, just wanted to confirm, the std. deviation if 6.8 is a value given in the question on 10th percentile calculation. Or is it something that is calculated from somewhere?
+Shirin Tejani That was a given -- it represents the true standard deviation of heights of adult American women. (In reality, the true standard deviation is unknown. The value given here is based on sample data, but is likely very close to the true value. In this question we are pretending that the true standard deviation is a known quantity.)
+jbstatistics Thanks! (PS - the videos are really great!)
+Shirin Tejani Thanks! I'm glad you find them helpful!
the last question really confused me i do not understand where we got the z= -1,28 from and dat was the reason i watched this video can someone please explain @jbstatistics
-1.28 is the 10th percentile of the standard normal distribution, and it can be found using software or a standard normal table. If you don't know how to use the standard normal table, then it would be helpful to watch one of my videos on using the standard normal table (e.g. czcams.com/video/-UljIcq_rfc/video.html).
jbstatistics thank you so much Sir, I have a statistic exam tomorrow and your quick response has helped me a lot thank you thank you and I do get it now ☺
what did you do to find 0.846 ???
ye im also wondering
Think about it, the whole area to the left of 1,22 is 76%
You need to subtract the area to the left of -1,72 (which is 4,2%) to get the total area of 72,15%
He is wrong
Hello Thank you for the videos
Can anyone please tell How we can calculate this expression " P(-1.72
Got it ,,we have some table with values!
@@huntersikari care to explain how to get area for -1.72 ?
@@pikapikacheww_ If using R you could write: pnorm(1.22) - pnorm(-1.72) to get the %. -1.72 and 1.22 are Z scores (x-mean/sd) for each side. I am terrible at explaining and 2 years late but I hope this helps someone.
how could we know that 0. 1 lies on left but not on the right part please explain i couldn't understand
I still did not get how you ended up with the 0.11 probability. can you please explain how can I calculate the probability of 0.11
Can I ask what if the question is
"Ten texpayers are selected at random. What is the probability that 3 of the taxpayers will get their refunds at least 16 weeks?" but before this question it alreasy state that the average is 12 and stdev 3..the variable is the amount of time..
Can I know how to calculate it.. Thank you
So what if you want to find the percent of women with the height between 150.5 and 170.5? Great video by the way :-)
We'd just express the probability as a percentage: 84.6%. (The probability that a randomly selected adult American woman has a height between those 2 values is the same as the proportion of adult American women with a height between those 2 values.) Cheers.
jbstatistics Thank you, but what calculations did you do to get that answer?
Valery Foote
We'd look up the appropriate area under the standard normal curve, using software or a standard normal table.
THANKK YOUU ♥️
i have a problem can you help me about this exersise
N(32 , 2 )
P(27
solve copying procedure as he did
for your first case, μ==E=32, σ=Var=2, then your Z1=(27-32)/2=-2.5, your Z2=(29-32)/2=-1.5. So problem now is P(-2.5
wait, why arent you dividing by the square root of 6.8 if 6.8 is the second moment? shouldn't you square root the second moment to get the standard deviation?
6.8 isn't the second moment. 6.8 is given as the standard deviation (sigma). sigma^2 = E[(X - mu)^2] = 6.8^2.
@@jbstatistics oh okay, understood. Thanks a bunch. The lesson is great.
I dont get where did you get 0.846 in q2, I know its from the table but where.. and how..
How would you calculate the probability of the exercise with the adult American female and her height if the equation said P(150.5 < X -10 < 170.5)?
Isolate X by adding 10 everywhere, then apply the same techniques from this video.
thank you !!
You are very welcome!
I think your z score is incorrect at 7:45 it should be .9984? Where are you getting a zscore of 1.02 to get .846?
The z-scores and area in that example are correct. A z-score of 1.02 does not appear in any way. The area between -1.72 and 1.22 under the standard normal curve is 0.846, and this can be found using software or a table. I have videos that illustrate how to use the table, if you haven't got that down yet. Cheers.
I see what was incorrect. I subtracted the initial 1.22-(-1.72) and then used that zscore at 2.94. But that is wrong thanks!
By the way youre videos clarify things very nicely.
Cheers m8
ur simply awsome thnx
+Fikri Saoudi You are very welcome!