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 :).
Sorry to be offtopic but does anyone know a trick to get back into an Instagram account? I was stupid forgot the password. I would appreciate any assistance you can offer me!
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.
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!
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!
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!
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.
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.
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.
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 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.
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
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.
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?
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?
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!
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.
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.
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
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, 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
+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.
sir plz help me to solve this... ** Given a normal distribution with a mean of 60 and sd of 14, n=100 what percentage of cases will lie between the score of50 and 65??
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.
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.
ah! i got it. +537yaya, +Jordan, +Alix, 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
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.
How do you find the -1.28 using software though? (ie. using R). I understand how to get the value from the standard normal table but I don't know the formula in R. Do you know the formula?
The R command qnorm(.1) returns -1.281552 (the 0.1th quantile of the standard normal distribution). qnorm(.1,0,1) would also work (returning the 0.1th quantile of a normal distribution with mean = 0 and SD = 1, which is of course the SND).
+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
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.)
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."
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. ruclips.net/video/-UljIcq_rfc/видео.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 ☺
@@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.
Hi can you help he solve this question. ..if a mean is 280 and sd is 12..in a normal distributed random variable. .what's the probability between 275 than 290
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
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.
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?
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
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
Sorry to be offtopic but does anyone know a trick to get back into an Instagram account?
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 👍
even 10 years later, you're still saving people from failing their exams last-minute :D
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.
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!
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!
video of 7 years old and it is saving my life for the statistics exam. Definitely awesome
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.
you actually saved me. Thank you for explaining this in a digestable manner. I wish I had found you earlier!
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😊🤓
Your video rescued us from the hell of our quiz in stats! Thank you so much!!
You are very welcome Amy!
My UND statistics class links directly to your videos as supplemental! This is about as good as it gets lol
Your UND prof is perceptive and wise :)
So glad I found this channel. Thanks so much for the step by step explanation. Your video was very helpful.
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
Best Statistics videos on RUclips.
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!
Thank you so much! I now can pass my class because of you!
One of the best videos out there. Thanks man !
Your videos are helping me to finally understand econometrics. Thank you so much.
You are very welcome Adam. All the best!
you can't imagine how helpful your videos have been ! thank you
You are very welcome!
Thanks for these videos. They're some of the more sensible ones out there ...
You are welcome Brolnox.
You make it so simple....thank you so much 🤩👍
You are very welcome!
Yes, videos from 2006 and up are still helping.
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!
I just watched the fast version in 2X and now watching this makes me feel I am watching it in -2X :D
Best stats videos on RUclips
Thanks!
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
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
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!
amazing, thank you so much. so easy with this explanation. you deserve an award.
Thanks for the very kind words!
Still relevant. Love this!
You are Incredible.Thanks For the Videos.
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.
You're a lifesaver. Thanks!
This is a great explaination!
Superb video...!!! Very useful!! Thank you.
You are very welcome!
thank you for all of your videos
You are very welcome!
Very well explained and clear, well done.
Thanks!
Perfect pace, good stuff!
Thanks!
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
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.
Could you or someone explain me with more details how to find the probability when there are two values like (a
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?
Really amazing video.Thanks a lot!
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?
Good Video
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!
Great explanation
VERY WELL EXPLAINED, THANK YOU
Thanks! nicely explained and really helpful .
What i was looking for. Thanks
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.
Thank you for this!
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.
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.
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
Hey!!! Can u suggest some websites to practice questions based on these concepts? Help appreciated
Thank you so much !
I dont get where did you get 0.846 in q2, I know its from the table but where.. and how..
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..
how could we know that 0. 1 lies on left but not on the right part please explain i couldn't understand
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.
Man you are amazing. you should teach in my University !!!
Thanks! You never know what the future might bring :)
Going through almost all your videos, haha.
god bless u for making these
Very useful ❤thanks for this😂 helping us in 2018 too! Phew
Thanks for useful video and if possible please make video for Random Process. Thanks again.
sir plz help me to solve this...
** Given a normal distribution with a mean of 60 and sd of 14, n=100
what percentage of cases will lie between the score of50 and 65??
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.
Good presentation
Thanks!
Hi can you elaborate more on how you found the area for the second example (around 7:42) with the P(-1.72
P(-1.72
jbstatistics Don't quite understand how you came up with 0.846 out of -1.72 and 1.22 at the 8:00 mark. Could you elaborate?
+jbstatistics please would you be able to explain this
+1, i don't understand either, please help to answer, thanks
ah! i got it. +537yaya, +Jordan, +Alix, 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
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.
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
where on earth did you get the -1.28? I cannot find any table that says this. Please, anyone?
It's the 10th percentile of the standard normal distribution, found with software or a standard normal table.
How do you find the -1.28 using software though? (ie. using R). I understand how to get the value from the standard normal table but I don't know the formula in R. Do you know the formula?
The R command qnorm(.1) returns -1.281552 (the 0.1th quantile of the standard normal distribution). qnorm(.1,0,1) would also work (returning the 0.1th quantile of a normal distribution with mean = 0 and SD = 1, which is of course the SND).
@@jbstatistics Thank you so much! That helps a lot!
why is youtube better than my university which I pay a lot of money?
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
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!
simply superb
Thanks!
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."
Thanks, this really helps!
You are very welcome.
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. ruclips.net/video/-UljIcq_rfc/видео.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 ☺
waTCHING IT IN 2022
Still good :)
in what video did you compute for the area?
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.
thanks this helps me a lot.
you're the best bro
Hi can you help he solve this question. ..if a mean is 280 and sd is 12..in a normal distributed random variable. .what's the probability between 275 than 290
I suggest you watch the example in this video that starts at 5:50. Cheers.
Your videos have been fucking helpful for my short summer class. THANK YOU!
You are very welcome!
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 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.
As I opened my Chinese Version book of probability in Gausion's distribution chapter.
I thought...w*f
but after your explaination...
Thx
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 do not know how did you get the 0.111
What software can I use
Simply Great
Thanks!
Thank you
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