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!
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 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!
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 I wonder, if x is not normally distributed (e.g. are quite skewed), are z-scores no longer valid?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?
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?
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.
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.
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..
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.
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??
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!
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.
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.
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.
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.
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).
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 ☺
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.
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.)
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
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
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.
@@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
+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
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.
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.
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)
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 👍
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.
you actually saved me. Thank you for explaining this in a digestable manner. I wish I had found you earlier!
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!
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😊🤓
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.
video of 7 years old and it is saving my life for the statistics exam. Definitely awesome
Thank you so much! I now can pass my class because of you!
So glad I found this channel. Thanks so much for the step by step explanation. Your video was very helpful.
One of the best videos out there. Thanks man !
Your video rescued us from the hell of our quiz in stats! Thank you so much!!
You are very welcome Amy!
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!
I just watched the fast version in 2X and now watching this makes me feel I am watching it in -2X :D
even 10 years later, you're still saving people from failing their exams last-minute :D
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!
Best Statistics videos on RUclips.
Your videos are helping me to finally understand econometrics. Thank you so much.
You are very welcome Adam. All the best!
Thanks! nicely explained and really helpful .
Really amazing video.Thanks a lot!
Thanks for these videos. They're some of the more sensible ones out there ...
You are welcome Brolnox.
This is a great explaination!
amazing, thank you so much. so easy with this explanation. you deserve an award.
Thanks for the very kind words!
Still relevant. Love this!
Yes, videos from 2006 and up are still helping.
VERY WELL EXPLAINED, THANK YOU
Perfect pace, good stuff!
Thanks!
Very useful ❤thanks for this😂 helping us in 2018 too! Phew
thank you for all of your videos
You are very welcome!
You are Incredible.Thanks For the Videos.
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!
Superb video...!!! Very useful!! Thank you.
You are very welcome!
You make it so simple....thank you so much 🤩👍
You are very welcome!
Thank you for this!
You're a lifesaver. Thanks!
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?
you can't imagine how helpful your videos have been ! thank you
You are very welcome!
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?
Very well explained and clear, well done.
Thanks!
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
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.
Could you or someone explain me with more details how to find the probability when there are two values like (a
Great explanation
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..
Hey!!! Can u suggest some websites to practice questions based on these concepts? Help appreciated
Thank you so much !
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
Thanks for useful video and if possible please make video for Random Process. Thanks again.
What i was looking for. Thanks
You are very welcome!
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??
thanks this helps me a lot.
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!
Good Video
Best stats videos on RUclips
Thanks!
I dont get where did you get 0.846 in q2, I know its from the table but where.. and how..
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
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!
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.
Thank you.
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.
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.
god bless u for making these
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.
Thanks, this really helps!
You are very welcome.
Thank you
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
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!
Man you are amazing. you should teach in my University !!!
Thanks! You never know what the future might bring :)
simply superb
Thanks!
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 ☺
how could we know that 0. 1 lies on left but not on the right part please explain i couldn't understand
in what video did you compute for the area?
thank you so much
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.
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.
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!
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
Simply Great
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
Good presentation
Thanks!
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
Going through almost all your videos, haha.
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.
THANKK YOUU ♥️
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.
i cant thank you enough.
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.
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
I do not know how did you get the 0.111
Your videos have been fucking helpful for my short summer class. THANK YOU!
You are very welcome!
As I opened my Chinese Version book of probability in Gausion's distribution chapter.
I thought...w*f
but after your explaination...
Thx
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.
thank you !!
You are very welcome!
What software can I use
thanks
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.
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)
ur simply awsome thnx
+Fikri Saoudi You are very welcome!