Probability Exponential Distribution Problems
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- Опубліковано 22 гру 2024
- This statistics video tutorial explains how to solve continuous probability exponential distribution problems. It explains how to do so by calculating the rate parameter from the mean. It also tells you how to graph the probability density function.
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"Thanks again for watching" I mean why??
Thank you, Professor!!!!!!!
We all truly appreciate you and your time!
Thank you so much man !! You have no idea how much you've helped me in learning statistics. The fact that you are providing all of this to us for free is just incomprehensible. Cheers man
I think you intended to say ‘incredible’ and not ‘incomprehensible’.
@@ktiwari31 no they meant what they meant as in its so shocking its hard to understand how he does it
NO THEY MEANT IT AS ITS SO HARD TO UNDERSTAND WHY THEY WOULD GIVE THIS INFO FOR FREE@@ktiwari31
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True.
This video on exponential distributions has caused my knowledge on exponential distributions grow exponentially. Thank you 🙏🙏🙏🧬🍻
I am so glad I came across this channel. Math generally isn't my best suit, but I am so happy youre helping me get there!
pretty sure this is statistics man
U r the best UA-cam channel for learning. Ican find any videos I search for in your channel.u r Really great
The last question where 4 < x < 7. I guess there is a small mistake in the equation. It should be e^(-0.8) instead of 1 - e^(-0.8). Since x is supposed to be greater than 4 and less than 7.
Correct me if I am wrong please.
Thank you!
i think you are right
Since the question asked BETWEEN 4 and 7 years. We can only use the differences between the area less than 7 and the area less than 4 to calculate the answer. (7:00) I believe e^-0.8 is the amount greater than 4. It makes no sense of differences between (the area less that 7) and (the area greater than4).
If I have any errors, please correct me. Thanks
Was thinking the exact same thing...
I also realised that mistake
I get what you're saying but in his case what he did is correct.
Clear and concise, despite of tiny mistake at the beginning
I believe you made a mistake towards the beginning when calculating the value for lambda. You wrote "lambda = 1/lambda" instead of "lambda = 1/mu"
Of course, the teacher has made a mistake. But it is up to the student to correct the mistake himself.
@@chinchang5117 wrong
@@chinchang5117 Ironically, you are the person who is wrong here. What you said applies to entire subjects, as in, the student needs to discover the things that the teacher could not. It does not mean that during teaching, the teacher can stack up minor mistakes without care.. Know when to apply your knowledge.
@@chinchang5117 u got finessed
@@chinchang5117 I am surprised that no one said anything about his username? so Disrespectful
Thanks!
Simple and straight forward,u have made my life so simple in med school 🙏thank you
You have been so helpful every time I feel a struggle. But today I am here just to hear your voice.
get a life
I recently discovered your channel and I must confess that it has been super resourceful. I have had trouble understanding some theories and concepts in statistics. Mind you, I have been done with Uni for years. However, because I want to purse a career in Data science which requires having fundamental knowledge of some area of statistics makes it an evitable challenge that I must overcome in order to excel in this desired field. Great Job!
God bless you, Thank God for leading me to your channel. Thank you for the videos
Amazing lesson! thank you! I found a difficulty in the material but you made it really easy!
In first minute lambda = 1/mu you have made a typo there.
was gonna say, thanks!
This is correct!! lambda = 1/mu
What would I do without this man🥺🥺🥺
Very useful! Thank you very much! You saved me hours of trial and error trying to figure out what this kind of problems were about!
Another Great Video, you explain things so well that's easy to understand!
Another helpful video, very excited for my Statistics test.
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You're really good at explaining things easily. Keep up the good word.
Great video, helped a lot, greetings from Sweden!
You have been a life saver dude
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As always, you're the best. TY.
Hi, brother.
I think you meant:
Now, the rate parameter is going to be 1 over miu, not one over lambda
1:00 its 1/MYU RIGHT
Hi for part (e) shouldn't x be Greater than 4 and less than 7, was that a mistake?
No, because when you solve for the area of the graph for x (less than) 7, you are finding the area of the graph from x=0 to x=7, meaning that the area between 4 and 7 is already accounted for, but the area between 0 and 4 is also counted in that. So you have to subtract the area from 0 to 4 to get only what is between 4 and 7. hopefully that made sense.
I just realized that this comment was from a year ago and you probably either no longer care about the solution or you figured it out haha. anyway, in case you were still wondering... there it is🤣
You never disappoint. Thank you
Thank you so much Sir, you have really helped me in studies.
Thank you for sharing this video. I was watching my class lecture and just did not get it. I also didn't realize I needed to use the "power of e" on the calculator to get the number.
5:07 shouldn't the func be 0.20e^-(0.20X) missing 0.20 only e^-(0.20X)
true!
No, because at around 5:07 he is calculating the area to the right of 10. This is different from F(x) which gives us only the height above the curve but not the area; two different calculations.
@@frankbarboza I am still confused. The PDF calculated here is: 0.20e^-(0.20X). But to calculate the probabilities, the equation used is: e^-(0.20X). Can someone explain me this?
@@teja1853 The PDF is used to calculate a single point (height) above the curve. When X is 0, the height above the curve is 0.20; when X is 10, the height above the curve is 0.02706 (I just substituted 10 for X in the equation you typed in your reply). The CDF is used to calculate the probability (the area). When calculating the area to the right of a critical point, use e^(-lambda multiplied by X); when calculating the area to the left of a critical point, use 1 minus e^(-lambda multiplied by X).
@@frankbarboza Thanks for explaining! I understand it now
Thank you for your explanation, I understand it better through your example. Previously I read the formulas but did not feel how this distribution get the specific value corresponding to X= time.
Statistics is very difficult and really confusing and too much information are given so please do explain the formula of getting lamda from the mean and also in one of your videos you calculate lamda=1/lamda ?? Is it 2/lamda or 1/mue ? Or mue=lamda time t ???
Hello your channel is amazing and I never go to any other tutors to understand my problem but please do explain when we use each distribution as there are many of questions they don’t tell you the distribution type so you have to guess and stay in doubt and also when you are calculating lamda or mue too many books has introduce it in different ways and it is so confusing so when we have the mean please clearly explain how we can calculate lamda thanks
you save my life.
its has been an amazing lesson thank you
Love you! ❤️❤️
btw you can also calculate the area under the curve with integrals since we know the formula
Why is there 1- in (1-e)
Hello, is it always stated that the problem will be of exponential distribution, and if not could you give an example of a problem? Thanks.
Always coz there are many ways of distribution....
6:21 shouldnt that be P(x>4) ?
i don't get the first question option c, why do we need to subtract with 1, isn't the probability of adding f(0) , f(1), f(2) right would be the answer?
hows this guy carrying my statistics as well
Clearly explained. Well done, and thank you!
please what is the mean for the exponential distribution when theta is given instead of lambda?
Another great video. Thank you.
I was, however, hoping you would include an example question that involves the less than and equals sign or more than and equals sign. I thought that if we wanted to know the probability of a laptop lasting a week or more (which each event being a day), we could simply make x=8 and use the greater than sign but it doesn't seem to be right.
Kinda late but just in case when the less than and equal or more than and equal signs are used in distributions that are continuous(i.e. not discrete) it is essentially irrelevant and it is calculated the same way as if it was just a regular bigger than or smaller than 'x'.
btw its x is greater than 4, not x is less than 4
How can one proof that exponential distribution of independent variables can lead to poisson distribution?
How are these videos so helpful!!!!
Thank you so much! This was really helpful.
I believe you made a mistake towards the end because
P (4
i know this is way too late but you forgot to put the 1- in the second part
so shouldve been 1- e...
@@mouinemhb4960 Late response. You don't need to put 1 in the second part, because, you are calculating (0-4) whereas in the first part you are calculating (7-inf). So, the (1 - ) is just to calcuate the compliance
yep your answer is correct.i also got 0.30407
@@L.O.L.Adventures but i think that the video is correct and this comment section is wrong. Since we need to find X b/w 4and 7 we need to find the left side are of 7 and substract it with the left side area of 4 to get the answer. So, P(X
I have a doubt is that shoukdn't we be doing integration using the given years/time as limits?
Please is theta the same as lambda when given a question?
But doesn't the problem say BETWEEN 4 and 7 years, meaning MORE than 4 and less than 7, not LESS then 4 and less than 7?
That's what I worked out. So i'm hoping he made a small error on this and we are correct
Yeah, what's up with that ?
I thought it that way too, but if i am correct then that would mean the answer is 0,3041.
7:00 he explains it here
Amazing tutor video, only thing I feel like you should be following more "standard practice" that mathematicians use, like when you show us P(x
bro there's a mistake in 1:04....lambda=1/mean....you wrote it as lambda=1/lambda....please see to it
I'm having trouble understanding the probability density function for the exponential distribution. What does it mean to say that the event at time=0 has a positive probabilty? I had thought that no event occurred at that time, so its probability should be zero. Why not?
Time=0 is when we started measuring. In this example I believe it is when all of the laptops are functioning properly, and as time progresses, they start failing with a rate given by the distribution function.
Thanks very much for your reply. I had already looked into this further. My question to you was about the continuous exponential density function, although I had not mentioned it because I usually think in terms of continuous densities. I have discovered that for any continuous density P, P(X) = 0 for all X. I had never thought about this before. It reflects the fact that in a continuous density, X must be defined exactly. If X is, say, weight, and we ask what is the probability that someone's weight is 150 lbs., the continuous density will give that probability for exactly 150 lbs. However, no one weighs exactly 150 lbs. so P(150) = 0. But the probability of someone weighing between 150 lbs and 150.1 lbs can be obtained from the cumulative probability density of P. So, for a continuous exponential distribution, the probability of an event occurring at time=0 is really 0 because, literally, no event can really take place at time=0. But if we really mean the short interval between time=0 and time=30 seconds, then we can get the correct probability from the CDF. I learned a great deal from your video and the extended search it led me to. Thanks again, Larry Schwartz
Did he mean to say that lamba is defined as one over mu? Can somebody clarify? Thank you
yeah thats what he meant
I have a question concerning (E)
You wrote that X is less than 4
I thought X should be greater than 4
Thank you very much !
Is it right to do P(47)?
THANK YOU SO MUCH! litterally about break sth over this stupid math problem
clear and helpful
are the formulas applicable whether the interval is inclusive or not?
Yes, they should be applicable, because in this situation you are calculating continuous values, and one single specific value has a 0 chance of occurring, since it is one out of an infinite number of values.
So "x ≤ a" and "x < a" would actually mean the same thing, since P(x = a) = 0.
solid follower here
wow that easy, thanks dude
life saver
Can u make a video about integral problems in physics because that’s my first test in ap physics and I haven’t taken calc yet. Keep up the great work 👍
He already uploaded.
this could be an obvious answer but why aren't we using integrals in these cases?
Like with the one less than 3, can't we just do integral [3, 0] f(x)?
Thank you for thi one!!!!
Thanks 👍
The income tax of man is exponentially distributed with f(x) = 1/3*e^(-x/3) ;x>0. what is the probability that his income will exceed rs. 17000? assume that the tax is levied at the rate of 15% on the income above rs. 15000.
Why are you saying y/lamda but where did you get what lambda = to?
Lambda = 1/mu (mu=the mean)
i dont understand why it was .20yrs ^-1 in the first problem
Thanks a lot
When I see an Orgo video, I know everythings gonna be alright
i have understood exponentially
THANK YOU
You are great
you are the best I the world j g
Amazing!
Thanks
thanks a lot!
gud explanation
I think there is no correction for the question between 4 and 7 years. Knowing that the area of P(4
This is just something I've noticed from the comment section (I know it's very late to point this out)
The last part is it not supposed to be x>4 instead x being less
What is P(X=x)?
For any continuous random variable, 0.
u cant get exactly one point cuz it use continuouse varibale that same for any continuouse nummer like P{X=3.32542623}=0
againnn u da bestttt
if no one got me, i know the OCT got me
Fantastic
in last answer you have written x
yeah he made a mistake
Part c where x
This is a continuos distribution bro
bravo!
I think P(4
There is a mistake at the end of the Video it should be x>4 and x
There is no mistake
thanks :)