Variance and standard deviation of a discrete random variable | AP Statistics | Khan Academy
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- Опубліковано 13 лип 2017
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Finding the variance and standard deviation of a discrete random variable.
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what formula was used for the variance?
Khan academy never fails
Thank you!
how to find E|X-1| in above problem?
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no divide by n?
why is it - not plus in v(x) , please someone answer me???
Whata are important terms?
Is it possible for the variance to be equal to 0?
Yes.
Do we not minus the end result of Ex^2P(x) by the mean squared?
I ended up here wondering the same thing, but a different formula was used in this video. Both give the same results though. Ex^2P(x)-sigma^2 and (x-sigma)^2*P(x) give the same results.
No not when you take into account how far away you are from the mean in the initial calculation. You only need to subtract the mean squares when you do (2.1)^2 • 0.1.
Man, you are God personified
I am seriously sitting here doing my finals and watching this
mean 0.1 + 0.15 + 0,4 + 0.25 + 0.1 = 1/ 5 = 0.20
this way you are calculating the mean of the probability alone which does not represent the mean we are calculating.
what you just did makes no sense
POEZJA!!!
My Stat tutor can't even put right topic in the right week. So he keeps editing and it drives me crazy. Fml
hey it should be 0.14 not 0.4, thus making the mean 1.58
probabilities must sum to 1, so the table in the video is correct.
@@justinnewman4474 true; was just gonna say this
sounds like andrew tate
you replace .4 to .8 in the mean calculation: all answers are off
we need to multiply the p(x) with x and add.
multiply .4 with 2 you get .8