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S3 sample nonsense

You can't combine the mean and the variance of a sample right?

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Reply 1
Original post by AlmostNotable
You can't combine the mean and the variance of a sample right?


Nope?
Original post by Zacken
Nope?


Nope. or Nope?
Reply 3
Original post by AlmostNotable
Nope. or Nope?


As in, if you have a population XDX \sim D and you take samples X1,X2,,XnDX_1, X_2, \cdots, X_n \sim D then the variance of the samples are fundamentally distinct from the mean of the samples. I'm unsure as to what you mean by 'combine'? When calculating the mean of the samples, you don't concern yourself with the variance and likewise the other way around.
Original post by Zacken
As in, if you have a population XDX \sim D and you take samples X1,X2,,XnDX_1, X_2, \cdots, X_n \sim D then the variance of the samples are fundamentally distinct from the mean of the samples. I'm unsure as to what you mean by 'combine'? When calculating the mean of the samples, you don't concern yourself with the variance and likewise the other way around.


No wait.

X~D(n,m)
How would I do this P(X1+X2+..+X50>t)=?

50X~(50n,m)
or
P(Xn>(t/50))
Reply 5
Original post by AlmostNotable
No wait.

X~D(n,m)
How would I do this P(X1+X2+..+X50>t)=?

50X~(50n,m)
or
P(Xn>(t/50))


I'm going to assume you mean normal here.

You would do Y=X1++X50N(50n,50m)Y = X_1 + \cdots + X_{50} \sim N(50n, 50m) so P(Y>t)=P(Z>t50n50m)P(Y > t) = \mathbb{P}\left(Z > \frac{t - 50n}{\sqrt{50m}}\right)
Original post by Zacken
I'm going to assume you mean normal here.

You would do Y=X1++X50N(50n,50m)Y = X_1 + \cdots + X_{50} \sim N(50n, 50m) so P(Y>t)=P(Z>t50n50m)P(Y > t) = \mathbb{P}\left(Z > \frac{t - 50n}{\sqrt{50m}}\right)


Not normal
Reply 7
Original post by AlmostNotable
Not normal


E(Y)=50nE(Y) = 50n and Var(Y)=50m\text{Var}(Y) = 50m then. But you don't know the distribution of YY.
(edited 7 years ago)
Original post by Zacken
YD(50n,50m)Y \sim D(50n, 50m) then. Then use whatever the CDF/PDF of DD is to find the probability.



If the population had a mean of m and a variance of n. A sample of 50 is taken, how do I find the P((all 50)>t)?

Population is not normal right?
Reply 9
Original post by AlmostNotable
If the population had a mean of m and a variance of n. A sample of 50 is taken, how do I find the P(sample>t)?

Population is not normal right?


I edited my answer, sorry. You don't. You can't talk about it in stuff like that generally. Sums of two random variables with distributions of DD don't necessarily have to have a DD distribution.
Original post by Zacken
I edited my answer, sorry. You don't. You can't talk about it in stuff like that generally. Sums of two random variables with distributions of DD don't necessarily have to have a DD distribution.


I am probably asking wrong. I'll make a question.

An average ticket for a game of poker is $2 with a standard deviation of $0.7. You want to play 100 games and have $220 to spend on tickets (can't spend winnings). What's the probability that you will run out?

How would I do this?
(edited 7 years ago)
Reply 11
Original post by AlmostNotable
I am probably asking wrong. I'll make a question.

A average ticket for a game of poker is $2 with a standard deviation of $1. You are going to play 100 games and have $220 to spend on tickets (can't spend winnings). What's the probability that you will run out?

How would I do this?


The expected value will be $200 and the standard deviation will be $10. Then, I think you'll need to invoke the CLT and assume it's approximately normally distributed to complete the question.
Original post by Zacken
The expected value will be $200 and the standard deviation will be $10. Then, I think you'll need to invoke the CLT and assume it's approximately normally distributed to complete the question.


So will the new thing be T~N(200,100)?

How can you just combine the random variables if you don't know how they are distributed?
(edited 7 years ago)
Reply 13
Original post by AlmostNotable
How did you get 10..


Answered your previous question, should now be $7 in accordance to your modified question.
Original post by Zacken
Answered your previous question, should now be $7 in accordance to your modified question.


I modified my second question to your answer of my modified question.
Reply 15
Original post by AlmostNotable
I modified my second question to your answer of my modified question.


E(X1+X2++Xn)=E(X1)+E(X2)++E(Xn)E(X_1 + X_2 + \cdots + X_n) = E(X_1) + E(X_2) + \cdots + E(X_n) holds for any random variable, doesn't depend on its distribution at all. Proof involves some wacky marginal stuff. Same for the variance (although variance depends on independence).
Original post by Zacken
E(X1+X2++Xn)=E(X1)+E(X2)++E(Xn)E(X_1 + X_2 + \cdots + X_n) = E(X_1) + E(X_2) + \cdots + E(X_n) holds for any random variable, doesn't depend on its distribution at all. Proof involves some wacky marginal stuff. Same for the variance (although variance depends on independence).


Okay now I am more confused, how did you get 7?
0.7*100=70/games = 0.7
Reply 17
Original post by AlmostNotable
Okay now I am more confused, how did you get 7?
0.7*100=70/games = 0.7


You said the standard deviation was 0.7, so the variance is 0.49. Adding up the variances gives 0.49 * 100 = 49 is the variance. So standard deviation is sqrt(49) = 7.
Original post by Zacken
You said the standard deviation was 0.7, so the variance is 0.49. Adding up the variances gives 0.49 * 100 = 49 is the variance. So standard deviation is sqrt(49) = 7.


Oh right that's how its done if we knew the mean/variance of our games but we have sampling here.
(edited 7 years ago)
Original post by Zacken
You said the standard deviation was 0.7, so the variance is 0.49. Adding up the variances gives 0.49 * 100 = 49 is the variance. So standard deviation is sqrt(49) = 7.


The question was

An average ticket for a game of poker is $2 with a standard deviation of $0.7. You want to play 100 games and have $220 to spend on tickets (can't spend winnings). What's the probability that you will run out?

The average of all games is $2, not the average of your games.

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