# S3 sample nonsenseWatch

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Thread starter 3 years ago
#1
You can't combine the mean and the variance of a sample right?
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3 years ago
#2
(Original post by AlmostNotable)
You can't combine the mean and the variance of a sample right?
Nope?
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Thread starter 3 years ago
#3
(Original post by Zacken)
Nope?
Nope. or Nope?
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3 years ago
#4
(Original post by AlmostNotable)
Nope. or Nope?
As in, if you have a population and you take samples 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.
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Thread starter 3 years ago
#5
(Original post by Zacken)
As in, if you have a population and you take samples 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))
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3 years ago
#6
(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 so 0
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Thread starter 3 years ago
#7
Not normal
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3 years ago
#8
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Thread starter 3 years ago
#9

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?
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3 years ago
#10
(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 don't necessarily have to have a distribution.
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Thread starter 3 years ago
#11
(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 don't necessarily have to have a 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? 0 reply 3 years ago #12 (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.
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Thread starter 3 years ago
#13
(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?
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3 years ago
#14
(Original post by AlmostNotable)
How did you get 10..
Answered your previous question, should now be $7 in accordance to your modified question. 0 reply Thread starter 3 years ago #15 (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.
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3 years ago
#16
(Original post by AlmostNotable)
I modified my second question to your answer of my modified question. 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).
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Thread starter 3 years ago
#17
(Original post by Zacken) 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
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3 years ago
#18
(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.
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Thread starter 3 years ago
#19
(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.
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Thread starter 3 years ago
#20
(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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