S3 sample nonsense Watch

Zacken
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#21
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#21
(Original post by AlmostNotable)
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.
I'm confused, you said the average ticket of a game is $2 and you play 100 games?
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AlmostNotable
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#22
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#22
(Original post by Zacken)
I'm confused, you said the average ticket of a game is $2 and you play 100 games?
Yes, average of the entire population is $2. Not the average of your games otherwise how would you run out?
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Zacken
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#23
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#23
(Original post by AlmostNotable)
Yes, average of the entire population is $2.
What I said still applies? If E(X_1) = \$2 for one game, then E(X_1 + \cdots + X_{100}) = 100 \times 2 = \$ 200
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AlmostNotable
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#24
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#24
(Original post by Zacken)
What I said still applies? If E(X_1) = \$2 for one game, then E(X_1 + \cdots + X_{100}) = 100 \times 2 = \$ 200
Yes that's correct, the variance is wrong. Wait I will find a real question and see if you get the right answer using your method.
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Zacken
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#25
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#25
(Original post by AlmostNotable)
Yes that's correct, the variance is wrong. Wait I will find a real question and see if you get the right answer using your method.
Sure.
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AlmostNotable
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#26
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#26
Never mind that was stupid, I got the right answer.
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Zacken
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#27
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#27
(Original post by almostnotable)
the average male drinks 2l of water when active outdoors with sd 0.7. You are planning a full trip for 50 men and will bring 110l of water. What is the probability that you will run out?

Correct answer is 2.17%
p(z < 100 - 110 / 7) = p(z < -1.43) = 1 - p(z < 1.43) = 0.0764. I disagree with your 'correct' answer.
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AlmostNotable
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#28
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#28
(Original post by zacken)
p(z < 100 - 110 / 7) = p(z < -1.43) = 1 - p(z < 1.43) = 0.0764. I disagree with your 'correct' answer.
0.7^2=0.49*50=49/2

p(z>(110-100)/(49/2)^(1/2))=p(z>2.02)?
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Zacken
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#29
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#29
(Original post by AlmostNotable)
0.7^2=0.49*50=49/2

p(z>(110-100)/(49/2)^(1/2))=p(z>2.02)?
Why the divide by 2?
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AlmostNotable
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#30
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#30
(Original post by Zacken)
Why the divide by 2?
50*0.49=24.5=49/2=24.5

but I feel like both our answers are wrong now.
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Zacken
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#31
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#31
(Original post by AlmostNotable)
50*0.49=24.5=49/2=24.5

but I feel like both our answers are wrong now.
Ah, yes. I agree with yours. The problem looked identical to the other problem so I multiplied the variance by 100 instead.
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AlmostNotable
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#32
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#32
(Original post by Zacken)
Ah, yes. I agree with yours. The problem looked identical to the other problem so I multiplied the variance by 100 instead.
When do I use variance/n?
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Zacken
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#33
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#33
(Original post by AlmostNotable)
When do I use variance/n?
When you're dealing with the distribution of the mean: \bar{X}
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