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    (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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    (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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    (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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    (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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    (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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    Never mind that was stupid, I got the right answer.
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    (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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    (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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    (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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    (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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    (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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    (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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    (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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