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S3 sample nonsense

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Reply 20
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Zacken
22
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?
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?
(edited 7 years ago)
Reply 22
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Zacken
22
Original post by AlmostNotable
Yes, average of the entire population is $2.


What I said still applies? If E(X1)=$2E(X_1) = \$2 for one game, then E(X1+⋯+X100)=100×2=$200E(X_1 + \cdots + X_{100}) = 100 \times 2 = \$ 200
Original post by Zacken
What I said still applies? If E(X1)=$2E(X_1) = \$2 for one game, then E(X1+⋯+X100)=100×2=$200E(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.
Reply 24
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Zacken
22
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.
Original post by Zacken


Never mind that was stupid, I got the right answer.
(edited 7 years ago)
Reply 26
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Zacken
22
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.
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)?
Reply 28
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Zacken
22
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?
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.
Reply 30
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Zacken
22
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.
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?
Reply 32
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Zacken
22
Original post by AlmostNotable
When do I use variance/n?


When you're dealing with the distribution of the mean: Xˉ\bar{X}

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