Bernoulli, Poisson & Normal Probability

AI Thread Summary
The discussion focuses on calculating the probability of a chocolate bar containing between 3 and 7 hazardous squares using Binomial, Poisson, and Normal distributions. The Binomial distribution is approached by summing the probabilities for 3 to 7 squares, while the Poisson distribution uses a mean (μ) of 10. For the Normal distribution, participants discuss the need for the mean and standard deviation, which are derived from the Binomial parameters. The correct method involves applying a continuity correction by integrating from 2.5 to 7.5 to find the probability. Overall, the conversation emphasizes the complexities of transitioning from discrete to continuous distributions and the importance of understanding dependencies in probability calculations.
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[SOLVED] Bernoulli, Poisson & Normal Probability

Homework Statement

Every chocolate bar contains 100 squares, with 10% of the individual squares presenting a health hazard to people consuming them.

(a) Using the Binomial, Poisson and Normal distributions, write down formulas for
the probability that a single chocolate bar has at least 3 but no more than 7 deadly
squares.

The attempt at a solution

For the binomial part, I've just done Pr(x=3) + Pr(x=4) +... Pr(x=7). Just wondering if that's what you would do?

I've done the same for the Poisson, using μ = 10.

I'm stuck on Normal distribution though.
I'm thinking of using
http://img294.imageshack.us/img294/5079/untitledzi8.jpg
but I don't know σ.
 
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You can look at this as a Bernoulli for each individual square with p=0.1 and q=0.9; then variance = pq for each square. That would be the first approximation. Then you need to formulate a way around the problem that squares in a given bar are mutually dependent; e.g. if you know the first 90 are safe then you know that the remaining 10 are deadly. At least that's how I interpret your question.
 
Hmm, what do you suggest? I realize it's not as simple as the bernoulli and poisson where you can just add up the individual Pr's...
 
When I re-read the question I realized that the question does not imply a dependence. If there can be as few as 3 deadly squares, then knowing that 90 are safe does not tell me the remaining 10 are deadly. This makes it much easier. Under binomial, mean = np = 10 and variance = npq = 9, which you can apply to a normal distribution.
 
Yeah so std dev is 3, mean is 10
I just slap it into the above formula? Initially I thought I should integrate from 3 to 7, but it seems as though I should be doing from 2.5 to 7.5? Does this sound correct?
 
edit: I am totally confused for normal dist now. Am I supposed to integrate at all or do I just use the formula I posted near the top? I mean to integrate that massive thing even though I know std dev and mean is out of my scope. Am I even on the right track?
 
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The normal approximation to the binomial distribution for given p (q= 1-p) and n has mean pn and standard deviation \sqrt{np(1-p)}. Use those in your formula for the standard z-score, and find the probability that x is between 2.5 and 7.5. (That's the "integer correction")
 
Can you guide me through it? I'm so damn lost.
 
So basically...

http://img520.imageshack.us/img520/1197/untitledzi8gt4.jpg
 
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  • #10
Correct; except the left hand side is FX(7.5) - FX(2.5).
 
  • #11
Thanks a bunch!
 
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