# Bernoulli, Poisson &amp; Normal Probability

1. Sep 12, 2007

### t_n_p

[SOLVED] Bernoulli, Poisson &amp; Normal Probability

The problem statement, all variables and given/known data

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 [Broken]
but I don't know σ.

Last edited by a moderator: May 3, 2017
2. Sep 12, 2007

### EnumaElish

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.

3. Sep 12, 2007

### t_n_p

Hmm, what do you suggest? I realise it's not as simple as the bernoulli and poisson where you can just add up the individual Pr's....

4. Sep 12, 2007

### EnumaElish

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.

5. Sep 12, 2007

### t_n_p

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?

6. Sep 13, 2007

### t_n_p

edit: im 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?

Last edited: Sep 13, 2007
7. Sep 13, 2007

### HallsofIvy

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")

8. Sep 13, 2007

### t_n_p

Can you guide me through it? I'm so damn lost.

9. Sep 13, 2007

### t_n_p

So basically...

http://img520.imageshack.us/img520/1197/untitledzi8gt4.jpg [Broken]

Last edited by a moderator: May 3, 2017
10. Sep 13, 2007

### EnumaElish

Correct; except the left hand side is FX(7.5) - FX(2.5).

11. Sep 13, 2007

### t_n_p

Thanks a bunch!