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September 27th, 2013, 02:59 PM   #1
Joined: May 2013

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Binomial distribution

So, this is what I have to calculate. If average 30% of 2100 visitors are buying ice cream and if an ice cream costs $100 calculate probability to earn at least $60 000 (more or equal to 60 000).

My assumption: we have binomial distribution with parameters B(2100, 0.3), n=2100, p = 0.3 and q = 1-p = 0.7.
If I put those values in formula for described distribution its too much complicated to calculate it. I was thinking about approximation with normal distribution and this is what I got so far:
n*p = 630
sqrt(n*p*q) = 21
P = 1-P(X<=600) = 1- phi(2827), and I'm stuck here.
I'll be happy to see your ideas.

Thank you
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September 28th, 2013, 01:16 PM   #2
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Re: Binomial distribution

600 is (10/7)x(standard deviation) < mean. Look up table of standard normal distribution to get the probability for < -10/7.
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