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September 27th, 2013, 02:59 PM  #1 
Newbie Joined: May 2013 Posts: 6 Thanks: 0  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 = 1p = 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 = 1P(X<=600) = 1 phi(2827), and I'm stuck here. I'll be happy to see your ideas. Thank you 
September 28th, 2013, 01:16 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,762 Thanks: 697  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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