Elementary probability
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Elementary probability
I'm studying for an exam and one of the practice questions is as follows: you're sending out mail and the probability is .4 that someone replies. How many letters do you need to send out so that you're reasonably certain at least 100 people will reply? It's about 292 according to my professors answer key but he doesn't show how to do it
EXPAND: (a+b)^n
(a+b)^n
(a+b)^n
(a+b)^n
GENERATION 191,184,382 : The first time you see this, copy it into your sig and divide the generation number by 2 if it's even, or multiply it by 3 then add 1 if it's odd. Social experiment.
(a+b)^n
(a+b)^n
(a+b)^n
GENERATION 191,184,382 : The first time you see this, copy it into your sig and divide the generation number by 2 if it's even, or multiply it by 3 then add 1 if it's odd. Social experiment.
Re: Elementary probability
That depends what you mean by "reasonably certain".
First, a simpler question: how many letters would you have to send out to expect 100 replies on average?
Second, in that scenario, what's the chance of getting less than 100? And how many responses are you "reasonably certain" you'll get?
Now do that last calculation again, but in the general case (ie, in terms of n).
First, a simpler question: how many letters would you have to send out to expect 100 replies on average?
Second, in that scenario, what's the chance of getting less than 100? And how many responses are you "reasonably certain" you'll get?
Now do that last calculation again, but in the general case (ie, in terms of n).
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Re: Elementary probability
After working backwards, it seems like you are intended to approximate the sum of a lot of i.i.d. binary variables as a normal distribution. The mean of a sum of independent variables is the sum of the means, and same with the variances.
If you have a weighted coin with a .4 probability of landing on "heads", then how many "heads" would you expect after N trials? What is the standard deviation and the variance of the number of "heads"? What are these quantities when N=292?
If you have a weighted coin with a .4 probability of landing on "heads", then how many "heads" would you expect after N trials? What is the standard deviation and the variance of the number of "heads"? What are these quantities when N=292?
 ThirdParty
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Re: Elementary probability
If the answer is intended to be "292", then the question is presumably intended to be "If each letter has an independent 40% probability of receiving a reply, how many letters do you need to send out so that you have at least a 98% chance of getting at least 100 replies?"Meteoric wrote:That depends what you mean by "reasonably certain".
Really? Why shouldn't he just use a binomial distribution?cyanyoshi wrote:After working backwards, it seems like you are intended to approximate the sum of a lot of i.i.d. binary variables as a normal distribution.
Re: Elementary probability
ThirdParty wrote:Really? Why shouldn't he just use a binomial distribution?cyanyoshi wrote:After working backwards, it seems like you are intended to approximate the sum of a lot of i.i.d. binary variables as a normal distribution.
When you have a lot of trials with p close to 0.5, the binomial distribution is very close to the normal distribution with the same mean and variance. A rule of thumb is that you can treat a binomial distribution as normal if both np and n(1p) are greater than 5. In this case, both np and n(1p) are clearly greater than 100, which is even better. You would get the same answer either way (±1), so why work harder than you have to?
Edit to add: Coming from an engineering background, I'd expect the error from approximating a binomial distribution as a normal distribution to be much less than the error that comes from the uncertainty in p. If p=0.404 or 0.396, the final answer should change by more than one. Of course it's not usually a good idea to assume a priori that a distribution is normal, but it may not be the worst thing to assume if you are stuck or could justify it somehow (like with the np > 5 thing).
 gmalivuk
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Re: Elementary probability
Because for this question the binomial distribution would require adding all the terms for 0 to 99 responses (and subtracting from 1), and for N=292 that includes numbers likeThirdParty wrote:Really? Why shouldn't he just use a binomial distribution?cyanyoshi wrote:After working backwards, it seems like you are intended to approximate the sum of a lot of i.i.d. binary variables as a normal distribution.
(0.4^92)*(0.6^200)*292!/(92!*200!).
Unless it's a programming class (and your job is to figure out a way to calculate that exact sum as quickly as possible), binomial distributions with numbers like n=292 or k=100 aren't intended to be calculated exactly.

Edit to add: This blog post explains that the error in the cumulative distribution is less than C(p^2 + q^2) /√(npq) for a constant C that is not known precisely, but is less than 0.4748. Call it 0.5 for simplicity, and with p=0.4, q=0.6, and n=292, the error is always less than 0.031.
Edit2: According to Wolfram Alpha, the exact error in the cumulative distribution from 0 to 99 responses (i.e. that there won't be at least 100 responses) to 292 mailings is 0.00193 and with the continuity correction (taking the CDF to 99.5 for the normal distribution), it's only 0.000715.
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