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The probability function of the binomial distribution is P(X = r) = (nCr)*p^r*(1-p)^(n-r) where nCr =n!/[r!(n-r)!]
Let n -> infinity while np = L, a constant, so that p = L/n
then
P(X = r) = lim as n -> infinity of n*(n-1)*...*(n-k+1)/r! * (L/n)^r * (1 - L/n)^(n-r)
= lim as n -> infinity of {n^r - O[(n)^(k-1)]}/r! * (L^r/n^r) * (1 - L/n)^(n-r)
= lim as n -> infinity of 1/r! * (L^r) * (1 - L/n)^(n-r) (cancelling out n^r and removing O(n)^(r-1) as being insignificantly smaller than the denominator, n^r)
= lim as n -> infinity of (L^r) / r! * (1 - L/n)^(n-r)
Now lim n -> infinity of (1 - L/n)^n = e^(-L)
and lim n -> infinity of (1 - L/n)^r = lim (1 - 0)^r = 1
lim as n -> infinity of (1 - L/n)^(n-r) = e^(-L)
So P(X = r) = L^r * e^(-L)/r! which is the probability function of the Poisson distribution with parameter L.
ref veeru
The mean and variance are equal in the Poisson distribution. The mean and std deviation would be equal only for the case of mean = 1. See related link.
Depends on the kind of binomials. Case 1: If both binomials have different terms, then use the distribution property. Each term of one binomial will multiply both terms of the other binomial. After distribution, combine like terms, and it's done. Case 2: If both binomials have exactly the same terms, then work as in the 1st case, or use the formula for suaring a binomial, (a ± b)2 = a2 ± 2ab + b2. Case 3: If both binomials have terms that only differ in sign, then work as in the 1st case, or use the formula for the sum and the difference of the two terms, (a - b)(a + b) = a2 - b2.
Depends on the kind of binomials. Case 1: If both binomials have different terms, then use the distribution property. Each term of one binomial will multiply both terms of the other binomial. After distribution, combine like terms, and it's done. Case 2: If both binomials have exactly the same terms, then work as in the 1st case, or use the formula for suaring a binomial, (a ± b)2 = a2 ± 2ab + b2. Case 3: If both binomials have terms that only differ in sign, then work as in the 1st case, or use the formula for the sum and the difference of the two terms, (a - b)(a + b) = a2 - b2.
In the ancient Greek system, the letter Lambda had a value of 30. Nowadays, the lower case lambda stands for the wave length. In statistics, it is the parameter of the Poisson distribution and represents the reciprocal of the frequency.
ref veeru
Normal Distribution is a key to Statistics. It is a limiting case of Binomial and Poisson distribution also. Central limit theorem converts random variable to normal random variable. Also central limit theorem tells us whether data items from a sample space lies in an interval at 1%, 5%, 10% siginificane level.
The mean and variance are equal in the Poisson distribution. The mean and std deviation would be equal only for the case of mean = 1. See related link.
This depends on what information you have. If you know the success probability and the total number of observations, you can use the given formulas. Most of the time, this is the case. If you have data or experience which allow you to estimate the parameters, it may sometimes happen that you work like this. This mostly happens when n is very large and p very small which results in an approximation with the Poisson distribution.
Depends on the kind of binomials. Case 1: If both binomials have different terms, then use the distribution property. Each term of one binomial will multiply both terms of the other binomial. After distribution, combine like terms, and it's done. Case 2: If both binomials have exactly the same terms, then work as in the 1st case, or use the formula for suaring a binomial, (a ± b)2 = a2 ± 2ab + b2. Case 3: If both binomials have terms that only differ in sign, then work as in the 1st case, or use the formula for the sum and the difference of the two terms, (a - b)(a + b) = a2 - b2.
Depends on the kind of binomials. Case 1: If both binomials have different terms, then use the distribution property. Each term of one binomial will multiply both terms of the other binomial. After distribution, combine like terms, and it's done. Case 2: If both binomials have exactly the same terms, then work as in the 1st case, or use the formula for suaring a binomial, (a ± b)2 = a2 ± 2ab + b2. Case 3: If both binomials have terms that only differ in sign, then work as in the 1st case, or use the formula for the sum and the difference of the two terms, (a - b)(a + b) = a2 - b2.
In the ancient Greek system, the letter Lambda had a value of 30. Nowadays, the lower case lambda stands for the wave length. In statistics, it is the parameter of the Poisson distribution and represents the reciprocal of the frequency.
The hyper-geometric distribution is a discrete probability distribution which is similar (in some respects) to the binomial distribution. Suppose you have a population of N which contains R successes. The Binomial describes the probability of r successes in n draws out on N with replacement.However, in many situations the draw is not replaced. In this case you get the hyper-geometric distribution.The function is given by:Prob(r successes in n draws out of N) = RCr/[N-RCn-r * NCn]With the binomial distribution the probability of success remains constant (=R/N) throughout. With the hypergeometric, the numerator for success reduces by one after each successful outcome whereas the denominator reduces by one whatever the outcome.
The hyper-geometric distribution is a discrete probability distribution which is similar (in some respects) to the binomial distribution. Suppose you have a population of N which contains R successes. The Binomial describes the probability of r successes in n draws out on N with replacement.However, in many situations the draw is not replaced. In this case you get the hyper-geometric distribution.The function is given by:Prob(r successes in n draws out of N) = RCr/[N-RCn-r * NCn]With the binomial distribution the probability of success remains constant (=R/N) throughout. With the hypergeometric, the numerator for success reduces by one after each successful outcome whereas the denominator reduces by one whatever the outcome.
Yes. You are measuring the number of 'successes', x, (in this case the number of heads) out of a number of 'trials', n, (in this case coin tosses) that has an assumed probability, p, (in this case 50% expressed as 0.5) of happening. This phenomenon follows a binomial distribution. Apply the binomial distribution to evaluate whether the the probability of x success from n trials with probability p of occurring is within a pre-determined 'acceptable' limit. Let's say you observe 54 heads in 100 tosses and you wonder if the coin really is fair. From the binomial distribution, the probability of getting *exactly* 54 heads from 100 tosses (assuming that the coin *is* fair & should have 0.5 chance of landing on either side) is 0.0580 or 5.8%. Note that this is not the same probability as 54 heads *in a row*. Most statisticians would agree that 5.8% is too large and conclude that the coin is fair.
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The yield of the reaction depends in this case only on the concentration of the limiting reactant.