Examples, solutions, videos, activities, and worksheets that are suitable for A Level Maths.
What are the conditions for which a Poisson Distribution can be used as an approximation to the Binomial distribution?
The binomial distribution tends towards the Poisson distribution when n → ∞ , p → 0 and λ = np stays constant.
Poisson approximation to the Binomial Distribution
This is the 6th in a series of
tutorials for the Binomial Distribution.
This tutorial shows you the conditions for which a Poisson Distribution can be used as an approximation to the Binomial distribution by comparing probability graphs of the distributions
Poisson Approximation to the Binomial Distribution [Example]
This is the 6th in a
series of tutorials for the Poisson Distribution.
This tutorial runs through an example comparing the actual value to the approximated value and compare the two methods of working.
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Poisson approximation to Binomial: S2 Edexcel January 2013 Q1
- [a] Write down the conditions under which the Poisson distribution can be used as an approximation to the binomial distribution.
The probability of any one letter being delivered to the wrong house is 0.01
On a randomly selected day Peter delivers 1000 letters.
[b] Using a Poisson approximation, find the probability that Peter delivers at least 4 letters to the wrong house.
Give your answer to 4 decimal places.
The Relationship Between the Binomial and Poisson Distributions
A look at the relationship between the binomial and Poisson distributions
[roughly, that the Poisson distribution approximates the binomial for large n and small p]. This video works through some calculations in an example, showing that the approximate probability from the Poisson can be quite close to the exact probability from the binomial distribution.
[The example used involves albinism. Albinism affects all races, but the rates of albinism vary a little around the world. In Europe and North America, roughly 1 in 20,000 people have some form of albinism].
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The Poisson distribution is actually a limiting case of a Binomial distribution when the number of trials, n, gets very large and p, the probability of success, is small. As a rule of thumb, if $n \ge 100$ and $np \le 10$, the Poisson distribution [taking $\lambda = np$] can provide a very good approximation to the binomial distribution.
This is particularly useful as calculating the combinations inherent in the probability formula associated with the binomial distribution can become difficult when $n$ is large.
To better see the connection between these two distributions, consider the binomial probability of seeing $x$ successes in $n$ trials, with the aforementioned probability of success, $p$, as shown below.
$$P[x]={}_nC_x p^x q^{n-x}$$Let us denote the expected value of the binomial distribution, $np$, by $\lambda$. Note, this means that
$$p=\frac{\lambda}{n}$$and since $q=1-p$,
Now, if we use this to rewrite $P[x]$ in terms of $\lambda$, $n$, and $x$, we obtain
$$P[x] = {}_nC_x \left[ \frac{\lambda}{n} \right]^x \left[ 1-\frac{\lambda}{n} \right]^{n-x}$$Using the standard formula for the combinations of $n$ things taken $x$ at a time and some simple properties of exponents, we can further expand things to
$$P[x] = \frac{n[n-1][n-2] \cdots [n-x+1]}{x!} \cdot \frac{\lambda^x}{n^x} \left[ 1 - \frac{\lambda}{n} \right]^{n-x}$$Notice that there are exactly $x$ factors in the numerator of the first fraction. Let us swap denominators between the first and second fractions, splitting the $n^x$ across all of the factors of the first fraction's numerator.
$$P[x] = \frac{n}{n} \cdot \frac{n-1}{n} \cdots \frac{n-x+1}{n} \cdot \frac{\lambda^x}{x!}\left[ 1 - \frac{\lambda}{n} \right]^{n-x}$$Finally, let us split the last factor into two pieces, noting [for those familiar with Calculus] that one has a limit of $e^{-\lambda}$.
$$P[x] = \frac{n}{n} \cdot \frac{n-1}{n} \cdots \frac{n-x+1}{n} \cdot \frac{\lambda^x}{x!}\left[ 1 - \frac{\lambda}{n} \right]^n \left[ 1 - \frac{\lambda}{n} \right]^{-x}$$It should now be relatively easy to see that if we took the limit as $n$ approaches infinity, keeping $x$ and $\lambda$ fixed, the first $x$ fractions in this expression would tend towards 1, as would the last factor in the expression. The second to last factor, as was mentioned before, tends towards $e^{-\lambda}$, and the remaining factor stays unchanged as it does not depend on $n$. As such, $$\lim_{n \rightarrow \infty} P[x] = \frac{e^{-\lambda} \lambda^x}{x!}$$
Which is what we wished to show.