# Fitting Poisson Distribution to data and calculating probabilities

Michael, the produce manager at the local Albertson's supermarket store, would like you to estimate the probability that a customer will purchase 5 or more bunches of bananas. He has collected the following data:

$$\begin{array}{c|c} \text{N=Number of bunches of bananas} & 0 & 1 & 2 & 3 & 4 & \text{5 or more}\\ \hline \text{Frequency} & 49 & 35 & 12 & 3 & 1 & 0\\ \\ \\ \end{array}$$

(a) Calculate the mean and the variance for the aforementioned data.

(b) Which distribution (Poisson, binomial, or negative binomial) would you use to model the data? Which parameter(s) would you choose?

(c) If 1000 customers visit the store, estimate the probability that someone will purchase 5 or more bunches of bananas.

To start, I first calculate the mean and variance:

(a) $E[N] = \frac{1}{100} (1 * 35 + \cdots+4*1) = 0.72$

$Var[N]=E[N^2]-E[N] = 1.26-0.72^2=0.7416$

(b) Since $E[N] \approx Var[N]$, this can be modeled as a Poisson distribution because in this distrubtion $E[N]=Var[N]=\lambda$, and here we can assign $\lambda=0.72$.

$\therefore \Pr(N=k)= \cfrac{\lambda^k}{k!} e^{-\lambda} \text{ for } k \ge 0$

(c) This part I'm a little confused. The probability of purchasing 5 or more bananas can be written as

$\Pr[N \ge 5] = \sum\limits_{k=5}^{\infty} \cfrac{\lambda^k}{k!} e^{-\lambda} = 1-\Pr[N<5] = 1-\sum\limits_{k=0}^{4} \cfrac{\lambda^k}{k!} e^{-\lambda} =1 - 0.999110 =0.000889$

So is it correct to say if there is 1 person, the probability that he will choose 5 or more bunches of bananas is 0.00089?? Hence, if we have 1000 people the probability they will choose 5 or more is 0.00089^1000. But this is wrong. Can someone please explain? How do I apply this probability to 1000 people? This is where I am stuck. Thanks in advance.

Oh by the way, does anyone know how to fix my tex code for the horizontal table? I totally winged it and don't know the proper way. Thanks again.

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The probability that a randomly chosen person will buy $5$ or more bunches is say $p$, where $p$ has been estimated to be $0.000889$ (I have not checked the calculation that led to this number: the basic method was right).
Thus the probability she will not buy $5$ or more is $1-p$. The probability that $1000$ people in a row will not buy $5$ or more is $(1-p)^{1000}$. So the probability at least one person will buy $5$ or more is $1-(1-p)^{1000}$.
Remark: The Poisson model is a model. On the assumption that we do indeed have independent Poissons with parameter $0.72$, the calculation is fully correct. However, it is possible that the store is in a village, so independence cannot be assumed.
Thanks @AndréNicolas, can you please explain what $p^{1000}$ means? If $p$ is the probability that a person buys 5 or more bunches of bananas, then I thought $p^{1000}$ means the probability that 1000 people buy 5 or more bunches of bananas. Still a little confused can you please clarify? What is the difference between $p^{1000}$ and $1-(1-p)^{1000}$? – user1527227 Jul 31 '13 at 0:33
Are you asking about what is the difference between $p^{1000}$ and $1-(1-p)^{1000}$? The first is incredibly tiny. The second is about $0.59$. The proability that a person buys $5$ or more is $p$. The probability she doesn't is $1-p$, about $0.99911$. The probability that $k$ people in a row are non-buyers is $(0.99911)^k$. For $k=1000$ the calculator gives about $0.41$. So the probability at least one of the $1000$ is a buyer is about $1-0.41$. – André Nicolas Jul 31 '13 at 0:40
$p^{1000}$ is the probability that all 1000 people buy 5 or more bananas, i.e. P(person 1 buys 5 or more AND person 2 buys 5 or more AND ...). The question asks for the probability that at least one person buys 5 or more, i.e. P(person 1 buys 5 or more OR person 2 buys 5 or more OR ...). We can say that P(at least one person buys 5 or more) is the complement of P(no-one buys 5 or more), i.e. P(at least one) + P(no-one) = 1. – ConMan Jul 31 '13 at 0:59
So P(no-one buys 5 or more) = P(person 1 doesn't buy 5 or more AND person 2 doesn't buy 5 or more AND ...) = P(person 1 doesn't) * P(person 2 doesn't) * ..., and we can say that the probability that a particular person doesn't buy 5 or more bananas is given by $1-p$, i.e. it's the complement of the probability that they do. So P(no-one buys 5 or more bananas) = $(1-p)*(1-p)*(1-p)*...=(1-p)^{1000}$. – ConMan Jul 31 '13 at 1:02