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.
Best Answer
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.