Question

On-site treatment of hazardous waste. The Resource Conservation and Recovery Act mandates the tracking and disposal of hazardous waste produced at U.S. facilities. Professional Geographer (February 2000) reported the hazardous-waste generation and disposal characteristics of 209 facilities. Only 8 of these facilities treated hazardous waste on-site. Use the hypergeometric distribution to answer the following:

a. In a random sample of 10 of the 209 facilities, what is the expected number in the sample that treat hazardous waste on-site? Interpret this result.

b. Find the probability that 4 of the 10 selected facilities treat hazardous waste on-site.

Short answer

a. Hence the expected number in the sample that treats hazardous waste on site is 0.383

b. The probability that 4 of the 10 selected facilities treat hazardous waste on site is 0.0002

Step-by-step solution

Given information 

Professional Geographer (February 2000) reported the hazardous-waste generation and disposal characteristics of 209 facilities. Only 8 of these facilities treated hazardous waste on-site.

That is, the total number of facilities,$N=209$.

Number of facilities which are hazardous waste on site,$r=8$

Computing the distribution of random variable X

a.

The probability of random variable X is given as:

${\rm P}\left(X\right)=\frac{\left(\begin{array}{l}r\\ x\end{array}\right)\left(\begin{array}{l}N-r\\ n-x\end{array}\right)}{\left(\begin{array}{l}N\\ n\end{array}\right)}[x=\max imum[0,n-(N-r),\mathrm{..}\min imum(r,n)]]$

Where

N is total number of elements

r is the number of success in N elements

n is the number of elements drawn

x is the number of successive drawn in the n elements.

Number of facilities which are hazardous waste on site,

Calculating the expected number in the sample that treat hazardous waste on site. 

A random sample of 10 of 209 facilities is taken.

That is $N=209$,$n=10$

Hence the probability distribution is defined as:

localid="1661966396068" ${\rm P}\left(x\right)=\frac{\left(\begin{array}{l}8\\ x\end{array}\right)\left(\begin{array}{l}209-8\\ 10-x\end{array}\right)}{\left(\begin{array}{l}209\\ 10\end{array}\right)},x=0,1,\cdots ,8$

The expected number in the sample that treat hazardous waste on-site is:

$\begin{array}{c}Mean=\frac{nr}{N}\\ =\frac{10\times 8}{209}\\ =0.38278\end{array}$

This indicates that the average number one can expect in the sample that treat hazardous waste on site is 0.38278.

Calculating the probability that 4 of the 10 selected facilities treat hazardous waste on site

b.

The probability that 4 of the 10 selected facilities treat hazardous waste on site is calculated as follows.

$\begin{array}{c}{\rm P}(X=4)=\frac{\left(\begin{array}{l}10\\ 4\end{array}\right)\left(\begin{array}{l}209-8\\ 10-4\end{array}\right)}{\left(\begin{array}{l}209\\ 10\end{array}\right)}\\ =\frac{\left(\begin{array}{l}10\\ 4\end{array}\right)\left(\begin{array}{l}201\\ 6\end{array}\right)}{\left(\begin{array}{l}209\\ 10\end{array}\right)}\\ =0.000169\end{array}$

That is approximately 0.0002

Hence the probability that 4 of the 10 selected facilities treat hazardous waste on site is 0.00002q