91影视

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,$r=8$

The probability of random variable X is given as:

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

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,

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" $巍\left(x\right)=\frac{\left(\begin{array}{l}8\\ x\end{array}\right)\left(\begin{array}{l}209鈭�8\\ 10鈭�x\end{array}\right)}{\left(\begin{array}{l}209\\ 10\end{array}\right)},\text{鈥�}x=0,1,鈰�,8$

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

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

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

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

$\begin{array}{c}巍(X=4)=\frac{\left(\begin{array}{l}10\\ 4\end{array}\right)\left(\begin{array}{l}209鈭�8\\ 10鈭�4\end{array}\right)}{\left(\begin{array}{l}209\\ 10\end{array}\right)}\\ =\frac{\left(\begin{array}{l}10\\ 4\end{array}\right)\left(\begin{array}{l}201\\ 6\end{array}\right)}{\left(\begin{array}{l}209\\ 10\end{array}\right)}\\ =0.000169\end{array}$

That is approximately 0.0002

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