Question

The Resource Conservation and Recovery Act mandates the tracking and disposal of hazardous waste produced at U.S. facilities. Professional Geographer (Feb. 2000) reported the hazardous waste generation and disposal characteristics of 209 facilities. Only eight of these facilities treated hazardous waste on-site.

Required:
a. In a random sample of 10 of the 209 facilities, what is the expected number in the sample that treats hazardous waste on-site? Interpret this result.
b. Find the probability that 4 of the 10 selected facilities treat hazardous waste on-site.

Answer 1

(a)
`E(X) = 0.383`

The expected number in the sample that treats hazardous waste on-site is 0.383.

(b)
`P(x = 4) = 0.000169`

There is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.

Explanation:

Professional Geographer (Feb. 2000) reported the hazardous waste generation and disposal characteristics of 209 facilities.

N = 209

Only eight of these facilities treated hazardous waste on-site.

r = 8

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

n = 10

The expected number in the sample that treats hazardous waste on-site is given by

`$ E(X) = (n * r)/(N) $`

`$ E(X) = (10 * 8)/(209) $`

`$ E(X) = (80)/(209) $`

`E(X) = 0.383`

Therefore, the expected number in the sample that treats hazardous waste on-site is 0.383.

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

The probability is given by

`$ P(x = 4) = \frac{\binom{r}{x} \binom{N - r}{n - x}}{\binom{N}{n}} $`

For the given case, we have

N = 209

n = 10

r = 8

x = 4

`$ P(x = 4) = \frac{\binom{8}{4} \binom{209 - 8}{10 - 4}}{\binom{209}{10}} $`

`$ P(x = 4) = \frac{\binom{8}{4} \binom{201}{6}}{\binom{209}{10}} $`

`$ P(x = 4) = (70 * 84944276340)/(35216131179263320)`

`P(x = 4) = 0.000169`

Therefore, there is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.