Respuesta :
Answer:
(a) [tex]E(X) = 0.383[/tex]
The expected number in the sample that treats hazardous waste on-site is 0.383.
(b) [tex]P(x = 4) = 0.000169[/tex]
There is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.
Step-by-step 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
[tex]$ E(X) = \frac{n \times r}{N} $[/tex]
[tex]$ E(X) = \frac{10 \times 8}{209} $[/tex]
[tex]$ E(X) = \frac{80}{209} $[/tex]
[tex]E(X) = 0.383[/tex]
Therefore, the expected number in the sample that treats hazardous waste on-site is 0.383.
b. Find the probability that 4 of the 10 selected facilities treat hazardous waste on-site
The probability is given by
[tex]$ P(x = 4) = \frac{\binom{r}{x} \binom{N - r}{n - x}}{\binom{N}{n}} $[/tex]
For the given case, we have
N = 209
n = 10
r = 8
x = 4
[tex]$ P(x = 4) = \frac{\binom{8}{4} \binom{209 - 8}{10 - 4}}{\binom{209}{10}} $[/tex]
[tex]$ P(x = 4) = \frac{\binom{8}{4} \binom{201}{6}}{\binom{209}{10}} $[/tex]
[tex]$ P(x = 4) = \frac{70 \times 84944276340}{35216131179263320}[/tex]
[tex]P(x = 4) = 0.000169[/tex]
Therefore, there is a 0.000169 probability that 4 of the 10 selected facilities treat hazardous waste on-site.
