Question

A smog alert has been called in a certain area of Los Angeles County in which there are 50 industrial rms. An inspector will visit 10 randomly selected rms to check for violations of regulations. a. If 15 of the rms are actually violating at least one regulation, what is the pmf of the number of rms visited by the inspector that are in violation of at least one regulation? b. If there are 500 rms in the area, of which 150 are in violation, approximate the pmf of part (a) by a simpler pmf.

Answer 1

Answer:

Part (A) The required PMF is: $P(X=x)=\frac{\binom{15}{x}\binom{50-15}{10-x}}{\binom{50}{10}}$

Part (B) $B(x;n,p)=\binom{10}{x}(0.3)^x(1-0.3)^{10-x}$

Step-by-step explanation:

Consider the provided information.

There are 50 industrial rms. An inspector will visit 10 randomly selected rms to check for violations of regulations.

Part (A)

15 of the rms are actually violating at least one regulation.

Let X is the number of firms violate at least one regulation from 10 randomly selected rms to check for violations of regulations out of 50 firms of which 15 5 of the rms are actually violating.

Therefore, $X\sim Hypergeom(n=10, M=15\ and\ N=50)$

We need to determine probability mass function.

$P(X=x)=Hyper(x; n=10, M=15, N=50)\\=\frac{\binom{15}{x}\binom{50-15}{10-x}}{\binom{50}{10}}$

Hence, the required PMF is: $P(X=x)=\frac{\binom{15}{x}\binom{50-15}{10-x}}{\binom{50}{10}}$

Part (B)  If there are 500 rms in the area, of which 150 are in violation, approximate the pmf of part.

Here N=500 so find the probability of p as shown below:

$p=\frac{150}{500} =0.3$ and n=10

$B(x;n,p)=B(x;10,0.3)\\=\binom{10}{x}(0.3)^x(1-0.3)^{10-x}$