Answer to Question #260634 in Statistics and Probability for Okay I'm

Question

Question 4 [16 Marks]

Research has shown that about 2% of all products produced by BVH Company are defective. A quality

inspector has just sampled 300 products from this company. Using the Poisson distribution, what is the

probability that:

(a) exactly 12 of them are defective? [4]

(b) at least 8 of them are defective?

Answer 1

We need to establish the relationship between binomial distribution and Poisson distribution because finding probabilities for number of defectives requires the use of binomial distribution.

Here,

$n=300$

$p=2\%=2/100=0.02$

Relationship between these two distributions is that parameter $\lambda$ for a Poisson distribution is the mean of Binomial distribution. This relationship can be written as,

$\lambda=np$ where $np$ id the mean of a Binomial random varible.

Express by $X$ the number of defectives produced by BVH company then, $X\sim Poisson(\lambda)$ and its probability mass function is given as,

$p(X=x)=e^{(-\lambda)}(\lambda)^x/{x}!$

The value of $\lambda$ is given as,

$\lambda=n*p=0.02*300=6$

a.

Probability of exactly 12 defectives is written as,

$p(X=12)=e^{-6}(6)^{12}/12!$

By running this command in $R$,

$dpois(12,6)= 0.01126448$

Therefore, the probability that there are exactly 12 defectives is 0.01126448

b.

The probability that there are at least 8 defectives is given as,

$p(X\geqslant8)=1-p(X\lt8)=1-\displaystyle\sum^7_{x=0}(e^{-6}6^x)/x!$

Again, we use run the following commands in $R$

$p(X\geqslant8)=1-ppois(7,6)=1-0.7439798= 0.2560202$

Therefore, the probability that there are at least 8 defectives is 0.2560202