Question

According to a​ study, nearly 80​% of workers in law libraries are satisfied with their job. Assume the true proportion of law librarians are satisfied with their job is 0.8. In a random sample of 20 law​ librarians, what is the probability that at most 4 are unsatisfied with their​ job?

Answer 1

Answer:

0.618

Step-by-step explanation:

This is a binomial distribution. We are to determine the probability of people unsatisfied. If we denote the probability of being satisfied with $q$, then $q=0.8$. Let $p$ be the probability of being unsatisfied. Then $p+q=1$

$p=1-q =1-0.8=0.2$

Since the random sample is 20 and $p$ is small enough, we can use the Poisson probability function

$P(X=x) = \dfrac{e^{-\lambda}\lambda^x}{x!}$

where $\lambda$ is the mean and is given by $\lambda=np=20\times0.2=4$.

Denoting the probability of at most 4 unsatisfied with $P(x\le4)$,

$P(x\le4)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)$

$P(x=0)= \dfrac{e^{-4}4^0}{0!} = e^{-4}$

$P(x=1)= \dfrac{e^{-4}4^1}{1!} = 4e^{-4}$

$P(x=2)= \dfrac{e^{-4}4^2}{2!} = 8e^{-4}$

$P(x=3)= \dfrac{e^{-4}4^3}{3!} = e^{-4}\dfrac{32}{3}$

$P(x=4)= \dfrac{e^{-4}4^4}{4!} = e^{-4}\dfrac{32}{3}$

$P(x\le4)=e^{-4}(1+4+8+\dfrac{32}{3}+\dfrac{32}{3})=0.018\times34.33 = 0.618$