Answer to Question #97210 in Statistics and Probability for mani

Question

20. a) The number of accidents occurring in a city in a day is a Poisson variate with mean 0.8. Find the probability that on a randomly selected day (i)There are no accidents (ii)There are accidents
b) The incidence of an occupational disease in an industry is such that the worker have 25% change of suffering from it. What is the probability that out of 5 workers, at the most two contract that disease?

Answer 1

a)

X = the number of accidents occurring in a city in a day.

$$P(X=x)={\lambda^x e^-\lambda \over x!}, x=0,1,2,...$$

(i) Find the probability that on a randomly selected day there are no accidents

$1\ day=>\lambda=0.8$

$$P(X=0)={0.8^0 e^{-0.8} \over 0!}=e^{-0.8}\approx0.449329$$

(ii) Find the probability that on a randomly selected day there are accidents

$$P(X>0)=1-P(X=0)=1-{0.8^0 e^{-0.8} \over 0!}=1-e^{-0.8}\approx$$ $$\approx0.550671$$

b)

X = the number of diseased workers.

Binomial distribution

$$b(x;n, p)=\binom{n}{x}p^x(1-p)^{n-x},x=0,1,2,...,n$$

Given that $n=5,p=0.25.$

Find the probability that out of 5 workers, at the most two contract that disease

$$P(X\leq2)=P(X=0)+P(X=1)+P(X=2)=$$

$$=\binom{5}{0}0.25^0(1-0.25)^{5-0}+\binom{5}{1}0.25^1(1-0.25)^{5-1}+$$

$$+\binom{5}{2}0.25^2(1-0.25)^{5-2}=$$

$$=0.75^5+5(0.25)(0.75)^4+10(0.25)^2(0.75)^3=$$

$$=0.2373046875+0.3955078125+0.263671875=$$

$$=0.896974375$$

The probability that out of 5 workers, at the most two contract that disease is $0.896974375$