Answer to Question #139664 in Statistics and Probability for siyam

Question

Twenty Companies were asked whether they provide retirement benefits to their employees.
Fourteen of the companies said they do provide retirement benefits to their employees and six
said they do not. Five companies are randomly selected from these 20. Find the probabilities
that:

i. Exactly two of them provide retirement benefits to their employees.
ii. None of them provide retirement benefits to their employees.
iii. At most one of them provides retirement.

b. The number of people arriving for treatment at an emergency room can be modeled by a
Poisson process with a rate parameter of five per hour
i. What is the probability that exactly four arrivals occur during a particular hour?
ii. What is the probability that at least four people arrive during a particular hour?
iii. What is the probability that exactly four arrivals occur during 45 minutes period?

Answer 1

Population proportion $p=\frac{14}{20}=0.7$

Sample size $n=5$

$$P(X=r) = \ _nC_r P^r(1-P)^{n-r}$$

i) p(X=2)

$$\ _5C_2 \ * 0.7^2(1-0.7)^{5-2}=0.1323$$

the probability is 0.1323

ii) p(x=0)

$$\ _5C_0 *0.7^0(1-0.7)^{5}= 0.00243$$

the probability is 0.00243

iii) $p \le 1$

$$= \ _5C_0 *0.7^0(1-0.7)^{5-0} + \ _5C_1 *0.7^1(1-0.7)^{5-1} =0.03078$$

the probability is 0.03078

PART b)

$$f(x) = \frac{e^{- \mu } \mu^x}{x!}= \frac{e^{-5} 5^x}{x!}$$

i) p(X=4)

$$\frac{e^{-5} 5^4}{4!} = 0.17547$$

the probability is 0.17547

ii) the probability that x is greater than or equal to 4

$$\int _4 ^\infin \frac{e^{-5} 5^x}{x!}dx=0.73498$$

iii) in 45 minute period,

$$Rate, \mu = \frac{45}{60}*5=3.75$$

$$\frac{e^{-3.75} *3.75^4}{4!}=0.19378$$

The probability is 0.19378