Question

Use the information in the b) Fore Information in the table aboven swer the follow a) For each of the two random var For each of the two random yariables state the s c) Construct a cross tabulation table (columns) d) What is the probability that a rand the following questions. v e Variabl a ndom Variables state whether it is a que 12 e two random yariables zi e between Sector rows) and compot her it is a qualitative or sale of measureme (11 rge company? company operate in the fishing [1] Company is a medium size and it [2] company is either a small company [2] is either small com randomly selectedcom (2) company is a small and large h e fishing at is the probability that a randomly selected company y that a randomly selected company is a large c Sector? What is the probability that a randomly selected compa operates in the agriculture sector? 8 What is the probability that a randomly selected company or a mining sector company or both? h) What is the probability that a randomly sele or a large company or both What is the probability that a randomly selected company company? 1 What is the probability that a randomly sele company? k) What is the probability that a randomly selected companys company? 1) What is the probability that a randomly selected company company if it is known that the con is a medium sized company m) If it is known that a company operate in the mining that it is a medium size company? 131 n) Is company size statistically independent of sector in which the company Motivate. O) If we define event company small and event company is large, a A and B mutually exclusive? Motivate. [2] randomly selected company is not a fishing secto 12] mpany is not a medium size [2] d company is a mining sector [3] mpany operate in the mining sector what is the probabm 3] the company operate? company is large; are the events (3) QUESTION 2 [18] insurance company found that 25% of all insurance malicies are terminated be maturity date. Assume that 10 polices are rundomly selected from the company database. Out of the 10 randomly selected policies; mly selected from the company's policy [1] a) What is the expected number of policies to be terminated before maturing b) What is the standard deviation for the number of policies terminate maturity? c) What is the probability that no policy will be terminated before maturity! [1] 22 d) What is the probability that all policies will be terminated before maturing? e) What is the probability that at least two policies will be terminated? 13] 1) What is the probability that more than 5 but less than eight policies will be terminated? [3] g) What is the probability that at most eight policies will be terminated? [3] Page 14 of 16

Answer 1

Binomial Distribution

question 2 : Let x be the random variable denoting the number of policies terminated before maturity. = x ~ Binomial (10, 0.25) a) We know that for binomial distribution, con ta E(x) = np Here, na 10, p=0.25. : E(X) -- 10 x 0:25 - 2.5 Expected number of policies to be terminated before maturing is 2.5. We know that Var (x) = np (1.6) Here, Var(x) = 10x 0.25% 0.75 = 1.875

Now, standard deviation = (1.875)/2 1 - 1.3693. And c) Required probability: P(x=0) Now, P(x+x) - (m) px (4-)*-* : P(X=0) - (10)(0.25)º(0.75). = 0.0563 Ang d) Required probability: P (x= 10). Now, P(X10) - ( 16 ) (0-25)590.45) =0.0000 Ang e) Required proobability: P (*2) Now, P(X72) - 1-{P(x=0)+P(x-1)} P(x34) = (40) (0.25) 4 (0.75)9 20:1877 .. P(X72) - 1-{0:0863 +0.1877} = 0.736 Ang f) Requined proobability: P (5<x<8) Now, P(55X48) - P(X-6) + P(X27) P(X+6) = ( 10 ) (0.25)6(0.75) = 0.0162

P(x-7) = ()(0.25)| (0.75)3 20.0031 P (@<x<8) - 0.0162 +0.0031 = 0.0193 Ang g) Required proobability: P(x+8) Now, P (*28) - 1-{P(X29) + P(x10)} P(x29) 2 ( 19 ) (0-25390.75) = 0.0000 : P(x38) - 1-{0'0000 + 0.0000} 21 Ang