A particular telephone number is used to receive both voice calls and fax messag

Question

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 20% of the incoming calls involve fax messages, and consider a sample of 20 incoming calls. (Round your answers to three decimal places.) What is the probability that at most 8 of the calls involve a fax message? What is the probability that exactly 8 of the calls involve a fax message? What is the probability that at least 8 of the calls involve a fax message? What is the probability that more than 8 of the calls involve a fax message? You may need to use the appropriate table in the Appendix of Tables to answer this question. Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of documented disabilities. Consider a random sample of 15 students who have recently taken the test. (Round your probabilities to three decimal places.) What is the probability that exactly 1 received a special accommodation? What is the probability that at least 1 received a special accommodation? What is the probability that at least 2 received a special accommodation? What is the probability that the number among the 15 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated? Suppose that a student who does not receive a special accommodation is allowed 3 hours for the exam, whereas an accommodated student is allowed 4.5 hours. What would you expect the average time allowed the 15 selected students to be? (Round your answer to two decimal places.)

Explanation / Answer

PROBABILITY OF FAX MESSAGE = 0.2

PROBABILITY OF VOICE CALLS = 0.8

N = 20

A)P(AT MOST 8 ARE FAX ) = P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)

P(0) = 20C0(0.2)^0*(0.8)^20 = 0.011

P(1) = 20C1*(0.2)^1*(0.8)^19 = 0.057

P(2) = 20C2*(0.2)^2*(0.8)^18 = 0.136

P(3) =  20C3*(0.2)^3*(0.8)^17 = 0.205

P(4) =  20C4*(0.2)^4*(0.8)^16 = 0.218

P(5) =  20C5*(0.2)^5*(0.8)^15 = 0.174

P(6)=  20C6*(0.2)^6*(0.8)^14 = 0.109

P(7) =  20C7*(0.2)^7*(0.8)^13 = 0.054

P(8) =  20C8*(0.2)^8*(0.8)^12 = 0.022

HENCE THE P(ATMOST 8) = 0.986

B)P(X =8) =

20C8*(0.2)^8*(0.8)^12 = 0.022

C)P( AT LEAST 8) = 1 - P( LESS THAN 8)

= 1 -P(AT MOST 8) +P(8) = 1 -0.986+0.022 = 0.036

D)P( MORE THAN 8) = P(AT LEAST 8) -P(8) = 0.036-0.022 = 0.014

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