8. (5 pts) On a particular production line, the likelihood that a light bulb is defective is 5%. Ten light bulbs are randomly selected. What is the probability that none of the light bulbs will be defective? A. 0.0105 B. 0.0746 C. 0.3151 D. 0.5987 9. (5 pts) Alison has all her money invested in two 10.(5 pts) The number of cars sold by a car mutual funds, A and B. She knows that there salesperson during each of the last 25 weeks is is a 40% chance that fund A will rise in price, the following: and a 60% chance that fund B will rise in price Number of cars sold Frequency given that fund A rises in price. There is also a 10 20% chance that fund B will rise in price. What 10 is the probability that at least one of the funds 5 will rise in price? What is the expected number of cars sold by A. 0.24 the salesperson during a week? B. 0.36 A. 15 C. 0.60 B. 1.0 D. 0.76. C. 0.8 D. 1.2 PART B (64%): Problem and open questions. Show all of your work and make sure you circle your answer. You may use Excel or Minitab to find the answers except question 14. If you use Minitab, please attach the associate graphs to the answer. If you use Excel, please copy the formula itself (for example " =BINOM.DIST(2,10,0.2,1)") to the answer. 11. It is known that 1% of the population has a certain disease. If an infected person is tested, then there is a 95% chance that the test is positive. If the person is not infected, then there is a 2% chance that the test gives an erroneous positive result (so called 'false positive"). (a) (5 pts) What is the probability that the test is positive? D: has a certain disease P(D)=0.01 P(Dc) = 1-0.01=0.99 +: The test is positive P(+| D)=0.95 P(+| Dc)=0.02 P(+) = P(+D)+P(+Dc) = P(+|D)*P(D)+P(+|Dc)*P(Dc} =0.95*0.01+0.02*0.99 = 0.0293 (b) (5 pts) Given that a person tests positive, what are the chances that he has the disease? P{D|+) = P(+|D)*P{DY/P{+) = P(+[D}*P{DY/(P(+|D)*P(D)+P(+| Dc)*P(Dc)) = 0.95*0.01/(0.95*0.01+0.02*0.99) = 0.3242=32.42% 12. During the regular business hours of a bank opens to the public, it is observed that, on average, one customer arrives the bank every three minutes. Please answer the following questions. (a) (5 pts) What is the probability that no customer arrives the bank within six minutes during the regular business hours? 1 customer every 3 minutes > 2 customers every 6 minutes P(X=0) =POISSON. DIST(0, 2 , 0)=0.1353 Alternatively: Y~ ~time between events. "exp(3) average = 3 -> lamba =1/3 P(Y>6) = 1-P(Y 1/3*10 customers every 10 minutes P(X=5) =POISSON.DIST(5,10/3,0)=0.1223 (d) (3 pts) On average, how long does it take to have 5 customers arrived the bank since it opens to the public during the regular business hours? 1 customer every 3 minutes - 5 customers every 3*5 = 15 minutes (e) (3 pts) Given one customer just arrived the bank, what is the probability that the next customer will arrive after exactly 2 minutes? P(212) = 1-P(X80) = P{Z>((80-63.25)/13.4322);= 1-4(1.2470)=0.1062 4000*0.1062 = 424.8