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BT Second ExamVersion3

# BT Second ExamVersion3 - Problem 1 2 3 4 5 6 7 8 Earned...

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Problem Value Earned 1 12 2 12 3 12 4 12 5 13 6 13 7 13 8 13 Total 100 0 Problem 1) The following data was collected based on a random sample 1000 tourists visiting a small souvenir shop in the Bahamas. Age 16-25 26-45 >=46 Total Bought Souvenirs 100 200 100 400 Didn't Buy Souvenirs 250 250 100 600 Total 350 450 200 1000 a) What is the probability that a randomly chosen person bought a souvenir? P(bought souvenir) = 0.4 b) What is the probability that a random person older than 45 bought a souvenir? P(bought souvenir | age >= 46) = 0.5 c) Based on your answers to a) and b) can you concluded that buying a souvenir is independent of age? Justify your answer. Buying a souvenir is not independent of age. If it was the case, we would have P(bought d) For each age group, compute the probability of buying a souvenir. P(bought souvenir | age 16-25) = 0.29 P(bought souvenir | age 26-45) = 0.44 P(bought souvenir | age >= 46) = 0.5

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t souvenir | age >= 46) = P(bought souvenir), which is not the case. The left side of the equation i
is .5 while the right side is .4: they are obviously not equal.

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Problem 2) A company receives a large shipment of computer processor chips. The company do quality control check by taking a random sample of 40 chips. If 35 or more of these a defective, the company accepts the shipment. a) Assume that 10% of the chips in the shipment are defective. What is the probability the company will reject the shipment? P(reject) = 1 - P(accept) = 1 - P(5 or less defective) Using a cumulative binomial distribution with p = 0.1 P(reject)= 0.21 b) Assume that 20% of the chips in the shipment are defective. What is the probability the company will reject the shipment? with p = 0.2 P(reject)= 0.84 c) Assume that 30% of the chips in the shipment are defective. What is the probability the company will reject the shipment? with p = 0.3 P(reject)= 0.99
p=0.1 p=0.2 p=0.3 oes a cumul cumul cumul are not 0 0.01 0 0 1 0.08 0 0 2 0.22 0.01 0 3 0.42 0.03 0 4 0.63 0.08 0 that 5 0.79 0.16 0.01 6 0.9 0.29 0.02 7 0.96 0.44 0.06 0.63 8 0.98 0.59 0.11 0.37 9 0.99 0.73 0.2 10 1 0.84 0.31 11 1 0.91 0.44 that 12 1 0.96 0.58 13 1 0.98 0.7 14 1 0.99 0.81 15 1 1 0.88 16 1 1 0.94 that 17 1 1 0.97 18 1 1 0.99 19 1 1 0.99 20 1 1 1 21 1 1 1 22 1 1 1 23 1 1 1 24 1 1 1 25 1 1 1 26 1 1 1 27 1 1 1 28 1 1 1 29 1 1 1 30 1 1 1 31 1 1 1 32 1 1 1 33 1 1 1 34 1 1 1 35 1 1 1 36 1 1 1 37 1 1 1 38 1 1 1 39 1 1 1 40 1 1 1

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Problem 3) A baseball team makes an average of 1.2 errors a game. Assuming the number of errors follows the Poisson distribution; a) What is the probability that the team makes no error in a game? See Poisson non cumulative with mu = 1.2 in column M P(0 error)= 0.3 0.3 b) What is the probability that they make more than five errors in five games? See Poisson cumulative with mu = 1.2 * 5 = 6 in column N P(error >5) = 1 - P(error <=5) = 0.55 0.45 0.55 c) What is the probability that they make less than five errors in 10 games?
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BT Second ExamVersion3 - Problem 1 2 3 4 5 6 7 8 Earned...

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