Question

Prey Question 14 A study by Bechtel et al., 2009, described in the Archives of Environmental & Occupational Health considered polycyclic aromatic hydrocarbons and immune system function in beef cattle. Some cattle were near major oil- and gas-producing areas of western Canada. The mean monthly exposure to PM1.0 (particulate matter that is c 1um in diameter) was approximately 7.1 ug/m with standard deviation 1.5. Assume that the monthly exposure i normaly dstributed a) What is the probability of a monthly exposure greater than 8ugm? Round your answer to three decimal places (eg.98.765) b) What is the probability of a monthly exposure between 4 and 7ug/m? Round your answer to three decimal places (eg. 98.765) (c) What is the monthly exposure level that is Round your answer to three decimal places (eg 98.765). with probability 0.05? (d) What value of mean monthly exposure is needed so that the probability of a monthly exposure more than 8ug/m is001 Round your answer to two decimal places (e.g. 98.76 ug/m FUJITSU Question 15 Requests for service in a queuing model follow a Poisson distribution with a mean of 8 per unit time a) What is the probability that the time until the first request is less than 4 time units? Round your answer to three decimal places (eg. 98.765). P(Xi) b) What is the probability that the time between the second and third requests is greater than 7.5 time units? Round your answer to three decimal places (e.g. 98.765). P(X2) the mean rate of requests such that the probability is 0.9 that there are no requests in 0.5 time units? Round your answer to three decimal places (e g. 98.765) P(XI)- eTextbook and Media

Answer 1

Question 14:

1.5 P(Z 0.6) 0.274 since area under standard normal table 1.5 1.5 P(-2.06667k) = 0.05 we have to find the value of k' 1-P(X P(X k) = 0.05 k) = 1-0.05-0.95 1.5 From Area under standard noramal table P(Z 1.645 0.95 From(1) and (2) (2) 1.6459.568 1.5 d) P(X > 8) = 0.01 we have to find the value of k' 1- P(X S8) 0.01 P(X 8) = 1-0.01 = 0.99 H | = 0.99 (1) 1.5 From Area under standard noramal table P(Z 2.326) = 0.99 (2) From(1) and (2) _= 2.326 =451