8. -5 pointsPODStat3 7.E.055. Thirty percent of all automobiles undergoing an em

Question

8. -5 pointsPODStat3 7.E.055. Thirty percent of all automobiles undergoing an emissions inspection at a certain inspection st the inspection. (Round all answers to three decimal places.) (a) Among 12 randomly selected cars, what is the probability that at most 4 fail the Among 12 randomly selected cars, what is the probability that between 4 and 8 (inclu fail to pass inspection? (c) Among 25 randomly selected cars, what is the mean value of the number that pass inspection, and what is the standard deviation of the number that pass inspection? Mean Standard deviation (d) What is the probability that among 25 randomly selected cars, the number that pass 1 standard deviation of the mean value? ou may need to use the appropriate table in Appendix A to answer this question. Need Help? Read It ointsPODStat3 7.E.077, time that it takes a randomly selected job applicant to perform a certain task has a distribu be approximated by a normal distribution with a mean value of 130 sec and a standard dev ec. The fastest 10% are to be given advanced training. What task times qualify individuals ing? (Round the answer to one decimal place.) seconds or less nay need to use the appropriate table in Appendix A to answer this question. 299 Com

Explanation / Answer

1 a) n = 12, p =0.3

This is binomial probability distribution, P(X = k) = nck * pk * (1-p)(n-k)

P(X <= 4) = 12c0 * 0.30 * 0.712 + 12c1 * 0.31 * 0.711 + 12c2 * 0.32 * 0.710 + 12c3 * 0.33 * 0.79 + 12c4 * 0.34 * 0.78

= 0.7237

b) P(4 <= X <=8) = 12c4 * 0.34 * 0.78 + 12c5 * 0.35 * 0.77 + 12c6 * 0.36 * 0.76 + 12c7 * 0.37 * 0.75 + 12c8 * 0.38 * 0.74 = 0.5058

c) mean = np = 25*0.3 = 7.5

std dev. = sqrt(npq) = sqrt(25*0.3*0.7) = 2.2913

d) question is not completely visible. I presume, Binomial to normal approximation needs to be used

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