Again, I keep pluging numbers into the formulas I can not comeup with the correc

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Again, I keep pluging numbers into the formulas I can not comeup with the correct solution. This is a four partquestion. Thank you for your help!!!! Q: A study showed that 75% of airline flights in and outof US airports were on-time arrivals, and 19% were on latedepartures. 300 flights are to be randomly identified fromall flights and their logs examined. 1- What is the mean (M) and standard deviation (SD) offlights in & out of US... that were on timearrivals?    - M= 225 & SD=7.5     - M= 200 & SD=6.5     - M= 225 & SD=6.5     - M= 200 & SD=7.5 2- What is the probability that more than 80% of the samplewill be on time arrivals. Ans. to choose from: 0.0912, 0.0291, 0.0219 or 0.0192 3- What is the mean (M) and standard deviation (SD)of flights in & out of US... that were late departures?     - M= 57 & SD=7.555     - M= 75 & SD=7.555     - M= 57 & SD=6.79     - M= 75 & SD=6.795 4- What is the probability that less tan 15% of thesamle will have departed late? Ans. to choose from:0.0329, 0.0923, 0.0239 or 0.0375 Again, Thank you for your time. I know it is long, Iwould have broken it up, but I think this is my last question forthe day. Thanks Again, I keep pluging numbers into the formulas I can not comeup with the correct solution. This is a four partquestion. Thank you for your help!!!! Q: A study showed that 75% of airline flights in and outof US airports were on-time arrivals, and 19% were on latedepartures. 300 flights are to be randomly identified fromall flights and their logs examined. 1- What is the mean (M) and standard deviation (SD) offlights in & out of US... that were on timearrivals?    - M= 225 & SD=7.5     - M= 200 & SD=6.5     - M= 225 & SD=6.5     - M= 200 & SD=7.5 2- What is the probability that more than 80% of the samplewill be on time arrivals. Ans. to choose from: 0.0912, 0.0291, 0.0219 or 0.0192 3- What is the mean (M) and standard deviation (SD)of flights in & out of US... that were late departures?     - M= 57 & SD=7.555     - M= 75 & SD=7.555     - M= 57 & SD=6.79     - M= 75 & SD=6.795 4- What is the probability that less tan 15% of thesamle will have departed late? Ans. to choose from:0.0329, 0.0923, 0.0239 or 0.0375 Again, Thank you for your time. I know it is long, Iwould have broken it up, but I think this is my last question forthe day. Thanks Again, Thank you for your time. I know it is long, Iwould have broken it up, but I think this is my last question forthe day. Thanks

Explanation / Answer

1) Mean = n*p = .75 * 300 = 225 Stdev = sqrt(n*p*(1-p)) = sqrt(300 * .75 * .25) = 7.5 2) p = .75 p_hat = .8 SE = sqrt(p * (1-p) / n) = sqrt(.75 * .25 / 300) = .025 z(.8) = (.8 - .75) / .025 = 2 Area to the left is .9772. Therefore, area to the right is 1- .9772 = .0228 3) mean = n * p = 300 * .19 = 57 sd = sqrt(n * p * (1-p)) = sqrt(300 * .19 * .81) = 6.79 4) p = .19 p_hat = .15 SE = sqrt(p * (1-p) / n) = sqrt(.19 * .81 / 300) = .0226 z(.15) = (.15 - .19) / .0226 = -1.77 Area to the left is therefore .0384. Note: On 2 and 4, use the one that is the closest to the answers Igave. It looks like your professor may have rounded beforecomputing the final answers along the way.

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