Question 80662: I hope someone can help me I am stuck.
Acme Airlines flies airplanes that seat 12 passengers. From experience, they have learned that, on average, 80% of the passengers holding reservations for a particular flight actually show up for the flight. If they book 13 passengers for a flight, what is the probability that 12 or fewer passengers holding reservations will actually show up for the flight?
a. less than 90% b. between 90 and 95%, inclusive c. more than 95% d. cannot be determined from information given
Let be a normally distributed random variable with and . The probability that is between 80 and 120 is (to the nearest whole percent)
a. 8 % b. 82 % c. 91% d. answer not listed
these two qusetion have been killing me all night thanks for your help Mirna
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Acme Airlines flies airplanes that seat 12 passengers. From experience, they have learned that, on average, 80% of the passengers holding reservations for a particular flight actually show up for the flight. If they book 13 passengers for a flight, what is the probability that 12 or fewer passengers holding reservations will actually show up for the flight?
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It's a binomial problem: n=13, p=0.8, 0<=x<=12
If you have a TI calculator use 1-binompdf(13,0.8,13)=0.945
If you do not have a TI use 1-(0.8)^13=0.945
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a. less than 90% b. between 90 and 95%, inclusive c. more than 95% d. cannot be determined from information given
ANSWER: "b" is the answer.
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Let be a normally distributed random variable with and . The probability that is between 80 and 120 is (to the nearest whole percent)
a. 8 % b. 82 % c. 91% d. answer not listed
Comment: A bunch of words are missing from your posting. You will have
to post this problem again.
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cheers,
Stan H.
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