Chapter 5-3 Find the indicated probability. In a study, 39% of adults questioned

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Chapter 5-3 Find the indicated probability. In a study, 39% of adults questioned reported that theirhealth was excellent. A researcher wishes to study the health ofpeople living close to a nuclear power plant. Among 13 peopleselected in this area, only 3 reported that their health wasexcellent. Find the probability that when 13 adults are selected, 3or fewer are in excellent health. A) .1256   B) .1877 C) .1210 D).0667 E)none I used the binomial formula for 0,1,2, and 3 people and addedthem together to get I got .188019 I also had a problem with : Air America has a policy of booking as many as 15 persons onan airplane that only holds 14. A past study showed that 85% of thebooked passenger actually showed up for the flight, Find theprobability that if they booked 15 passangers that they would nothave enough available seats. Back of book gives an answer of .08374 which means I amsetting up my problem wrong because my answer is way off. I amusing 15 trials 14 successes probabitlity .85 prob of failure .15.What am I doing wrong? Chapter 5-3 Find the indicated probability. In a study, 39% of adults questioned reported that theirhealth was excellent. A researcher wishes to study the health ofpeople living close to a nuclear power plant. Among 13 peopleselected in this area, only 3 reported that their health wasexcellent. Find the probability that when 13 adults are selected, 3or fewer are in excellent health. A) .1256   B) .1877 C) .1210 D).0667 E)none I used the binomial formula for 0,1,2, and 3 people and addedthem together to get I got .188019 I also had a problem with : Air America has a policy of booking as many as 15 persons onan airplane that only holds 14. A past study showed that 85% of thebooked passenger actually showed up for the flight, Find theprobability that if they booked 15 passangers that they would nothave enough available seats. Back of book gives an answer of .08374 which means I amsetting up my problem wrong because my answer is way off. I amusing 15 trials 14 successes probabitlity .85 prob of failure .15.What am I doing wrong?

Explanation / Answer

I think you must have had a math error. You did itcorrectly. I get .1877. Part 2: p=.85 n=15; No seats available oif 15 out of the15 bookings showed. Not 14 out of 15. P(No seats available)=P(X=15)=(.85)15=.087354

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