A newspaper reported that what airline passengers like to do most on long flight

Question

A newspaper reported that what airline passengers like to do most on long flights is rest or sleep; in a survey of 3697 passengers, almost 70% did so. Suppose that for a particular route the actual percentage is exactly 70%, and consider randomly selecting ten passengers n-10 p-0.7). Hint: use technology or the appendix table of binomial probabilities. Round your answers to three decimal places.) (a) What is p(9)? p(9) = P[9)- (b) Calculate x s 6). (e) Caloulate the probability that more than half of the selected passengers rested or slept.

Explanation / Answer

Ans:

Binomial distribution:

n=10,p=0.7

a)P(k=9)=10C9*0.79*0.31

=0.121

using excel function:

P(k=9)=BINOMDIST(9,10,0.7,FALSE)=0.121

b)P(k<=6)=P(k=0)+P(k=1)....+P(k=6)=BINOMDIST(6,10,0.7,TRUE)=0.350

c)P(k>5)=1-P(k<=5)=1-BINOMDIST(5,10,0.7,TRUE)=1-0.15=0.85

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