21. Suppose that the probability that a flight leaving from Cleveland to Chicago

Question

21. Suppose that the probability that a flight leaving from Cleveland to Chicago arrives on time 75% of the time. Assume that this flight follows a binomial distribution and that a sample of 18 flights is selected. Approximately, how many trips would you expect to arrive on time out of 18 randomly selected flights, give or take how many flights? Round to the nearest whole flight.

22. Suppose that the probability that a flight leaving from Cleveland to Chicago arrives on time 75% of the time. Assume that this flight follows a binomial distribution and that a sample of 18 flights is selected. What is the probability that less than half of the flights out of 18 randomly selected flights will arrive in Chicago on time? Round to the nearest thousandth.

Explanation / Answer

21. Here we have p = 0.75 , n = 18

So x ~ Bin(n,p )

So expected trips = E(x) = n*p = 0.75 * 18 = 13.5

22. Here x ~ Bin ( n,p )

p ( X = x ) = nCx * px * ( 1- p ) n-x   

Here we need to find the probability that less than half of the flights out of 18 flights.

So p ( x < 9) = p( x = 0 ) + p( x=1 ) + p( x =2 ) + p ( x = 3 ) + p( x = 4 ) + p ( x = 5 ) + p ( x = 6 ) + p ( x = 7 ) + p( x = 8)

=  18C0 * 0.750 * ( 1- 0.75 ) 18-0 + 18C1 * 0.751 * ( 1- 0.75 ) 18-1 + 18C2 * 0.752 * ( 1- 0.75 ) 18-2 +18C3 * 0.753 * ( 1- 0.75 ) 18-3 +18C4* 0.754 * ( 1- 0.75 ) 18-4 +18C5 * 0.755 * ( 1- 0.75 ) 18-5 +18C6 * 0.756 * ( 1- 0.75 ) 18-6 +18C7 * 0.757 * ( 1- 0.75 ) 18-7 +18C8 * 0.758 * ( 1- 0.75 ) 18-8

= 0.005

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