Anonymous
Anonymous asked in Education & ReferenceHomework Help · 9 years ago

Not sure how to solve using binomial probability distribution table or binomial formula?

Suppose 15 flights are randomly selected and the number of on time flights is recorded. Round to four decimal places.

Probability that at least 14 flights are on time

Probability that fewer then 14 flights are on time

Probability that between 12 and 14 flights inclusive are on time

What is the expected number of on time flights

Please show me what to do to solve this? I have no clue.

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  • 9 years ago
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    I'll assume that 9 out of 10 flights are on time. We need a p value to make these computations:

    Binomial Sampling Distribution:

    P(n) = C(N,n) * p^n * (1-p)^(N-n)

    P(0)= 1 * ( 1/ 1 ) * ( 1 / 1000000000000000 ) =

    0.0000000000 Total: 0.0000000000

    P(1)= 15 * ( 9/ 10 ) * ( 1 / 100000000000000 ) =

    0.0000000000 Total: 0.0000000000

    P(2)= 105 * ( 81/ 100 ) * ( 1 / 10000000000000 ) =

    0.0000000000 Total: 0.0000000000

    P(3)= 455 * ( 729/ 1000 ) * ( 1 / 1000000000000 ) =

    0.0000000003 Total: 0.0000000003

    P(4)= 1365 * ( 6561/ 10000 ) * ( 1 / 100000000000 ) =

    0.0000000090 Total: 0.0000000093

    P(5)= 3003 * ( 59049/ 100000 ) * ( 1 / 10000000000 ) =

    0.0000001773 Total: 0.0000001866

    P(6)= 5005 * ( 531441/ 1000000 ) * ( 1 / 1000000000 ) =

    0.0000026599 Total: 0.0000028465

    P(7)= 6435 * ( 4782969/ 10000000 ) * ( 1 / 100000000 ) =

    0.0000307784 Total: 0.0000336249

    P(8)= 6435 * ( 43046721/ 100000000 ) * ( 1 / 10000000 ) =

    0.0002770056 Total: 0.0003106305

    P(9)= 5005 * ( 387420489/ 1000000000 ) * ( 1 / 1000000 ) =

    0.0019390395 Total: 0.0022496701

    P(10)= 3003 * ( 3486784401/ 10000000000 ) * ( 1 / 100000 ) =

    0.0104708136 Total: 0.0127204836

    P(11)= 1365 * ( 31381059609/ 100000000000 ) * ( 1 / 10000 ) =

    0.0428351464 Total: 0.0555556300

    P(12)= 455 * ( 282429536481/ 1000000000000 ) * ( 1 / 1000 ) =

    0.1285054391 Total: 0.1840610691

    P(13)= 105 * ( 2541865828329/ 10000000000000 ) * ( 1 / 100 ) =

    0.2668959120 Total: 0.4509569811

    P(14)= 15 * ( 22876792454961/ 100000000000000 ) * ( 1 / 10 ) =

    0.3431518868 Total: 0.7941088679

    P(15)= 1 * ( 205891132094649/ 1000000000000000 ) * ( 1 / 1 ) =

    0.2058911321 Total: 1.0000000000

    If 9 out of 10 flights are expected to arrive on time, there is a 0.3432 probability that in these 15 flights that exactly 14 will be on time.

    Probability that at least 14 flights are on time

    0.3431518868 + 0.2058911321

    Probability that fewer then 14 flights are on time

    0.4509569811

    Probability that between 12 and 14 flights inclusive are on time

    0.1285054391 + 0.2668959120 + 0.3431518868

    What is the expected number of on time flights

    if p=9/10, then 9/10*15 = 1.5*9 = 13.5 flights (note: you didn't provide p in the problem)

    Please show me what to do to solve this? I have no clue

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