SOLUTION: According to flightstats.com, American Airlines flights from Dallas to Chicago are on time 78.2% of the time. Suppose 15 of these flights are randomly selected and the number of

Algebra ->  Probability-and-statistics -> SOLUTION: According to flightstats.com, American Airlines flights from Dallas to Chicago are on time 78.2% of the time. Suppose 15 of these flights are randomly selected and the number of      Log On


   



Question 1197156: According to flightstats.com, American Airlines flights from Dallas to Chicago are on time 78.2% of the time.
Suppose 15 of these flights are randomly selected and the number of on-time flights is recorded.

Calculate the probability that 9 or fewer of the selected flights were on time. Enter your answer as a decimal rounded to four decimal places.

Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!
.

It is a standard binomial distribution problem.

Number of trials is 15; number of success trials <= 9; the probability of each success trial is p=0.782.


The standard formula is

    P = P(n=15; k <= 9; p= 0.782) = sum%28C%5B15%5D%5Ek%2A0.782%5Ek%2A%281-0.782%29%5E%2815-k%29%2Ck=0%2C9%29 = sum%28C%5B15%5D%5Ek%2A0.782%5Ek%2A0.218%5E%2815-k%29%2C+k=0%2C15%29.


     To facilitate my calculations, I used online calculator at this site  https://stattrek.com/online-calculator/binomial.aspx

     It provides nice instructions  and  a convenient input and output for all relevant options/cases.


          The resulting number is P = 0.08719  (rounded).    ANSWER

Solved.

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If you want to see many similar  (or different)  solved problems,  look into the lessons
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    - Typical binomial distribution probability problems
    - How to calculate Binomial probabilities with Technology (using MS Excel)
    - Solving problems on Binomial distribution with Technology (using MS Excel)
    - Solving problems on Binomial distribution with Technology (using online solver)
in this site.

After reading these lessons,  you will be able to solve such problems on your own,
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