SOLUTION: According to flightstats.com, American Airlines flights from Dallas to Chicago are on time 80% of the time. Suppose 15 flights are randomly selected, and the number of on-time flig

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Question 1201749: According to flightstats.com, American Airlines flights from Dallas to Chicago are on time 80% of the time. Suppose 15 flights are randomly selected, and the number of on-time flights is recorded.
Note: Write your answer rounded to four decimal places.
(a) Find the probability that exactly 11 flights are on time.
0.1136
(b) Find the probability that fewer than 11 flights are on time.
(c) Find the probability that at least 11 flights are on time.
(d) Find the probability that between 9 and 11 flights, inclusive, are on time.
(e) Compute the mean and standard deviation of the random variable X, the number of on-time flights in 15 trials of the probability experiment.
Mean
, Standard deviation
.

Answer by ikleyn(52799)   (Show Source): You can put this solution on YOUR website!
.
According to flightstats.com, American Airlines flights from Dallas to Chicago
are on time 80% of the time. Suppose 15 flights are randomly selected,
and the number of on-time flights is recorded.
Note: Write your answer rounded to four decimal places.
(a) Find the probability that exactly 11 flights are on time.
0.1136
(b) Find the probability that fewer than 11 flights are on time.
(c) Find the probability that at least 11 flights are on time.
(d) Find the probability that between 9 and 11 flights, inclusive, are on time.
(e) Compute the mean and standard deviation of the random variable X, the number of on-time flights in 15 trials of the probability experiment.
Mean
Standard deviation
~~~~~~~~~~~~~~~~~~~


            It is a typical binomial distribution probability problem.

            Go to web-site https://stattrek.com/online-calculator/binomial.aspx
            and use free of charge online calculator there, specially developed for such problems.

            The number of trials is n= 15; the probability p of each success individual trial is 0.8.


(a) p = P(n=15;  k=11; p=0.8) = 0.1876.


(b) p = P(n=15; k<11; p=0.8) = 0.16424.


(c) p = P(n=15; k>=11; p=0.8) = 0.83576.


(d) p = P(n=15; 9<= k <=11; p=0.8) = P(n=15; k<=11; p=0.8) - P(n=15; k<9; p=0.8) = 0.35184 - 0.01806 = 0.33378.


(e) Mean = n*p = 15*0.8 = 1.2

    SD =  =  = 1.5492  (rounded).

Solved, with all necessary explanations.

----------------

To see a variety similar solved problems,  look into the lessons
    - Simple and simplest probability problems on Binomial distribution
    - 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,
which is your  PRIMARY  MAJOR  GOAL  visiting this forum  (I believe).



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