Question

In the following binomial probability distribution, an October2016 CNN/USA Today/Gallup poll reported that 61% of adult Americans were satisfied with the job that the nations major airlines were doing. Ten adult Americans are selected at random and the results are recorded. Find the probbability that at least eiht adults are satisfied. Find the probability that fewrer than three adults are satisfied.

Answer 1

Let the number of Americans satisfied be denoted by n where n will follow binomia distribution.

Then Probability n = k for any k in this case belonging from 0 to 10 since 10 adults where selected in random is by the formula for binomial distribution

P(n=k) = where p = 0.61 since 61 percent of Americans are satisfied with their job

For the first problem we need to calculate P(n>=8) which is probability that n is atleast 8

P(n>=8 ) = P(n=8) + P(n=9) + P(n=10) =

                    

                     45 * 0.61⁸ * 0.39² + 10 * 0.61⁹ * 0.39 + 1 * 0.61¹⁰ * 1 = 0.1839 = 0.184 approximately

And for the second question we need to calculate probability n is less than 3 that is it can take value 0 ,1 or 2

P(n < 3 ) = P(n=0) + P(n=1) + P(n=2)

                =

                = 1 * 1 * .39¹⁰ + 10 * 0.61 * .39⁹ + 45 * 0.61² * 0.39⁸ = 0.0103 approximately