Question

A CNN/USA Today/Gallup poll in April 2005 reported that 75% of adult Americans were satisfied with the job the major airlines were doing. Suppose 10 adult Americans are selected at random and the number who are satisfied is recorded. Find the probability that 5 or more (5,6,7,8,9,10 then add all together) are satisfied with the airline.

Answer 1

Let X is a random variable shows the number of people satisfied with the airline. Here X has binomial distribution with parameters n=10 and p=0.75.

The pdf of X is

Following table shows the probabilities for x=5,6,...10:

X	P(X=x)
5	0.0583992
6	0.145998001
7	0.250282288
8	0.281567574
9	0.187711716
10	0.056313515
Total	0.980272293

So the probability that 5 or more (5,6,7,8,9,10 then add all together) are satisfied with the airline is

P(X >= 5) = 0.9803