Based on data from the national center of health, 26% of adults are overweight. A random sample of 15 adults is taken. Calculate the probability that three of four of 15 people selected are overweight?
Calculating three or four of 15 adults
This is an example of a binomial event and probability distribution. For each person you select, they are either overweight or not-only two possible outcomes. If we define 'success' as selecting an overweight person, the probability of success when we select someone is denoted p, and is given as 26%, or, in decimal form, .26.
p=.26 <-- probability of success
q=1-.26=.74 <-- probability of 'failure', choosing someone who is not overweight. p+q=1
n=15 <-- Number of people we are selecting
We need to add the probabilities of getting exactly 3 out of 15 and getting exactly 4 out of 15. We'll use x to represent the number of successes in our selection of 15 adults:
Our expression of the total probability of selecting either 3 or 4 people out of 15: P(x=3) + P(x=4)
The binomial probability formula: P(X=x) = nCx * px * qn-x
nCx is our notation for a combination, which mathematically equals [ n! / ( x! * (n-x)! ) ]
The probability of picking 3 or four people out of 15 is the sum of the following two expressions:
P(X=3) = 15C3 * (0.26)3 * (0.74)15-3 = 455 * (0.26)3 * (0.74)12 = 0.216
P(X=4) = 15C4 * (0.26)4 * (0.74)15-4 = 1365 * (0.26)4 * (0.74)11 = 0.227
P(X=3) + P(X=4) = 0.216 + 0.227 = 0.443
The probability of selecting three or four overweight people in a random sample of 15 is 0.443.