Suppose that more than a decade ago high levels of lead in the blood put 81 percent of children at risk. A concerted effort was made to remove lead from the environment. Suppose, according to a survey, only 5 percent of children in the United States are at risk of high blood-lead levels. In a random sample of 150 children taken more than a decade ago, what is the probability that 110 or more had high blood-lead levels?
To find the probability that 110 or more children out of a random sample of 150 had high blood-lead levels, we need to use the binomial probability formula. The binomial probability distribution calculates the probability of getting a certain number of successes (children with high blood-lead levels) in a fixed number of trials (children in the sample), given a known success rate (81% in the past).
The formula for calculating the probability of getting exactly k successes in n trials, with a success rate of p, is:
P(k) = C(n, k) * (p^k) * ((1 - p)^(n - k))
Where:
- P(k) is the probability of getting exactly k successes
- C(n, k) is the binomial coefficient, calculated as n! / (k!(n - k)!)
- p is the probability of success
- k is the number of successes
- n is the number of trials
In this case:
- p = 0.81 (81% or 0.81 in decimal)
- n = 150 (the number of children in the sample)
- We want to find the probability of having 110 or more successes (k ≥ 110)
To find the probability of getting 110 or more children with high blood-lead levels, we need to sum up the individual probabilities of getting exactly k, k + 1, k + 2, ..., n successes. So we need to calculate:
P(110) + P(111) + P(112) + ... + P(150)
Let's calculate that using the formula.