More than a decade ago, high levels of lead in the blood put 88% of children at risk. A concerted effort was made to remove lead from the environment. Now, suppose only 12% of children in the United States are at risk of high blood-lead levels.
(a) In a random sample of 194 children taken more than a decade ago, what is the probability that 50 or more had high blood-lead levels? (Round your answer to three decimal places.)
To find the probability that 50 or more children had high blood-lead levels out of a random sample of 194 children, we can use the binomial distribution formula.
The binomial distribution formula is given by:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability of getting exactly k successes,
- C(n, k) is the number of combinations of n items taken k at a time,
- p is the probability of success on a single trial,
- n is the number of trials, and
- k is the number of successes.
In this case:
- n = 194 (sample size)
- p = 0.88 (probability of high blood-lead levels)
- k = 50 (minimum number of successes)
We want to find the probability of getting 50 or more successes, so we sum up the probabilities for all values of k from 50 to n.
P(X >= 50) = P(X = 50) + P(X = 51) + ... + P(X = n)
To calculate this, we can use a cumulative binomial probability distribution table or a statistical software. Using a cumulative binomial probability distribution table or software, the probability is found to be 0.999. Therefore, the probability that 50 or more children had high blood-lead levels in a random sample of 194 children taken more than a decade ago is 0.999 (rounded to three decimal places).
To find the probability that 50 or more children had high blood-lead levels in a random sample of 194 taken more than a decade ago when 88% of children were at risk, we can use the binomial probability formula.
The binomial probability formula is:
P(x) = (nCx) * p^x * q^(n-x)
Where:
P(x) is the probability of getting exactly 'x' successes,
n is the number of trials,
p is the probability of success in each trial,
q is the probability of failure in each trial, and
(nCx) is the binomial coefficient, also known as the number of combinations.
In this case, we want to find the probability of 50 or more children having high blood-lead levels, so we need to calculate the sum of probabilities for x = 50, 51, 52, ..., up to the maximum possible value.
First, let's calculate the probability of success in each trial. Given that 88% of children were at risk, the probability of a child having high blood-lead levels is p = 0.88.
The probability of failure (a child not having high blood-lead levels) is q = 1 - p = 1 - 0.88 = 0.12.
Now, we can calculate the probability using the binomial probability formula.
P(50 or more) = P(x≥50) = P(x=50) + P(x=51) + P(x=52) + ...
To calculate each individual probability, we substitute the values into the formula:
P(x) = (194Cx) * (0.88)^x * (0.12)^(194-x)
Now, we need to calculate each individual probability and sum them up.
P(50 or more) = P(x=50) + P(x=51) + P(x=52) + ...
For x = 50:
P(x=50) = (194C50) * (0.88)^50 * (0.12)^(194-50)
= (194! / (50!(194-50)!)) * (0.88)^50 * (0.12)^(194-50)
To find each individual probability, we will substitute the corresponding values into this formula.
We need to calculate this for each value starting from x = 50 up to the maximum possible value, which is 194.
After calculating each probability, we will sum them up to find the desired probability.
n=194
P(r >/= 50)
p= .12 (converted percent to decimal)
q= 1-p = 1-.12 = .88
P(x) = r-.5 if r is to the left(greater than) therefore,
P(x>/= 49.5)
You need P(z>/= ?)
So:
mean (mu) = np = (194)(.12) = 23.28
stan. dev. (sigma) = sqrt(npq) = (194)(.12)(.88) = 4.526
z= (x - mu)/sigma = (49.5 - 23.28)/4.526 = 5.79
So: P(z >/= 5.79), now you look up the probability on your z-chart and there's your answer.