In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities.
More than a decade ago, high levels of lead in the blood put 87% of children at risk. A concerted effort was made to remove lead from the environment. Now, suppose only 13% of children in the United States are at risk of high blood-lead levels.
(a) In a random sample of 212 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.)
(b) In a random sample of 212 children taken now, what is the probability that 50 or more have high blood-lead levels? (Round your answer to three decimal places.)
What might happen if the mucus in your noze,throat,and lungs suddenly dried up?
To determine if it is appropriate to use the normal approximation to the binomial, we need to check if the conditions for using the normal approximation are satisfied:
1. The random variable x (number of children with high blood-lead levels) follows a binomial distribution.
2. The number of trials n is large enough.
3. The success probability p remains the same for each trial.
(a) For the first scenario, we are examining a random sample of 212 children taken more than a decade ago. Since the sample size is large (n = 212) and the probability of success (p = 0.87) remains constant for each child tested, we can use the normal approximation to the binomial.
To estimate the probability that 50 or more children had high blood-lead levels from the sample, we can use the normal distribution. We will convert the binomial distribution into a normal distribution using the mean (μ) and standard deviation (σ).
The mean (μ) of the binomial distribution can be calculated using the formula: μ = n * p = 212 * 0.87 = 184.44
The standard deviation (σ) of the binomial distribution can be calculated using the formula: σ = sqrt(n * p * (1 - p)) = sqrt(212 * 0.87 * 0.13) = 6.34
Next, we calculate the z-score for 50 or more children: z = (x - μ) / σ = (50 - 184.44) / 6.34 = -20.37
Now, we can use a standard normal distribution table or a calculator to find the probability that the z-score is less than -20.37. However, since the normal distribution is symmetric, the probability of getting 50 or more children with high blood-lead levels is essentially zero (very close to 0).
Therefore, the probability that 50 or more children had high blood-lead levels more than a decade ago is approximately 0.000.
(b) Now, let's consider the second scenario where we are examining a random sample of 212 children taken now. Since the conditions still apply and the sample size is large (n = 212) with a constant probability of success (p = 0.13), we can again use the normal approximation to the binomial.
Using the same process as above, we determine the mean (μ) of the binomial distribution to be 212 * 0.13 = 27.56, and the standard deviation (σ) to be sqrt(212 * 0.13 * 0.87) = 5.91.
To estimate the probability that 50 or more children have high blood-lead levels from the sample, we calculate the z-score: z = (x - μ) / σ = (50 - 27.56) / 5.91 = 3.79
We can now use a standard normal distribution table or a calculator to find the probability that the z-score is greater than 3.79, which corresponds to the probability of having 50 or more children with high blood-lead levels in the current sample.
From the standard normal distribution table or the calculator, we find that the probability is approximately 0.000.
Therefore, the probability that 50 or more children have high blood-lead levels in a random sample of 212 children taken now is approximately 0.000.