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.
Suppose more than a decade ago, high levels of lead in the blood put 90% of children at risk. A concerted effort was made to remove lead from the environment. Now, suppose only 8% of children in the United States are at risk of high blood-lead levels. In a random sample of 196 children taken more than a decade ago, what is the probability that 50 or more had high blood-lead levels? (Round your answer to four decimal places.)

To solve this problem, we'll break it down into two steps:

1. **Check if it's appropriate to use the normal approximation to the binomial distribution.**
2. **Use the normal approximation to estimate the probability.**

### Step 1: Checking if it's appropriate to use the normal approximation

The binomial distribution can be approximated by a normal distribution when the sample size \(n\) is large and both \(np\) and \(n(1 - p)\) are sufficiently large (typically greater than 5). 

Here, we have:
- \(n = 196\) (sample size)
- \(p = 0.90\) (probability of children at risk more than a decade ago)

We calculate:
- \(np = 196 \times 0.90 = 176.4\)
- \(n(1 - p) = 196 \times 0.10 = 19.6\)

Since both \(np = 176.4\) and \(n(1 - p) = 19.6\) are greater than 5, it's appropriate to use the normal approximation.

### Step 2: Normal Approximation to Estimate the Probability

The binomial distribution \(X \sim \text{Binomial}(n=196, p=0.90)\) can be approximated by a normal distribution \(X \sim N(\mu, \sigma^2)\) with:
- Mean \( \mu = np = 176.4 \)
- Standard deviation \( \sigma = \sqrt{np(1 - p)} = \sqrt{196 \times 0.90 \times 0.10} = \sqrt{17.64} \approx 4.2\)

Now, we want to find the probability that 50 or more children had high blood-lead levels, i.e., \(P(X \geq 50)\).

Since we are approximating a discrete binomial distribution with a continuous normal distribution, we apply a continuity correction by subtracting 0.5 from 50. So we calculate \(P(X \geq 49.5)\).

Next, we standardize the value \(X = 49.5\) using the z-score formula:

\[
z = \frac{X - \mu}{\sigma} = \frac{49.5 - 176.4}{4.2} = \frac{-126.9}{4.2} \approx -30.21
\]

The z-score of \(-30.21\) is extremely low, which corresponds to a probability very close to 0. This means the probability that 50 or more children had high blood-lead levels more than a decade ago is practically 1.

### Final Answer:
The probability is approximately **1.0000**.

Would you like further clarification on any part of the solution?

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Here are 5 related questions that expand on this problem:
1. How would the probability change if only 80% of children were at risk more than a decade ago?
2. What is the probability that fewer than 40 children had high blood-lead levels, using the same parameters?
3. Can the normal approximation be used if the sample size were 100 children instead of 196? Why or why not?
4. How would you calculate the exact binomial probability without using the normal approximation?
5. What is the difference between continuity correction and z-score standardization?

**Tip:** Always verify if the normal approximation is valid by checking the conditions \(np\) and \(n(1 - p)\) before using it.

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. Suppose more than a decade ago, high levels of lead in the blood put 90% of children at risk. A concerted effort was made to remove lead from the environment. Now, suppose only 8% of children in the United States are at risk of high blood-lead levels. In a random sample of 196 children taken more than a decade ago, what is the probability that 50 or more had high blood-lead levels? (Round your answer to four decimal places.)

This problem explains how to check for the validity of using a normal approximation to the binomial distribution and estimate probabilities. It focuses on calculating the probability of children having high blood-lead levels using a random sample of 196, highlighting key statistical concepts like continuity correction and z-scores.

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