The scenario describes a binomial probability problem. We want to find the probability that at most 50 people will test positive, even though none of the 1,600 individuals have taken drugs. Let's break down the problem:

• Accuracy of test: 97%, meaning there is a 3% chance of a false positive.
• Number of trials (n): 1600 (since the test is given to 1600 people).
• Probability of success (p): The "success" here is a false positive, which has a probability of 3% or 0.03.
• X: Number of people who test positive.

We can model the number of false positives using a binomial distribution: $X \sim \text{Binomial}(n = 1600, p = 0.03)$

We are asked to find the probability that at most 50 people test positive: $P(X \leq 50)$

Since $n = 1600$ is large, we can approximate the binomial distribution with a normal distribution using the Central Limit Theorem (CLT). The mean $\mu$ and variance $\sigma^2$ of the binomial distribution are given by: $\mu = np = 1600 \times 0.03 = 48$ $\sigma^2 = np(1-p) = 1600 \times 0.03 \times 0.97 = 46.56 \quad \text{so} \quad \sigma = \sqrt{46.56} \approx 6.82$

Thus, we approximate $X$ with a normal distribution: $X \sim N(\mu = 48, \sigma = 6.82)$

To find $P(X \leq 50)$, we first convert 50 to a z-score using: $z = \frac{X - \mu}{\sigma} = \frac{50 - 48}{6.82} \approx 0.29$

Now, we look up the z-score $z = 0.29$ in a standard normal distribution table or use a calculator to find the cumulative probability: $P(Z \leq 0.29) \approx 0.6141$

Thus, the probability that at most 50 people test positive is approximately $0.6141$, or 61.41%.

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Would you like more details on any part of the calculation?

Here are 5 related questions:

1. How does increasing the accuracy of the test affect the probability of false positives?
2. What is the probability that exactly 50 people will test positive?
3. How can we model this problem without the normal approximation?
4. How does the size of the sample (1600) influence the approximation to normal distribution?
5. What would the result be if the accuracy were 99% instead of 97%?

Tip: When working with large binomial problems, using the normal approximation is often more practical than direct binomial calculations.