Suppose it is known that 60% of students at a particular university are smokers. A
sample of 500 students from the university is selected at random. What is the approximate
probability that at least 280 of these students are smokers?

To determine the probability that at least 280 out of 500 students are smokers, we can use the normal approximation to the binomial distribution. Here's a step-by-step explanation:

1. **Define the binomial distribution parameters:**
   - \( n = 500 \) (the number of trials)
   - \( p = 0.60 \) (the probability of success, i.e., the student is a smoker)
   - \( q = 1 - p = 0.40 \) (the probability of failure, i.e., the student is not a smoker)

2. **Calculate the mean (μ) and standard deviation (σ) of the binomial distribution:**
   \[
   \mu = n \cdot p = 500 \cdot 0.60 = 300
   \]
   \[
   \sigma = \sqrt{n \cdot p \cdot q} = \sqrt{500 \cdot 0.60 \cdot 0.40} = \sqrt{120} \approx 10.95
   \]

3. **Use the normal approximation to the binomial distribution:**
   Since \( n \) is large, the binomial distribution can be approximated by a normal distribution \( N(\mu, \sigma^2) \).

4. **Convert the problem to a standard normal distribution:**
   We want to find the probability that at least 280 students are smokers, i.e., \( P(X \geq 280) \).

   First, we apply the continuity correction by subtracting 0.5:
   \[
   P(X \geq 280) \approx P\left(X \geq 279.5\right)
   \]

   Then, we standardize this value:
   \[
   Z = \frac{X - \mu}{\sigma} = \frac{279.5 - 300}{10.95} \approx \frac{-20.5}{10.95} \approx -1.87
   \]

5. **Find the probability using the standard normal distribution:**
   Using the standard normal distribution table or a calculator, we find the area to the right of \( Z = -1.87 \).

   \[
   P(Z \geq -1.87) = 1 - P(Z \leq -1.87) \approx 1 - 0.0307 = 0.9693
   \]

So, the approximate probability that at least 280 of the 500 students are smokers is about 0.9693, or 96.93%.

Would you like more details or have any questions? 

Here are five questions you might find interesting:
1. How do you calculate the probability of exactly 280 students being smokers?
2. What is the impact of a larger sample size on the approximation?
3. How would the probability change if the percentage of smokers was different?
4. What are the assumptions behind using the normal approximation for the binomial distribution?
5. Can we use other methods to solve this problem, and how do they compare?

**Tip:** When using normal approximation for binomial distributions, always remember to apply the continuity correction for more accurate results.

Learn how to calculate the approximate probability that at least 280 out of 500 students are smokers using the normal approximation to the binomial distribution. Detailed step-by-step solution included.

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