The problem is asking to determine whether you can use a normal distribution to approximate a binomial distribution. Here's a breakdown of the key points and the steps needed to solve this problem:

Problem Breakdown:

• Survey Data: 62% of students (probability $p = 0.62$) are likely to re-enroll if new programs are offered.
• Sample Size: 39 students are randomly selected (sample size $n = 39$).
• Events to Calculate:
  1. The probability that exactly 24 students will re-enroll.
  2. The probability that at least 29 students will re-enroll.
  3. The probability that at most 23 students will re-enroll.

Step 1: Check if the Normal Approximation is Appropriate

We can approximate the binomial distribution by a normal distribution if both of the following conditions hold:

1. $n \cdot p \geq 5$
2. $n \cdot (1 - p) \geq 5$

Let’s check these conditions:

• $n \cdot p = 39 \cdot 0.62 = 24.18$
• $n \cdot (1 - p) = 39 \cdot (1 - 0.62) = 14.82$

Both values are greater than 5, so the normal approximation is valid.

Step 2: Find the Mean and Standard Deviation

For a binomial distribution:

• Mean $\mu = n \cdot p = 39 \cdot 0.62 = 24.18$
• Standard Deviation $\sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{39 \cdot 0.62 \cdot 0.38} \approx 3.06$

Step 3: Use the Normal Approximation

To calculate the required probabilities using the normal approximation, we will apply the continuity correction (adding or subtracting 0.5 when converting from binomial to normal).

1. Exactly 24 Students:

   • Apply continuity correction: $P(23.5 \leq X \leq 24.5)$
   • Convert to z-scores: $z_1 = \frac{23.5 - 24.18}{3.06} \approx -0.22, \quad z_2 = \frac{24.5 - 24.18}{3.06} \approx 0.10$ Use the standard normal table or a calculator to find the probabilities corresponding to these z-scores.

2. At least 29 Students:

   • Apply continuity correction: $P(X \geq 28.5)$
   • Convert to z-score: $z = \frac{28.5 - 24.18}{3.06} \approx 1.41$ Find the probability for $z \geq 1.41$.

3. At most 23 Students:

   • Apply continuity correction: $P(X \leq 23.5)$
   • Convert to z-score: $z = \frac{23.5 - 24.18}{3.06} \approx -0.22$ Find the probability for $z \leq -0.22$.

Would you like to see the detailed calculations for these probabilities?

Additional Questions:

1. What would happen if the sample size were larger, say 100 students? How would that affect the normal approximation?
2. How do we interpret the continuity correction when approximating binomial probabilities using the normal distribution?
3. Could this problem be solved using the binomial distribution directly? How would the results compare?
4. If 62% changed to 50%, how would that affect the calculations and conclusions?
5. What are some other situations where normal approximation to binomial distribution might not be valid?

Tip:

When applying normal approximations to binomial distributions, always remember to check both conditions involving $n \cdot p$ and $n \cdot (1 - p)$ to ensure the approximation is valid.