To determine whether a normal distribution can be used to approximate the binomial distribution, we need to check two conditions:

1. np ≥ 5
2. nq ≥ 5

Where:

• $n = 42$ is the number of trials (college students surveyed),
• $p = 0.62$ is the probability of success (the student will likely re-enroll),
• $q = 1 - p = 0.38$ is the probability of failure.

Step 1: Check if $np \geq 5$ and $nq \geq 5$

$np = 42 \times 0.62 = 26.04$ $nq = 42 \times 0.38 = 15.96$

Since both $np = 26.04$ and $nq = 15.96$ are greater than 5, the normal approximation can be used.

Thus, the correct answer is:

C. Yes, because both np ≥ 5 and nq ≥ 5.

Step 2: Apply the Normal Approximation

For normal approximation, we use the fact that the binomial distribution can be approximated by a normal distribution with:

• Mean $\mu = np = 26.04$,
• Standard deviation $\sigma = \sqrt{npq} = \sqrt{42 \times 0.62 \times 0.38} \approx 3.13$.

We will also apply the continuity correction to approximate the binomial probabilities using the normal distribution.

(a) Probability that exactly 26 students re-enroll ($P(X = 26)$):

Using the continuity correction, this becomes: $P(25.5 \leq X \leq 26.5)$

We standardize these values to Z-scores using: $Z = \frac{X - \mu}{\sigma}$

For $X = 25.5$: $Z = \frac{25.5 - 26.04}{3.13} \approx -0.17$

For $X = 26.5$: $Z = \frac{26.5 - 26.04}{3.13} \approx 0.15$

Now, we look up the Z-scores in the standard normal table:

• For $Z = -0.17$, the cumulative probability is approximately 0.4325.
• For $Z = 0.15$, the cumulative probability is approximately 0.5596.

Thus, the probability that exactly 26 students re-enroll is: $P(25.5 \leq X \leq 26.5) = 0.5596 - 0.4325 = 0.1271$

(b) Probability that at least 31 students re-enroll ($P(X \geq 31)$):

Using the continuity correction, this becomes: $P(X \geq 30.5)$

Standardizing: $Z = \frac{30.5 - 26.04}{3.13} \approx 1.42$

For $Z = 1.42$, the cumulative probability is approximately 0.9222.

Thus, the probability of at least 31 students re-enrolling is: $P(X \geq 31) = 1 - 0.9222 = 0.0778$

(c) Probability that at most 25 students re-enroll ($P(X \leq 25)$):

Using the continuity correction, this becomes: $P(X \leq 25.5)$

Standardizing: $Z = \frac{25.5 - 26.04}{3.13} \approx -0.17$

For $Z = -0.17$, the cumulative probability is approximately 0.4325.

Thus, the probability of at most 25 students re-enrolling is: $P(X \leq 25) = 0.4325$

Identifying Unusual Events:

An event is typically considered unusual if its probability is less than 0.05. Therefore, the probability of at least 31 students re-enrolling ($P = 0.0778$) is not considered highly unusual but is less likely. None of the events in this case fall under the "unusual" category based on a strict threshold of 0.05.

Would you like to see more details or ask any questions?

Relative Questions:

1. How do you calculate Z-scores for a normal approximation?
2. What is the continuity correction in normal approximation of binomial distributions?
3. Why do we use the normal distribution as an approximation for binomial distributions?
4. When is a binomial distribution a better option than using a normal approximation?
5. How does the sample size affect the use of normal approximation?

Tip:

Always apply the continuity correction when using the normal distribution to approximate binomial probabilities to improve accuracy.