Determine whether you can use a normal distribution to approximate the binomial distribution. If you​ can, use the normal distribution to approximate the indicated probabilities and sketch their graphs. If you​ cannot, explain why and use a binomial distribution to find the indicated probabilities. Identify any unusual events. Explain.
In a survey of U.S. college​ students, 62​% said that if their college offers new programs tailored to the new​ economy, it will make them more likely to​ re-enroll. You randomly select 42 college students. Find the probability that the number who are more likely to enroll if their college offers new programs tailored to the new economy is​ (a) exactly 26​, ​(b) at least 31​, and​ (c) at most 25.
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Part 1
Determine whether a normal distribution can be used to approximate the binomial distribution. Choose the correct answer.
A.
​No, because npless than5.
B.
​No, because nqless than5.
C.
​Yes, because both npgreater than or equals5 and nqgreater than or equals5.

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.

Determine whether you can use a normal distribution to approximate the binomial distribution. If you​ can, use the normal distribution to approximate the indicated probabilities and sketch their graphs. If you​ cannot, explain why and use a binomial distribution to find the indicated probabilities. Identify any unusual events. In a survey of U.S. college​ students, 62​% said that if their college offers new programs tailored to the new​ economy, it will make them more likely to​ re-enroll. You randomly select 42 college students. Find the probability that the number who are more likely to enroll if their college offers new programs tailored to the new economy is​ (a) exactly 26​, ​(b) at least 31​, and​ (c) at most 25.

Explore how to use the normal approximation to solve a binomial distribution problem based on a survey where 62% of U.S. college students are more likely to re-enroll with new programs. Learn about the conditions for applying normal approximation and calculate probabilities for exactly 26, at least 31, and at most 25 students re-enrolling.

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