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To solve this problem, let's break down each part:

### Given:
- Proportion of business executives willing to switch companies: \( p = 0.51 \)
- Sample size: \( n = 74 \)

### Part (a): Distribution of \(\hat{p}\)
The sampling distribution of the sample proportion \(\hat{p}\) can be approximated by a normal distribution if the sample size is large enough (which it is in this case, \( n = 74 \)). The distribution is given by:

\[
\hat{p} \sim N\left(p, \frac{p(1-p)}{n}\right)
\]

Substitute the values:
- \( p = 0.51 \)
- \( n = 74 \)

1. Mean (\( p \)):
   \[
   p = 0.51
   \]

2. Standard deviation (\( \sqrt{\frac{p(1-p)}{n}} \)):
   \[
   \sigma_{\hat{p}} = \sqrt{\frac{0.51 \cdot (1 - 0.51)}{74}} = \sqrt{\frac{0.51 \cdot 0.49}{74}}
   \]

Let's calculate this:

\[
\sigma_{\hat{p}} \approx 0.0582
\]

So, the distribution of \(\hat{p}\) is:

\[
\hat{p} \sim N(0.51, 0.0582)
\]

---

### Part (b): Probability that under 49% will be willing to switch
We need to find \( P(\hat{p} < 0.49) \).

Standardize using the Z-score formula:
\[
Z = \frac{\hat{p} - p}{\sigma_{\hat{p}}} = \frac{0.49 - 0.51}{0.0582}
\]

\[
Z \approx \frac{-0.02}{0.0582} \approx -0.3437
\]

Now, use a Z-table or calculator to find the probability for \( Z < -0.3437 \).

\[
P(Z < -0.3437) \approx 0.3665
\]

So, the probability that under 49% of executives will be willing to switch companies is approximately **0.3665**.

---

### Part (c): Probability that between 53% and 56% will be willing to switch
We need to find \( P(0.53 < \hat{p} < 0.56) \).

1. Find the Z-scores for \( \hat{p} = 0.53 \) and \( \hat{p} = 0.56 \).

   For \( \hat{p} = 0.53 \):
   \[
   Z = \frac{0.53 - 0.51}{0.0582} \approx 0.3437
   \]

   For \( \hat{p} = 0.56 \):
   \[
   Z = \frac{0.56 - 0.51}{0.0582} \approx 0.8582
   \]

2. Find the probabilities using the Z-table or calculator:

   \[
   P(Z < 0.8582) \approx 0.8040
   \]
   \[
   P(Z < 0.3437) \approx 0.6343
   \]

3. Subtract the probabilities:

   \[
   P(0.53 < \hat{p} < 0.56) = 0.8040 - 0.6343 \approx 0.1697
   \]

So, the probability that between 53% and 56% of executives will be willing to switch companies is approximately **0.1697**.

---

### Final Answers:
1. **Distribution**: \(\hat{p} \sim N(0.51, 0.0582)\)
2. **Probability that under 49% will switch**: 0.3665
3. **Probability that between 53% and 56% will switch**: 0.1697

Would you like a more detailed explanation of any part of the solution, or do you have further questions?

---

### Related Questions:
1. How does the sample size affect the standard deviation of \(\hat{p}\)?
2. What is the impact of changing \( p \) to 0.6 on the probabilities?
3. How would the probabilities change if the sample size increased to 100?
4. What does the Central Limit Theorem say about the distribution of \(\hat{p}\)?
5. How would we interpret the results in the context of the headhunter's research?

### Tip:
When using the normal approximation for a proportion, ensure that both \( np \) and \( n(1-p) \) are greater than 5 for accuracy. This rule checks if the sample size is large enough for the normal approximation to be valid.

Suppose that 51% of all business executives are willing to switch companies if offered a higher salary. If a headhunter randomly contacts a simple random sample of 74 executives, and show your answers to 4 decimal places.

a. What is the distribution of p-hat?

b. What is the probability that under 49% will be willing to switch companies?

c. What is the probability that between 53% and 56% will be willing to switch companies?

Solve a statistics problem involving the distribution of a sample proportion for business executives switching companies. Learn how to calculate the probability that under 49% or between 53% and 56% will switch, using normal distribution and Z-scores.

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