To solve this problem, we'll use the Central Limit Theorem (CLT) for proportions, which states that for sufficiently large $n$, the sampling distribution of the sample proportion $\hat{p}$ will be approximately normal with:

$\mu_{\hat{p}} = p \quad \text{and} \quad \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}.$

Step 1: Identify the given values

• $p = 0.40$ (population proportion)
• $n = 400$ (sample size)
• $39$ and $0.42$ are the sample proportions between which we want to find the probability.

Step 2: Compute the standard deviation of the sample proportion

$\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.40(1-0.40)}{400}} = \sqrt{\frac{0.40 \cdot 0.60}{400}} = \sqrt{\frac{0.24}{400}} = \sqrt{0.0006} \approx 0.02449$

Step 3: Convert sample proportions to z-scores

The z-score is calculated using the formula: $z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}}$

For $\hat{p} = 0.39$: $z_1 = \frac{0.39 - 0.40}{0.02449} = \frac{-0.01}{0.02449} \approx -0.408$

For $\hat{p} = 0.42$: $z_2 = \frac{0.42 - 0.40}{0.02449} = \frac{0.02}{0.02449} \approx 0.817$

Step 4: Find the probability between the z-scores

Using a standard normal distribution table or a calculator, we find the area under the curve:

• $P(Z < -0.408) \approx 0.3416$
• $P(Z < 0.817) \approx 0.7939$

The probability between $z_1$ and $z_2$ is: $P(-0.408 < Z < 0.817) = P(Z < 0.817) - P(Z < -0.408) = 0.7939 - 0.3416 = 0.4523$

Final Answer:

The probability that the sample proportion falls between 0.39 and 0.42 is approximately: $\boxed{0.452}$

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Would you like further details or clarifications? 😊

5 Related Questions:

1. How does the Central Limit Theorem ensure that the sampling distribution is normal?
2. What happens if the sample size $n$ is smaller than 30 in a similar problem?
3. How do you calculate the z-scores for proportions in different cases?
4. What are the conditions for using the normal approximation for proportions?
5. How does increasing the sample size affect the standard error $\sigma_{\hat{p}}$?

One Tip:

Always check that $n \cdot p \geq 10$ and $n \cdot (1-p) \geq 10$ before using the normal approximation for proportions to ensure it's valid.