Question 1186293
Here's how to solve this problem using the normal approximation to the binomial distribution:

1.  **Check if the normal approximation is appropriate:**

    *   n * p = 229 * 0.38 = 86.82 >= 10
    *   n * (1-p) = 229 * (1 - 0.38) = 229 * 0.62 = 142.18 >= 10

    Since both conditions are met, the normal approximation is reasonable.

2.  **Calculate the mean (μ) and standard deviation (σ) of the sample proportion:**

    *   μ = p = 0.38
    *   σ = sqrt[p * (1-p) / n] = sqrt[0.38 * 0.62 / 229] ≈ 0.0322

3.  **Calculate the z-scores:**

    The z-score formula for proportions is: z = (p̂ - p) / σ

    *   For p̂ = 0.3: z = (0.3 - 0.38) / 0.0322 ≈ -2.4845
    *   For p̂ = 0.4: z = (0.4 - 0.38) / 0.0322 ≈ 0.6211

4.  **Find the probabilities using the z-table or calculator:**

    *   P(z < -2.4845) ≈ 0.0065 (This is the area to the left of -2.4845)
    *   P(z < 0.6211) ≈ 0.7327 (This is the area to the left of 0.6211)

5.  **Calculate the probability between 0.3 and 0.4:**

    P(0.3 < p̂ < 0.4) = P(z < 0.6211) - P(z < -2.4845)
    P(0.3 < p̂ < 0.4) ≈ 0.7327 - 0.0065
    P(0.3 < p̂ < 0.4) ≈ 0.7262

Therefore, the probability that the sample proportion is between 0.3 and 0.4 is approximately **0.7262**.