Question 1186019
Here's how to solve this binomial probability problem using the normal approximation:

**1. Check for Normal Approximation Applicability:**

*   np = 90 * 0.40 = 36 ≥ 10
*   n(1-p) = 90 * 0.60 = 54 ≥ 10

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

**2. Calculate Mean and Standard Deviation:**

*   Mean (μ) = np = 36
*   Standard Deviation (σ) = sqrt(np(1-p)) = sqrt(90 * 0.40 * 0.60) = sqrt(21.6) ≈ 4.65

**3. Apply Continuity Correction:**

Because we're approximating a discrete distribution (number of TVs) with a continuous one (normal), we adjust the values:

*   **a. Less than 32:**  Use 31.5 as the upper limit.  P(x < 32) becomes P(x < 31.5)
*   **b. Between 38 and 42:** Use 37.5 as the lower limit and 42.5 as the upper limit. P(38 ≤ x ≤ 42) becomes P(37.5 < x < 42.5)

**4. Calculate Z-scores:**

The z-score formula is: z = (x - μ) / σ

*   **a. For x < 31.5:** z = (31.5 - 36) / 4.65 ≈ -0.97
*   **b. For x < 37.5:** z = (37.5 - 36) / 4.65 ≈ 0.32
*   **c. For x < 42.5:** z = (42.5 - 36) / 4.65 ≈ 1.40

**5. Find Probabilities from Z-table or Calculator:**

*   **a. P(x < 31.5) = P(z < -0.97) ≈ 0.1660** (This is the area to the left of z = -0.97)
*   **b. P(37.5 < x < 42.5) = P(0.32 < z < 1.40) = P(z < 1.40) - P(z < 0.32)**
    *   P(z < 1.40) ≈ 0.9192
    *   P(z < 0.32) ≈ 0.6255
    *   P(37.5 < x < 42.5) ≈ 0.9192 - 0.6255 ≈ 0.2937

**Answers:**

*   **a. The probability that less than 32 televisions need repairs is approximately 0.1660.**
*   **b. The probability that between 38 and 42 televisions need repairs is approximately 0.2937.**