SOLUTION: 2. It has been estimated that 40% of the televisions that the manufacturer makes need repairs in the first three years operation. A new hotel buys 90 televisions form the company.

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Question 1186019: 2. It has been estimated that 40% of the televisions that the manufacturer makes need repairs in the first three years operation. A new hotel buys 90 televisions form the company. Find the probability that in the first three years operation:
a. Less than 32 of the television needs repairs
b. Between 38 and 42 of the televisions, needs repairs
Answer by CPhill(2264) About Me  (Show Source):
You can put this solution on YOUR website!
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.**