Question 1186410: 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(1987)   (Show Source):
You can put this solution on YOUR website! 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 = 90 * 0.40 = 36 >= 10
* n * (1-p) = 90 * 0.60 = 54 >= 10
Since both conditions are met, the normal approximation is reasonable.
**2. Calculate the mean (μ) and standard deviation (σ) of the binomial distribution:**
* μ = n * p = 90 * 0.40 = 36
* σ = sqrt(n * p * (1-p)) = sqrt(90 * 0.40 * 0.60) = sqrt(21.6) ≈ 4.65
**3. Apply the continuity correction:**
Since we are dealing with a discrete variable (number of televisions) and approximating it with a continuous distribution (normal), we need a continuity correction.
* **a. Less than 32 televisions:** We want P(x < 32). With the continuity correction, we use P(x < 31.5).
* **b. Between 38 and 42 televisions:** We want P(38 <= x <= 42). With the continuity correction, we use P(37.5 < x < 42.5).
**4. Calculate the 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 the probabilities using the z-table or calculator:**
* **a. P(x < 31.5):** Look up the probability corresponding to z = -0.97. P(z < -0.97) ≈ 0.1660.
* **b. P(37.5 < x < 42.5):** Look up the probabilities for z = 1.40 and z = 0.32. P(z < 1.40) ≈ 0.9192 and P(z < 0.32) ≈ 0.6255. Then subtract the smaller probability from the larger: P(0.32 < z < 1.40) = P(z < 1.40) - P(z < 0.32) ≈ 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**.
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