SOLUTION: The lifetime of Evergo bulbs is normally distributed with a mean of 400 hours and a standard deviation of 50 hours. (I) What percentage of the bulbs will burn out in less than 36

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Question 1198608: The lifetime of Evergo bulbs is normally distributed with a mean of 400 hours and a standard deviation of 50 hours.
(I) What percentage of the bulbs will burn out in less than 360 hours?
(II) How many hours will 80% of Evergo bulbs burn out?
2. Supposed that a computerized procedure generates integer values between 1 and 5 (i.e 1,2,3,4, and 5) in such a way that each value has the same chance of selection. Let X represent a generated value.
a) Describe the probability distribution of X.
b) What is the mean and variance of the random variable X?

Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
**1. Evergo Bulbs**
**(I) Percentage of bulbs burning out in less than 360 hours**
1. **Standardize the value:**
* z = (X - μ) / σ
* z = (360 - 400) / 50 = -0.8
2. **Find the probability using a standard normal distribution table or calculator:**
* P(X < 360) = P(Z < -0.8) ≈ 0.2119
Therefore, approximately **21.19%** of the bulbs will burn out in less than 360 hours.
**(II) Hours for 80% of bulbs to burn out**
1. **Find the z-score corresponding to the 80th percentile:**
* Using a standard normal distribution table or calculator, find the z-score that corresponds to a cumulative probability of 0.80.
* z ≈ 0.84
2. **Use the z-score formula to find the corresponding bulb lifetime:**
* X = μ + zσ
* X = 400 + 0.84 * 50
* X = 400 + 42
* X = 442 hours
Therefore, 80% of Evergo bulbs will burn out in approximately **442 hours**.

**2. Computerized Procedure**
**(a) Describe the probability distribution of X**
* X represents the generated integer value.
* Since each value (1, 2, 3, 4, 5) has an equal chance of selection, this follows a **discrete uniform distribution**.
**(b) Mean and Variance of X**
* **Mean (μ):**
* For a discrete uniform distribution, the mean is:
* μ = (minimum value + maximum value) / 2
* μ = (1 + 5) / 2 = 3
* **Variance (σ²):**
* For a discrete uniform distribution, the variance is:
* σ² = [(maximum value - minimum value + 1)² - 1] / 12
* σ² = [(5 - 1 + 1)² - 1] / 12
* σ² = 24 / 12 = 2
* **Standard Deviation (σ):**
* σ = √σ² = √2 ≈ 1.4142
**Therefore, for the random variable X:**
* Mean (μ) = 3
* Variance (σ²) = 2
* Standard Deviation (σ) ≈ 1.4142

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