A consumer buys n light bulbs, each of which has a lifetime that has a mean

of 800 hours, a standard deviation of 100 hours, and a normal distribution. A light bulb is replaced by another as soon as it burns out. Assuming independence of the lifetimes, find the smallest n so that the succession of light bulbs produces light for at least 10,000 hours with a probability of 0.90.

To find the smallest value of n that produces light for at least 10,000 hours with a probability of 0.90, we can use the concept of the central limit theorem and the normal distribution.

The lifetime of each light bulb follows a normal distribution with a mean of 800 hours and a standard deviation of 100 hours. Since the light bulb replacements are independent, the total lifetime of n light bulbs will follow a normal distribution with a mean of n * 800 hours and a standard deviation of √(n * 100^2) = 10 * n hours.

To find the probability that the total lifetime is at least 10,000 hours, we need to calculate the area under the normal distribution curve to the right of 10,000 hours.

1. Standardize the value of 10,000 hours using the formula: z = (x - mean) / standard deviation
- In this case, x = 10,000 hours, mean = n * 800 hours, and standard deviation = 10 * n hours.
- So, z = (10,000 - n * 800) / (10 * n)

2. Use a standard normal table or a calculator to find the cumulative probability to the right of z.
- The cumulative probability to the right of z can be represented as P(X > z).

3. Find the smallest value of n that satisfies the condition: P(X > z) ≥ 0.90, where X is the total lifetime of n light bulbs.

Now, let's solve this equation step by step:

P(X > z) ≥ 0.90

Step 1: Substitute the value of z:
P((10,000 - n * 800) / (10 * n)) ≥ 0.90

Step 2: Convert the inequality to an equality:
P((10,000 - n * 800) / (10 * n)) = 0.90

Step 3: Look up the standard normal distribution table or use a calculator to find the z-score that corresponds to a cumulative probability of 0.90. Let's say this value is z_0.90.

Step 4: Substitute the value of z_0.90 into the equation:
(10,000 - n * 800) / (10 * n) = z_0.90

Step 5: Solve for n:
10,000 - n * 800 = 10 * n * z_0.90
10,000 - n * 800 = 9 * n * z_0.90
10,000 - n * 800 = 9 * n * z_0.90

Step 6: Simplify and solve for n:
10,000 = 9 * n * z_0.90 + n * 800
10,000 = n * (9 * z_0.90 + 800)
n = 10,000 / (9 * z_0.90 + 800)

By following these steps, you can find the value of n that satisfies the condition of producing light for at least 10,000 hours with a probability of 0.90.