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 solve this problem, we need to find the smallest value of n such that the sum of the lifetimes of n light bulbs is at least 10,000 hours with a probability of 0.90.
Given:
Mean lifetime of a light bulb (μ) = 800 hours
Standard deviation of a light bulb (σ) = 100 hours
First, we need to convert the problem into a standard normal distribution. We can do this by using the formula z = (x - μ) / σ, where x is a random variable from the original normal distribution and z is its corresponding standard normal variate.
In this case, we want to find the value of z such that the cumulative probability to the left of z is 0.90. Using a standard normal distribution table or a calculator, we can find that the z-score corresponding to a cumulative probability of 0.90 is approximately 1.28.
Now, we can use the z-score formula to find the value of x that corresponds to a z-score of 1.28. Rearranging the formula, we have x = z * σ + μ.
x = 1.28 * 100 + 800 = 1408 hours
Therefore, the sum of the lifetimes of n light bulbs should be at least 1408 hours to guarantee a probability of 0.90 of achieving at least 10,000 hours of light.
To find the smallest value of n, we divide 10,000 by the mean lifetime of a light bulb:
n = 10,000 / 800 = 12.5
Since we cannot have a fractional number of light bulbs, we round up to the nearest whole number. Therefore, n should be at least 13.
Therefore, the smallest value of n such that the succession of light bulbs produces light for at least 10,000 hours with a probability of 0.90 is 13.