Question

A consumer buys n light bulbs, each of which has a life time that has a mean of 800 hours and a standard deviation of 100 hours. A light bulb is replaced by another as soon as it burns out. Assuming independence of life times, find the smallest value of n so that the succession of light bulbs produces light bulbs for at least 10,000 hours with a probability of 0.9. Do you think it is necessary to know the probability distribution of the light bulbs? Explain.

Answer 1

Answer on Question #46752 – Math – Statistics and Probability

A consumer buys n light bulbs, each of which has a life time that has a mean of 800 hours and a standard deviation of 100 hours. A light bulb is replaced by another as soon as it burns out. Assuming independence of life times, find the smallest value of n so that the succession of light bulbs produces light bulbs for at least 10,000 hours with a probability of 0.9. Do you think it is necessary to know the probability distribution of the light bulbs? Explain.

Solution

Let's use The Central Limit Theorem. Suppose that $X_{1}, X_{2}, \ldots, X_{n}$ are independent and all having a distribution $W(\mu; \sigma^{2})$. Then if $n$ is sufficiently large:

$$S_{n} = X_{1} + X_{2} + \dots + X_{n} \approx N(n\mu; n\sigma^{2}).$$

Now let $X_{1}, X_{2}, \ldots, X_{n}$ be the lifetimes of the $n$ bulbs used in succession. Find $n$ such that

$$P(X_{1} + X_{2} + \dots + X_{n} > 10000) = 0.9 \text{ and } X_{1} + X_{2} + \dots + X_{n} \mapsto N(800n; 100^{2}n)$$ $$\begin{array}{l} P(X_{1} + X_{2} + \dots + X_{n} > 10000) = P\left(\frac{X_{1} + X_{2} + \dots + X_{n} - 800n}{100\sqrt{n}} > \frac{100 - 8n}{\sqrt{n}}\right) = P\left(Z > \frac{100 - 8n}{\sqrt{n}}\right) \\ = 0.9. \end{array}$$

From table of standard normal distribution, $\frac{100 - 8n}{\sqrt{n}} = -1.28$, and $n = 14$ by solving a quadratic equation.

It is not necessary to know the probability distribution of the light bulbs if we can use assumption that $n$ is sufficiently large. Then by The Central Limit Theorem we can work with normal distribution:

$$X_{1} + X_{2} + \dots + X_{n} \mapsto N(800n; 100^{2}n).$$

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