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A company manufactures a brand of light bulb with a lifetime in months that is normally distributed with mean 3 and variance 1. A consumer buys a number of these bulbs with the intention of replacing them successively as they burn out. The light bulbs have independent lifetimes.

What is the smallest number of bulbs to be purchased so that the succession of light bulbs for at least 40 months with probability at least 0.9772?

The chance that we only use one light bulb is $P(Z>\frac{40-3}{1})$. I don't know how to go about finding the probability that we use more than one light bulb. I don't know how to distribute the lifetime of each of the light bulb so that they add up more than 40. For example, if we use two light bulbs, we want the sum of their life to be more than 40 months. But we could have the first light bulb lives for 10 months and the second lives for more than 30 months or the first lives for more than 25 months and the second lives for 15 months. There are an infinite number of cases.

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2 Answers 2

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Let i.i.d. lifetimes $T_i \sim N(3,1)$. Then, $S_n= T_1+T_2+...+T_n \sim N(3n,n)$, where n is the variance of $S_n$.

We want $P(S_n > 40 ) > 0.9772$.

We have, $P(S_n>40)=P(Z=\frac{S_n-3n}{\sqrt{n}} > q) > 0.9772$ implies $q=2=\frac{ 40-3n}{\sqrt{n}}$. Thus, $n \ge 40$.

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  • $\begingroup$ Sorry, why $S_n$ ~ $N(3n,n)$? $\endgroup$
    – user614287
    May 5, 2019 at 1:48
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    $\begingroup$ Because $E(X+Y)=EX+EY$ and $S_n$ is a sum of $n$ rv-s with mean $3$. As for the variance $n$ it's the variance of the sum of n iid r-v-s of variance $1$. $\endgroup$
    – dnqxt
    May 5, 2019 at 1:52
  • $\begingroup$ thanks! umm also why would the joint distribution of the $T_i$ be normal? $\endgroup$
    – user614287
    May 5, 2019 at 1:54
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    $\begingroup$ It's not a joint dist but a distribution of a sum of normal rv-s which is in turn normal itself (it's one-dimensional). $\endgroup$
    – dnqxt
    May 5, 2019 at 1:55
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Let each individual i.i.d. lifetimes be defined as $T_i \sim N(3,1)$. Then, let the sum of i.i.d. lifetimes be denoted $S_n$. So that, $S_n= T_1+T_2+...+T_n \sim N(3n,n)$, where $n$ is the variance of $S_n$. (CLT)

We have that, \begin{align} \ P(S_n>40)> 0.9772 \end{align} So, \begin{align} \ 1- P(S_n<40) > 0.9772 \end{align} So, \begin{align} \ P(S_n<40) > 0.0228 \end{align} Then standardizing, \begin{align} \ P(\frac{S_n-3n}{\sqrt{n}}<\frac{40-3n}{\sqrt{n}}) > 0.0228 \end{align} Then, \begin{align} \ P(Z<\frac{40-3n}{\sqrt{n}}) > 0.0228 \end{align} Therefore, \begin{align} \ \phi(\frac{40-3n}{\sqrt{n}}) > 0.0228 \end{align} If we let, \begin{align} \phi(\frac{40-3n}{\sqrt{n}}) = 0.0228 \\ \phi(-2) = 0.0228 \end{align} So that, \begin{align} \ \frac{40-3n}{\sqrt{n}} = -2 \end{align} Now let, \begin{align} \ m &= \sqrt{n} \end{align} So, \begin{align} \ 3m^2 -2m -40 &= 0 \end{align}

The roots are 4 and –10/3. So $n$ is either 16 or 100/9.

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