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The lifetime of a brand of lamp is exponentially distributed with parameter $\lambda=0.01\;h^{-1}$. How many lamps are needed so that there's a 95% probability that a room is lit for 1,000 hours, if

a) when a lamp burns, the next one is used?

b) all lamps are turned on simultaneously?


a) The lifetimes $X_1,…,X_n$ are independent, and I know that the sum of i. i. d. exponential variables has a gamma distribution.

$$X_1+...+X_n\triangleq Y\overset\perp\sim \mathrm{Gamma}(n,\lambda)\\ \mathbb P(Y\ge 1000) \ge 0.95\\ \mathbb P(2\lambda Y\gt 2000\lambda) \ge 0.95\\ 2\lambda Y\triangleq T\sim \mathrm{Gamma}(n,0.5)\equiv \mathrm{\chi^2}(2n)\\ \mathbb P(T\gt 20) \ge 0.95\quad,$$ and a chi-squared table (and the restriction that $2n$ is even) gives $n=16$. Is there a way to answer this without a table (or just a normal table)?


b) Since the previous item uses one lamp at a time, I assume one is enough to light the room. Therefore, I need the largest lifetime to be 1,000 h or more.

$$\max\{X_1,...,X_n\}\triangleq Z\\ \begin{align}\mathbb P(Z\ge1000) &= 1-\mathbb P(Z\le1000) =\\ &=1 - \mathbb P(X_1\le1000,...,X_n\le1000)\overset{\mathrm{i.i.d.}}=\\ &=1 - [\mathrm{cdf_{Exp(0.01)}}(1000)]^n = 1 - (1-e^{-0.01\cdot1000})^n \end{align}\\ 1 - (1-e^{-10})^n \ge 0.95\\ (1-e^{-10})^n \le 0.05\\ n\log(1-e^{-10}) \le \log 0.05\\ n\log[(1-e^{-10})^{-1}] \ge \log 20\\ 4.5401\cdot10^{-5}\cdot n\ge 2.9957\\ n\ge 65983.9\quad,$$ and so I need 65,984 lamps, a suspiciouly large number, even though I want one of them to last ten times its average lifetime. Is something wrong?

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    $\begingroup$ This appears to be standard bookwork. It should probably have the self-study tag (please edit). I believe you're already complying with what the self-study tag wiki asks, already though. $\endgroup$
    – Glen_b
    Jan 7 '14 at 23:50
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    $\begingroup$ The calculations are simple. In (a), the sum of $n$ iid variables with means $100$ and variances $10^4$ has mean $100n$ and standard deviation $100\sqrt{n}$. Using a Normal approximation, you would like $100n-1.65\times 100\sqrt{n}\ge 1000,$ whence $n\gt 16.$ In (b), noting that $(1-e^{-10})^{e^{10}}\approx e$ and recalling $0.05\approx e^{-3}$ (because a useful fact to memorize is $e^3\approx 20$) you immediately obtain $n\approx 3e^{10}$ = $3e\times e^9$ = $3e\times (e^3)^3\approx 3e\times{20}^3$ = $3e\times 8000=24000e$. From $e\approx 2.72$ it follows the answer is around $65000.$ $\endgroup$
    – whuber
    Jan 8 '14 at 2:15
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I believe your answers are correct. [Of course, practically speaking, if one lamp alone is bright enough to light a room, around 66 thousand of them will probably set it alight well before the 1000 hour mark - certainly if they're incandescent bulbs. Better hope they're led lamps I guess. Even if the lights were individually very cool, it would be a very unpleasant place to be.]

Is there a way to answer this without a table

Well, the issue is to evaluate an integral at a given d.f., or several, in order to find the smallest df that has the desired coverage; you could numerically evaluate integrals in any number of ways - various forms of numerical quadrature, expansions for the incomplete gamma function, or built in functions in a package - like R, which is how I checked your answer:

> pgamma(10,15:17,lower.tail=FALSE)
[1] 0.9165415 0.9512596 0.9729584

You might try to write some kind of expansion for the integral in terms of the d.f., $\nu$ and solve for $\nu$, and I imagine you could do this with success, but that sounds like hard work to me - you could solve it more quickly by searching over a variety of $\nu$ values.

(or just a normal table)?

Certainly - you can use a Wilson-Hilferty transformation. Specifically, the cube root of a gamma (or chi-square) random variable is approximately normal. In the case of the chi-square, it has the rather nice form that if $X\sim\chi^2_\nu$ then $(X/\nu)^\frac{1}{3}$ is approximately $N(1-\frac{2}{9\nu},\frac{2}{9\nu})$, and this approximation is pretty good, probably good enough to deal with this problem. You'd probably still want to double check you had the right value once you found it (occasionally it might need to go up or down by 1 if you want the smallest coverage that gives at least 0.95, rather than just getting quite close)

So the 5th percentile of a standard normal is at about -1.645. Hence the 5th percentile of the cube-root of a chi-square on its df is about $1-\frac{2}{9\nu}-1.645\sqrt{\frac{2}{9\nu}}$, so the 5th percentile of the cube root of a chi-square is about $\nu^\frac{1}{3}(1-\frac{2}{9\nu}-1.645\sqrt{\frac{2}{9\nu}})$, so you'd solve

$\nu^\frac{1}{3}(1-\frac{2}{9\nu}-1.645\sqrt{\frac{2}{9\nu}})\geq 20^\frac{1}{3}$

for $\nu$ (in the sense that you find the smallest even $\nu$ for which it's true). That's your $2n$.

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