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A company uses a portable high-intensity flashlight. Batteries and bulbs burn out quickly.

The lifetime of batteries has Exponential distribution with mean 10 hours. The bulbs have

lifetimes that are Normally distributed with mean 32 and standard deviation 5. Assume batteries

and bulbs are randomly sampled.

Find the probabilities for the following events. Where appropriate you may approximate

probabilities.

a) A battery lasts over 11 hours.

b) A sample of 20 batteries has a sample mean over 11 hours.

c) A sample of 200 batteries has a sample mean over 11 hours.

I'm not sure how to attack these questions because I've only learned approximating with the normal distribution to the binomial, any suggestions?

  • 1
    I got a 0 probability for a) so I think there must've been something wrong in the calculation there, I put in 1-pexp(11,10) into R. However, with b) and c) I'm not sure how to solve it with the sample numbers in there. – 2017-02-20
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    In R function `pexp` the second argument is the _rate_, not the mean. So use `1/10`. // Since you're using R, I'll mention that sums and averages of exponential samples have _gamma_ distributions; have you studied that? In R, the 3rd argument of `pgamma` is also _rate_ $\lambda = 1/\beta,$ where $\beta$ is scale parameter. – 2017-02-20
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    Wow I can't believe I forgot about that thanks, haven't taken stats in a little bit as you can probably tell. – 2017-02-20
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    Suggest you [_look at this_](http://math.stackexchange.com/questions/155296/distribution-of-the-sample-mean-of-a-exponential). Including the answer and the Wikipedia ref. – 2017-02-20
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    Sorry, there's an error in that link, see my Answer to (b). I have posted a correction note at the link. Please doublecheck. – 2017-02-20
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    BTW, using an exponential distribution to model battery life seems quite strange. In reliability theory, the no-memory property of the exponential dist'n is sometimes phrased as "Used is as good as new." For any battery I know of, used is _not_ as good as new. – 2017-02-20

1 Answers 1

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Because the link that I provided in a Comment has incorrect information, I am posting this Answer to (b).

You have $n = 20$ observations from $\mathsf{Exp}(\lambda = 1/10),$ and you seek $P(\bar X > 11).$

The individual observations have $\mu =E(X_i) = 1/\lambda = 10,$ variance $\sigma^2 = 1/\lambda^2 = 100,$ and SD $\sigma = 1/\lambda.$

The mean of $n = 20$ such observations has $\bar X \sim \mathsf{Gamma}(n,n\lambda).$ Hence $E(\bar X) = \frac{n}{n\lambda} = \frac{1}{\lambda} = 10,$ variance $V(\bar X) = \frac{1}{n\lambda^2} = \frac{\sigma^2}{n} = \frac{100}{20} = 5,$ and $SD(\bar X) = \frac{1}{\lambda\sqrt{n}} = \frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{20}} = \sqrt{5} = 2.2361.$

These results can be derived using moment generating functions and standard formulas for means and variances of random variables. Hence $P(\bar X > 11) = 0.306027,$ as computed in R statistical software below:

n = 20;  lam = 0.1;  1 - pgamma(11, n, n*lam)
## 0.306027

Of course, part (c) can be done similarly in R. Also, in (c) a normal approximation based on $n=200$ is reasonably accurate.

n = 200; lam = .1;  mu = 1/lam;  sg = 1/(lam*sqrt(n))
1 - pgamma(11, n, n*lam)
## 0.08180569                # exact
1 - pnorm(11,  mu, sg)
## 0.0786496                 # norm aprx

Note: Just as a 'reality check', when claiming an error elsewhere, I took a million samples of size $n = 20$ from $\mathsf{Exp}(rate = \lambda = 0.1)$ and found the corresponding million averages with the following results, agreeing with the theoretical results for (b) above, within the margin of sampling error.

 a = replicate(10^6, mean(rexp(20, .1)))
 mean(a > 11)     
 ## 0.306304       # aprx P(Avg > 11) = 0.306027
 mean(a);  sd(a)
 ## 9.99888        # aprx 10
 ## 2.236212       # aprx sqrt(5)

Below is a histogram of the one million sample means along with the density function of $\mathsf{Gamma}(20, 2).$

enter image description here