Approximation of Exponentially and Normally Distributed Probabilities

Question

PROBLEM

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 battery lasts over $11$ hours.
• A sample of $20$ batteries has a sample mean over $11$ hours.
• A sample of $200$ batteries has a sample mean over $11$ hours.

I am not sure how to solve these questions because I've only learned approximating with the normal distribution to the binomial. Any suggestions?

Answer 1

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 $\mathsf{Gamma}(20, 2).$