Ytukyg

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?

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 $bar X sim mathsf{Gamma}(n,nlambda).$ Hence $E(bar X) = frac{n}{nlambda} = frac{1}{lambda} = 10,$
variance $V(bar X) = frac{1}{nlambda^2} = frac{sigma^2}{n} =
frac{100}{20} = 5,$
and $SD(bar X) = frac{1}{lambdasqrt{n}} = frac{sigma}{sqrt{n}} =
frac{10}{sqrt{20}} =
sqrt{5} = 2.2361.$

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