Central limit theorem example

Question

I am not able to make progress with the following exercise.

Assignment:

The advertising board is lit by one 100W bulb. Using the central limit theorem determine the minimum number of bulbs needed for advertising board lighting for 20 000 hours with probability at least 0.9. Bulb life has the exponential distribution and average life of the 100W bulb is 600 hour.

My progress:

$X_ \text{ ..... } \text{total time of bulb life}$

$X_i \sim Exp(\lambda)$

$Ex= \frac{1}{\lambda} \implies \lambda = \frac{1}{Ex} $

$ Ex = 600$

$ \lambda = \frac{1}{600} $

$Dx = \frac{1}{\lambda^2} $

$Dx = \frac{1}{(\frac{1}{600})^2} = 360000$

$X \sim N(\mu,\sigma^2)$

$ \mu = \frac{1}{600}; \sigma^2=360000$

Now I am don't how to continue. I expected some equation like this: $ P(X > 20000) \geq 0.9$

Can you please give me any hints on how to continue?

Thanks

Answer 1

$ S = X_1+X_2+...+X_n \xrightarrow[]{d} N( n\mu, n\sigma^2), $ where $\mu=1/\lambda=EX_i,$ and $\sigma^2 = 1/\lambda^2=Var(X_i).$

Then, $$P(S>T_0) = 1-P(S<T_0) = 1-\Phi(\frac{T_0 - n\mu}{\sqrt{n}\sigma})>0.9, $$ where $T_0=20,000.$

Thus, $$\Phi(\frac{T_0 - n\mu}{\sqrt{n}\sigma})<0.1 \implies \frac{T_0 - n\mu}{\sqrt{n}\sigma} < q_{0.1}, $$ where $q_{0.1} = -1.281552$ is the tenth percentile of the standard normal (corresponding to the probability of standard normal equal to $0.1$.

Solving for $n$ in the quadratic equation $T_0 - n\mu = \sqrt{n}\sigma q_{0.1}$ yields $n=42$.

Answer 2

Adopting your notation, the central limit theorem intuitively says that for very large integers $n$, the distribution of $$ \frac1{\sqrt n} \sum_{i=1}^n X_i $$ is approximately $N(\mu,\sigma^2)$. I believe the exercise is asking you to say that the answer $n$ lightbulbs that you need is large enough to use the approximation: $$ \frac1{\sqrt n} \sum_{i=1}^n X_i \sim N(\mu,\sigma^2) \tag{1}$$ This will be useful, because you can write the event of $n$ light bulbs lasting $20,000$-hours in terms of $X_1,\ldots,X_n$.

Try to do this, and then rewrite it in terms of the scaled distribution in (1). You will instead see that you want $P(X\geq f(n)) \geq 0.9$, where $f(n)$ is an expression that depends on $n$ — not only $20,000$.