I have a few exercises concerning Central Limit Theorem in which I stumble upon the same problem.
Here is one of them:
A random numbers generator generates numbers with exponential distribution with parameter $\lambda$. What should $\lambda$ be and how many numbers should we generate so that the arithmetic mean of those numbers is $1$ with accuracy $0,01$ and the probability of this event is at least $0,9.$
This is how I do it:
$X_i = k_i$ = the number generated in the $i$-th draw, $S_n = \sum_{i=1}^n X_i$
$EX_i = \frac{1}{\lambda}$ and $\sigma = \frac{1}{\lambda}$
Standarization $$Z_n = \frac{S_n - \frac{n}{\lambda}}{\sqrt{n} \frac{1}{\lambda}} = \frac{\lambda S_n - n}{\sqrt{n}}$$
We need to estimate the following:
$$P(|\frac{S_n}{n}-1|<0,01) = P(\sqrt{n}(-0,01 \lambda + \lambda -1) < Z_n < \sqrt{n}(0,01 \lambda + \lambda -1)) = \Phi(\sqrt{n}(1,01 \lambda -1)) - \Phi(\sqrt{n}(-0,99 \lambda +1)) = 0,9$$
And I do not know how to find $n$ or $\lambda$.
Could you explain to me what I am doing wrong?
Let $M_N$ be the RV representing the mean of samples of size $N$. This has mean $E(M_N)=1/\lambda$ and variance $\mbox{Var}(M_N)=(1/\lambda^2)/ N$.
As we want $E(M_N)=1/\lambda=1$ we need $\lambda=1$.
Now assume the required sample size is large so we can use the CLT to justify a normal approximation for the distribution of $M_N \sim N(1,1/N)$
Now the question is what is the smallest $N$ such that $$ \mbox{Pr}(|m_N-1|\le 0.01) \ge 0.9 $$ So we are asking for the smallest $N$ such that the absolute value of a Standard Normal RV is less than or equal $0.01\times \sqrt{N}$and noting that a two-sided critical value for a $90\%$ interval has $z_{crit}=\pm 1.645$. That is that $0.01 \sqrt N \ge \Phi^{-1}(0.95)=1.645$
Hence we require the smallest $N$ for which $\sqrt N\ge1.645/0.01$