The actual required CPU-time of a workstation session is assumed (due to a long-term study) as a random variable with unknown expected value $\mu$ and known variance $\sigma^2=6.25[s^2]$. How many independent measurement of CPU-time are needed such that with probability of at least $0.9$, you have that the difference $|\bar{X}-\mu|$ is less than $0.1$?
This is task from old exam with different teacher. I ask my teacher and he say he don't show this example in class but give me two method for solve it: Central limit theorem and Chebyshev's inequality. This already help me much and I try solve like this:
I use central limit theorem:
We have that $$\sqrt{n} (\bar{X}-\mu) \text{ for } n \rightarrow \infty \text { is approximately standard normally distributed}$$
Now assume we have very big $n$ such that approximation mistake is very small so we don't need to keep it in mind. Then we can do
$$P(|\bar{X}-\mu|) < \varepsilon) = P(\sqrt{n}|\bar{X}-\mu|<\sqrt{n} \varepsilon)=P(-\sqrt{n}\varepsilon< \sqrt{n}(\bar{X}-\mu)< \sqrt{n} \varepsilon) \\ \approx \Phi(\sqrt{n} \varepsilon)- \Phi(-\sqrt{n}\varepsilon) = 2\Phi(\sqrt{n}\varepsilon)-1$$
Now insert $\varepsilon= 0.1$ in inequality $P(|\bar{X}-\mu|<\varepsilon)\geq 0.9$ then we have
$$2\Phi(0.1 \cdot \sqrt{n})-1 \geq 0.9 \Leftrightarrow \Phi(0.1 \cdot \sqrt{n}) \geq 0.95$$
And from this step I don't know how finish it :/
Can you tell me if this is correct till here and how to finish it? I really think I almost got it but something is missing..
But actually I'm more interested to know how you would do this with Chebyshev? I tried it already but also not sure if correct (see my comment of karakfa's answer).
Chebyshev inequality won't help you here. With some overall assumptions you can use the normal approximation and work from the z-table.
The sample size you'll come up with should be
$$ n = (\frac{z_{\alpha/2} \sigma}{E})^2 $$
for 90% CI, $z_{\alpha/2} \approx 1.65$, and $E$ is given as $0.1$, $\sigma^2=6.25$
$ n \approx 1700$