2
$\begingroup$

Given a normal distribution with mean $\mu$ and unknown $\sigma$, what's the probability that, a sample of size $N$ will have mean in $[k-\epsilon, k+\epsilon]$ (as a function of $\sigma$)?

I can see that this is related to the problem of checking whether or not a coin is fair, but I can't see how to use that approach here.

$\endgroup$
2
  • $\begingroup$ @copper.hat Touche'. Hopefully now the answer is not zero :) $\endgroup$
    – kaharas
    Dec 23, 2014 at 18:08
  • $\begingroup$ I corrected [$k$ - $\epsilon$, $k$+$\epsilon$] so that it says $[k-\epsilon,k+\epsilon]$, and in the process I took the liberty of changing the vowel $u$ to the consonant $\mu$. ${}\qquad{}$ $\endgroup$ Dec 23, 2014 at 18:11

2 Answers 2

2
$\begingroup$

The distribution of the mean $\overline X$ of an i.i.d. sample from $N(\mu,\sigma^2)$ is $N(\mu, \sigma^2/n)$. So we have \begin{align} & \Pr(k-\varepsilon<\overline X < k+\varepsilon) = \Pr\left( \frac{k-\varepsilon-\mu}{\sigma/\sqrt{N}} < \frac{\overline X-\mu}{\sigma/\sqrt{N}}< \frac{k+\varepsilon-\mu}{\sigma/\sqrt{N}} \right) \\[10pt] = {} & \Pr\left( \frac{k-\varepsilon-\mu}{\sigma/\sqrt{N}} < Z < \frac{k+\varepsilon-\mu}{\sigma/\sqrt{N}} \right) = \Phi\left(\frac{k+\varepsilon-\mu}{\sigma/\sqrt{N}}\right) - \Phi\left(\frac{k-\varepsilon-\mu}{\sigma/\sqrt{N}}\right) \end{align} where $Z\sim N(0,1)$ and $\Phi$ is the c.d.f. of $N(0,1)$.

$\endgroup$
1
$\begingroup$

If $X$ is ${\cal N}(\mu_X, \sigma_X^2)$, $Y$ is ${\cal N}(\mu_Y, \sigma_Y^2)$ and $X,Y$ are independent then $X+Y$ is ${\cal N}(\mu_X+\mu_Y, \sigma_X^2+\sigma_Y^2)$.

If $X_1,...,X_N$ are iid. with distribution ${\cal N} (\mu, \sigma^2)$ then it is straightforward to obtain the distribution of $\sum_k X_k$. It is also straightforward to scale the distribution to get the distribution of ${1 \over N} \sum_k X_k$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .