Let $X$ be a random variable with distribution $N(\mu,\sigma^2)$. Draw a random sample of size $n$.
A homework problem asks me to compute the following probability "for general $n$":
$$P(\bar{X}-\mu>n^{-1/3})$$
I know $\bar{X}-\mu\sim N(0,\sigma^2/n)$. So my probability is
$$P(z>n^{-1/3}/(\sigma/n^{1/2}))=P\left(z>\frac{n^{1/6}}{\sigma}\right)$$
I know that as $n\to\infty$ this probability tends to zero.
I could also say
- when $n=\sigma^6$, $P=.16$
- when $n=(2\sigma)^6$, $P=0.025$
- when $n=(3\sigma)^6$, $P=0.0015$
But as for calculating $P$ "for general $n$," I'm not sure what to do except write down the integral of the standard normal pdf:
$$P\left(z>\frac{n^{1/6}}{\sigma}\right)=(2\pi)^{-1/2}\int_{n^{1/6}/\sigma}^\infty e^{-x^2/2}\,dx$$
Any thoughts on what the problem might want me to do beyond this?