0
$\begingroup$

My question is:

If a random variable has a normal distribution, what are the possibilities it will take on a value within one standard deviation of the mean?

How do you approach this? I don't care about the solution, but, rather, how to get to it.

1 Answers 1

0

Hint: You are asking for $\mathrm P(\mu-\sigma\leqslant X\leqslant\mu+\sigma)$, assuming that the distribution of $X$ is normal $N(\mu,\sigma^2)$. But then $X=\mu+\sigma X_0$ where the distribution of $X_0$ is standard normal $N(0,1)$. Hence...

  • 0
    could you explain the part about X=μ+σX0 for X02011-11-11
  • 1
    @nubela If $X_0$ follows standard normal distribution $\mathcal{N}(0,1)$, then $X = \mu + \sigma X_0$ follows normal distribution with mean $\mu$ and variance $\sigma^2$, thus $\mathcal{N}(\mu, \sigma)$.2011-11-11
  • 0
    It is a standard fact that if $X_0$ is gaussian $N(0,1)$ then $aX_0+b$ is gaussian $N(b,a^2)$. Without knowing how you got introduced to gaussian distributions, it is difficult to answer usefully your question. What is a normal distribution to you? A specified density, perhaps? If you tell me this, I could complete my answer.2011-11-11
  • 1
    Thats all that I got. I have the solution though. The solution is really 2F(1) - 1. Actually I think I kinda got it now, thanks!2011-11-11