# Probability that a normal random variable is within one standard deviation of its mean

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.

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Hint: You are asking for $\mathrm P(\mu-\sigma\leqslant X\leqslant\mu+\sigma)$ for $X$ with gaussian distribution $N(\mu,\sigma^2)$. But then $X=\mu+\sigma X_0$ for $X_0$ with gaussian distribution $N(0,1)$. Hence...
@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 $\mathcal{N}(\mu, \sigma)$. – Sasha Nov 11 '11 at 21:14
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. – Did Nov 11 '11 at 21:17