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For a normal curve, how much of the area lies within 1.5 standard deviations of the mean? I already know about the 68–95–99.7 rule, and see that it should be between 68% and 95%. I also know that it should be closer to 95%, so I estimate it to be around 80%. Can someone give me a hint on how to find the answer?

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    $\begingroup$ Look it up in a table or use a programming solution. For example this table gives $P(X \le 1.5) \approx 0.9332$ though you want $P(-1.5 \le X \le 1.5)$ $\endgroup$
    – Henry
    Jun 22, 2018 at 22:57

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You're close. It's about 87%. And see: probit.

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  • $\begingroup$ Combining henry's comment and rulnick's answer, I wanted to figure out how to get this answer using R, to compute $P(-1.5 < X < 1.5)$: > pnorm(1.5) - pnorm(-1.5) => [1] 0.8663856 > 1 - 2 * pnorm(-1.5) => [1] 0.8663856 This page has an interesting simplification, $1 - 2 * P(X < -1.5)$, as well. $\endgroup$
    – dossy
    Oct 18, 2021 at 15:03
  • $\begingroup$ Could you show your work or explain further, please? $\endgroup$
    – 2540625
    Sep 12, 2022 at 20:29
  • $\begingroup$ For a random variable with density $f$, mean $\mu$, standard deviation $\sigma$, the area under $f$ within $k$ standard deviations of the mean is $\int_{\mu-k\sigma}^{\mu+k\sigma} f(u)\,du$, or $F(\mu+k\sigma)-F(\mu-k\sigma)$, where $F$ is its distribution. For a normal r.v., $F(u)=\Phi((u-\mu)/\sigma)$, where $\Phi$ is the standard normal distribution, so the area of interest is $\Phi(k)-\Phi(-k)$, or $1-2\Phi(-k)$ since, due to $f$'s symmetry, $\Phi(k)=1-\Phi(-k)$. Here, $k=1.5$. $\endgroup$
    – rulnick
    Sep 14, 2022 at 5:59

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