I am still quite new to the whole idea of probability and statistics and am not sure how to do this question.

The random variable R, also normally distributed, has a standard deviation of 3.59 with unknown mean is unknown. Find the greatest possible value of $P(-3.74<R<5.82)$.

I tried to integrate the normal distribution function using Maclaurin's expansion but it just got very messy and I'm sure there is a better way of doing it.

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    $\begingroup$ Normal distribution have it mode in the same place that it mean and median, and it density function is symmetric respect this point. $\endgroup$ – Masacroso Feb 17 '16 at 13:43

Look at the shape of the normal probability density function. It's unimodal and symmetric. So the probability of coverage is maximized if the mean of the distribution, $\mu$, coincides with the middle of your interval, that is, $1.04$. This result is independent of the standard deviation, $\sigma$. However, the attained probability does depend on $\sigma$.


$$g(x)=\frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(x - \mu)^2}{2 \sigma^2}}$$

$$P(a<x<b)=\int_a^b{g(x)dx}=\int_a^b{\frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(x - \mu)^2}{2 \sigma^2}}}dx$$ Where $a=-3.74$, $b=5.82$ and $\sigma=3.59$

What you are interested in is finding $\mu$ that maximizes $P$ $$\frac{\partial P(a<x<b)}{\partial \mu} =0$$


As Luis Mendo mentioned, since the Normal distribution is centralized and symmetric, the maximum would be in the middle of the interval regardless of the standard deviation.


  • $\begingroup$ You mean find $\mu$ right? and the derivative should also be w.r.t. $\mu$, we know the standard deviation, but not the mean. $\endgroup$ – EHH Feb 17 '16 at 11:55
  • $\begingroup$ Right, fixed it... $\endgroup$ – Uri Goren Feb 17 '16 at 11:57
  • $\begingroup$ I thin you may be missing a $-$ so $\mu=\frac{a+b}{2}=\frac{-3.74+5.82}{2}$ $\endgroup$ – Henry Feb 17 '16 at 16:17

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