4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
4
$\begingroup$

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 $\sigma$.

$\endgroup$
2
$\begingroup$

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

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

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

EDIT

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.

$$\mu=\frac{a+b}{2}=\frac{3.74+5.82}{2}$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.