Probability of Normal Distribution with Unknown Mean

Question

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.

Answer 1

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

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

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

Answer 2

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 $\sigma$.