Consider two parameters $\sigma > 0 $ and $\mu \in \mathbb{R}$ for a random variable $X$ with the following density:
$$f(x) = \frac{1}{\sqrt{2\pi} \sigma} \exp(\frac{-1}{2 \sigma^2} x^2)$$
Let $0 < a < b$.
I have $5$ choices given and need to find out which of these are true.
As far as I know, (c) is false, because the left side represents the area under the curve while the right side does the opposite. I also think that (e) is true because probability is equal to area, but I don't know about the other ones. I think that only $e$ is true while all the other ones are false
(a) $P(-b \leq X \leq -a) = P(a \leq X \leq b)$
(b) $P(-b \leq X \leq -a) = P(-a \leq X \leq b)$
(c) $P(a < X < b) = P(-a \leq X \leq b)$
(d) $P(a < X < b) = P(-a \leq X \leq b)$
(e) $P(a < X < b) = P(a \leq X \leq b)$
What you have here is the density of a random variable $X$ that follows the Normal / Gaussian distribution. If there is really no $\mu$ in your function, then that just means that your distribution is centered. It's kind of weird to mention $\mu \in \Bbb{R}$ in the setup then, but I guess it's not prohibited.
Usually, $\mu$ is the parameter for the expected value (or the mean), then we would write $X \sim \mathcal{N}(\mu, \sigma^2)$, and $X$ would have density
$$f(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{(x-\mu)^2}{\sigma^2}\right).$$
So in your case, we actually have $\mu = 0$, or put differently $X \sim \mathcal{N}(0, \sigma^2)$. Here is a visualization:
What is important to note is that this density function is symmetric around the origin. This should help to answer your questions.
a) This is true by the above mentioned symmetry. To see this, draw some points $a,b,-a,-b$ and convince yourself that the area under $f(x)$ between $a$ and $b$ is the same as the area between $-b$ and $-a$.
b) This is not true. Consider $b = 2$, $a = 1$, then the area under $f(x)$ between $-2$ and $-1$ is clearly not the same as the area between $-1$ and $2$.
c) This is not true, it's essentially the same case as b)
d) I might be wrong but I think this one is identical to c)?
e) This is true. Here we are using the fact that if we have a continuous distribution, meaning a random variable with density, singletons have no mass. What this means is that $P(X=c) = 0$ for any $c \in \Bbb{R}$. Therefore it doesn't matter if we include the endpoints or not, and so $P(a<X<b) = P(a\leq X \leq b).$
I hope this helps!