Hint:
Write,
$$ \tag{1}\textstyle
P[\,X>31\,] =P\bigl[\,Z>{31-\mu\over\sigma}\,\bigr]=.2743\Rightarrow {31-\mu\over\sigma} = z_1
$$
$$\tag{2}\textstyle
P[\,X<39\,] =P\bigl[\,Z<{39-\mu\over\sigma}\,\bigr]=.9192\Rightarrow {39-\mu\over\sigma} =z_2 ,
$$
where $Z$ is the standard normal random variable.
You can find the two values $z_1$ and $z_2$ from a cdf table for the standard normal distribution. Then you'll have two equations in two unknowns. Solve those for $\mu$ and $\sigma$.
For example, to find $z_1$ and $z_2$, you can use the calculator here. It gives the value $z$ such that $P[Z<z]=a$, where you input $a$.
To use the calculator for the first equation first write
$$\textstyle P\bigl[\,Z<\underbrace{31-\mu\over\sigma}_{z_1}\,\bigr]=1-P\bigl[\,Z>{31-\mu\over\sigma}\,\bigr] =1-.2743=.7257.$$
You input $a=.7257$, and it returns $z_1\approx.59986$.
To use the calculator for the second equation,
$$\textstyle P\bigl[\,Z<\underbrace{39-\mu\over\sigma}_{z_2}\,\bigr]= .9192,$$
input $a=.9192$, the calculator returns $z_2\approx1.3997$.
So, you have to solve the system of equations:
$$
\eqalign{
{31-\mu\over\sigma}&=.59986\cr
{39-\mu\over\sigma}&=1.3997\cr
}
$$
(The solution is $\sigma\approx 10$, $\mu\approx 25$.)
Best Answer
It's hard to be sure what your question means. Suppose $X$ is normally distributed and has expected value $\mu$ and standard deviation $\sigma$. Let $Z = \dfrac {X-\mu} \sigma,$ so that $X = \mu + \sigma Z.$ Then $Z$ has expected value $0$ and standard deviation $1$ and is normally distributed. The one member of this family of distributions that has expected value $0$ and standard deviation $1$ is the one we call the "standard" one.
The distribution of $X$ is $$ \frac 1 {\sqrt{2\pi}} e^{-(1/2)\Big((x-\mu)/\sigma\Big)^2} \, \frac{dx} \sigma. $$ If you let $z= \dfrac{x-\mu}\sigma$, then we have $dz=\dfrac{dx}\sigma$ and the measure becomes $$ \frac 1 {\sqrt{2\pi}} e^{-(1/2) z^2} \, dz. $$ The expected value of that is $$ \int_{-\infty}^\infty z \left( \frac 1{\sqrt{2\pi}} e^{-(1/2)z^2} \, dz \right) $$ and this is clearly $0$ because an odd function is integrated over an interval symmetric about $0$. The variance takes some work. Let's call it $\tau^2:$ $$ \tau^2 = \operatorname{var}(Z) = \int_{-\infty}^\infty z^2 \left( \frac 1{\sqrt{2\pi}} e^{-(1/2)z^2} \, dz \right). $$ We have $$ \operatorname{E}(X) = \operatorname{E}(\mu+\sigma Z) = \mu + \sigma\operatorname{E}(Z) = \mu+\sigma\cdot0 = \mu, $$ and $$ \operatorname{var}(X) = \operatorname{var}(\mu+\sigma Z) = \sigma^2 \operatorname{var}(Z) = \sigma^2\tau^2. $$ If your question is how did we conclude that $\tau^2=1$, say so and maybe I'll post some more on that.