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 $\sigma\approx 10$, $\mu\approx 25$.)
First, the standard deviation is not the average distance to the mean, that is always zero. It is however, a value to measure how far the points are from the mean or not. Assuming the values are normally distributed, we know that ~68% of the values are between $\mu-\sigma$ and $\mu+\sigma$, for example.
Suppose we weigh potatoes with average weight 100 g and stadard deviation 5 g. What does hold for the average of the average weight of a group of 4 potatoes?
I hope you see that the average of the average weight is still 100 g. But what is the standard deviation of this average weight? That is where you use the formula
$$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{4}} =2.5$$
Feel free to ask if you still don't understand.
Proof that the average distance between the actual data and the mean is
$0$:
$$\frac{\sum^n_{i=1} (x_i-\mu)}{n} = \frac{(\sum^n_{i=1} x_i)-\mu n}{n} = \frac{\sum^n_{i=1} x_i}{n}-\mu = \mu - \mu = 0$$
Best Answer
In this particular case, it means that you draw $d$ times a $N(0,1)$ (real) random variable, and these random variables are independent.
It means that $\varepsilon=(\varepsilon_1,\dots,\varepsilon_d)$, the $\varepsilon_i$ are independent, and are normally distributed, with variance $1$ and mean $0$.
If you don't have the identity matrix but another symmetric positive $d\times d$ matrix $\Sigma_{ij}$, then you'd have $\Sigma_{ij}=Cov(\varepsilon_i,\varepsilon_j)$. You can also have a non-zero mean vector $\mu=(\mu_1,\dots,\mu_d)$, with $\mu_i=E(\varepsilon_i)$. In this case you'd write $\epsilon\sim N(\mu,\Sigma)$.