Normal distribution involving $\Phi(z)$ and standard deviation The random variable X has normal distribution with mean $\mu$ and standard deviation $\sigma$. $\mathbb{P}(X>31)=0.2743$ and $\mathbb{P}(X<39)=0.9192$. Find $\mu$ and $\sigma$. 
 A: 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 $\sigma\approx 10$, $\mu\approx 25$.)
