Probability problem without the usual information A random variable $\xi$ has the normal distribution with expectation value of $25$. The probability of such random variable being in the interval $(10,15)$ is $0.07$.
The question now is how can one get the probability of the random variable being in another interval?
I wonder how we can do this without having the deviation in this case. In particular, I'm looking at the interval $(30,35)$.
 A: If $\xi$ is a normally-distributed random variable with expectation $25$
and variance $\sigma^2$, then
$$Y = \frac{\xi - 25}{\sigma}$$
is a standard normal random variable.
Moreover,
$$\begin{eqnarray}
P(10 < \xi < 15)
 &=& P\left(\frac{10 - 25}{\sigma} < Y < \frac{15 - 25}{\sigma} \right) \\
 &=& P\left(-\frac{15}{\sigma} < Y < -\frac{10}{\sigma} \right) \\
 &=& \Phi\left(-\frac{10}{\sigma}\right)
      - \Phi\left(-\frac{15}{\sigma}\right) \\
 &=& \Phi\left(\frac{15}{\sigma}\right)
      - \Phi\left(\frac{10}{\sigma}\right).
\end{eqnarray}$$
Note that it is a mere coincidence that the same numbers show up in the
numerators inside the $\Phi$ function as at the endpoints of the
original interval $(10,15)$; this is due to the fact that $10 + 15 = 25.$
So we need to find $\sigma$ such that
$$\Phi\left(\frac{15}{\sigma}\right)
      - \Phi\left(\frac{10}{\sigma}\right) = 0.07.$$
I think this requires numerical methods to solve, but you can get
a good approximation.
(It's possible that when you actually look at a table of
cumulative probability for the standard normal distribution,
the answer will "jump out" at you. I have not looked.)
Once you have found $\sigma$ to a reasonable accuracy,
you have a complete description of the normal random variable $\xi$
and you can find the probability that it falls in some other interval.
A: This is not an answer but it is too long for a comment.
Following David K's answer, for $0 \leq x\leq2.2$, there is a very simple approximation (proposed by Shah in 1985) which is good to two decimal places (largest error $\approx 0.005$) is given by $$\Phi(x)\approx 0.1x(4.4-x)$$ Applying it to $$\Phi\left(\frac{15}{\sigma}\right)
      - \Phi\left(\frac{10}{\sigma}\right) = 0.07$$ we then arrive to a quadratic equation $$7 \sigma^2-220 \sigma+1250=0$$ for which the solution to be retained is $\sigma=\frac{5}{7} \left(22+\sqrt{134}\right)\approx 23.9827$.
Considering the more general case of $$\Phi\left(\frac{a}{\sigma}\right)
      - \Phi\left(\frac{b}{\sigma}\right) = c$$ the approximate solution is obtained solving  $$50 c \sigma ^2-  22(a-b)\sigma+5 (a^2- b^2)=0$$
Edit
For the same range, Shah formula can be slightly improved using  $$\Phi(x)\approx x\left(\frac{131}{299}-\frac{20 }{201}x\right) $$ but this will not change the results significantly.
