I have a stochastic variable with binomial distribution $T\sim B(n,\frac{1}{4})$
I need to estimate $n : P(0.30 < \frac{T}{n} < 0.70) > 0.95$
I have tried to resolve this problem with
$P(0.30 < \frac{T}{n} < 0.70) = P(T <0.70n) - P(T < 0.30n)$
but I don't know how can I proceed because I don't know the $\operatorname{Img}(T)$
No such $n$ exists. For sufficiently "large" $n$, we can use a normal approximation to the binomial and show that the function $$R(n) = \Phi\left(\frac{9\sqrt{n}}{5\sqrt{3}}\right) - \Phi\left(\frac{\sqrt{n}}{5\sqrt{3}}\right)$$ where $\Phi(z) = \Pr[Z \le z]$ is the standard normal CDF, has derivative $$r(n) = \frac{e^{-27n/50}(9-e^{8n/15})}{10\sqrt{6\pi n}}$$ with critical point at $n = \frac{15}{4} \log 3 \approx 4.1198$. This is a global maximum on $n > 0$, but $$R(4.1198) \approx 0.389891 < 0.95,$$ with very little hope of the exact binomial probability getting that large. But we make a table anyway:
$$\begin{array}{c|c|c|c} n & \lceil 0.3n \rceil & \lfloor 0.7n \rfloor & \Pr[0.3n < T < 0.7n] \\ \hline 1 & 1 & 0 & 0 \\ 2 & 1 & 1 & \frac{3}{8} \\ 3 & 1 & 2 & \frac{9}{16} \\ 4 & 2 & 2 & \frac{27}{128} \\ 5 & 2 & 3 & \frac{45}{128} \\ 6 & 2 & 4 & \frac{945}{2048} \\ 7 & 3 & 4 & \frac{945}{4096} \\ 8 & 3 & 5 & \frac{10395}{32768} \\ \end{array}$$
The maximum probability occurs at $n = 3$, but $9/16 < 0.95$.