I know that a distribution is a log normal with mean = 4744 and mode = 3777. I need to calculate the probability of X being lower than the mean.
In order to do so I need to calculate the CDF of my log normal distribution, and I know that I need to calculate the standard deviation to have the CDF. The problem is that I don't know how to get the standard deviation knowing the mean and the mode. I know what are the formulas of mean, standard deviation and mode, from these three I can substitute and reach the formula to get the standard deviation from the mode and the mean but I am not able to solve it.
I tried in python with simpy to solve the equation that I extrapolated by substitution from the formula of the Mode to get to the standard deviation. This is the code, but when I try to run it python gets stucked in an infinite running time:
x = symbols('x')
mode = 3777
mean = 4744
expr = mode -np.e**(mean - x)
sol = solve(expr)
Any ideas or suggestions to solve this problem? Maybe instead of trying to calculate the standard deviation having mean and mode, there is a way to calculate directly the CDF without std but using just mean and mode?
Let $X \sim \operatorname{Lognormal}(\mu, \sigma)$, so that $$Y = \log X \sim \operatorname{Normal}(\mu, \sigma).$$ Then the mean of $X$ is not $\mu$ but instead $$\operatorname{E}[X] = \operatorname{E}[e^Y] = M_Y(1) = e^{\mu + \sigma^2/2} \tag{1}$$ where $$M_Y(t) = \operatorname{E}[e^{tY}] = e^{\mu t + \sigma^2 t^2/2}$$ is the moment-generating function of $Y$.
Let $m_X$ be the mode of $X$. Then $m_X$ is a critical point of the lognormal density; i.e., it is a solution to the equation $$0 = f'_X(x) = \frac{d}{dx}\left[\frac{f_Y(\log x)}{x} \right] = \frac{f'_Y(\log x) - f_Y(\log x)}{x^2}$$ where $f_X$ is the lognormal density and $f_Y$ is the normal density. Since the normal density is $$f_Y(y) = \frac{1}{\sqrt{2\pi} \sigma} e^{-(y-\mu)^2/(2\sigma^2)}$$ and its derivative is $$f'_Y(y) = f_Y(y) \frac{d}{dy}\left[-\frac{(y-\mu)^2}{2\sigma^2}\right] = \left(-\frac{y-\mu}{\sigma^2}\right) f_Y(y),$$ it follows that $$0 = f_Y'(\log x) - f_Y(\log x) = -\left( \frac{\log x - \mu}{\sigma^2} + 1 \right) f_Y(\log x).$$ Since $f_Y$ is strictly positive, this implies the critical point must satisfy $\log x - \mu = -\sigma^2$ or $$m_X = e^{\mu - \sigma^2}. \tag{2}$$
Since both $(1)$ and $(2)$ are the antilogarithms of some function of the parameters $\mu$ and $\sigma$, it is convenient to let $$\alpha = \mu + \sigma^2/2, \quad \beta = \mu - \sigma^2$$ be the logarithms of the mean and mode. Then we can recover the parameters by solving the system for $\mu$ and $\sigma^2$: $$\mu = \frac{2\alpha + \beta}{3}, \quad \sigma^2 = \frac{2(\alpha - \beta)}{3}. \tag{3}$$ Then the probability that $X$ does not exceed its mean is $$\begin{align} \Pr[X \le e^\alpha] &= \Pr[e^Y \le e^\alpha] \\ &= \Pr[Y \le \alpha] \\ &= \Pr\left[\frac{Y - \mu}{\sigma} \le \frac{\alpha - \mu}{\sigma}\right] \\ &= \Pr\left[Z \le \frac{\sigma}{2}\right] \\ &= \Phi\left(\sqrt{\frac{\alpha - \beta}{6}}\right), \tag{4} \end{align}$$
where $Z$ is standard normal and $\Phi$ is the cumulative distribution function of the standard normal.
Let us apply this to your example. You have $\operatorname{E}[X] = 4744$ and $m_X = 3777$. Hence $$\alpha = \log(4744), \quad \beta = \log(3777),$$ and $$\sqrt{\frac{\alpha - \beta}{6}} \approx 0.194915.$$ Therefore the desired probability is $$\Phi(0.194915) \approx 0.57727.$$