μ and σ calculations for a random variable with a normal distribution 
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Calculating mu and sigma (μ and σ) of a normal random variable 

If I have a random variable X with parameters μ and σ unknown. It is known that $P (X \ge 75) = 0.7764$ and $P (X \ge 83) =  0.7291$ . With this information Is it possible to determine the μ and σ values ​​for $X$?
 A: The title says that $X$ has normal distribution, but the text does not.  If we do not know the distribution of $X$, we cannot find the mean $\mu$ and the standard deviation $\sigma$ from the given data.  If we know that $X$ has normal distribution, we can. So from now on we assume that $X$ has normal distribution.
Recall that if $X$ has normal distribution with mean $\mu$ and standard deviation $\sigma$, then $\frac{X-\mu}{\sigma}$ has standard normal distribution. Thus 
$$P(X \le \mu +a\sigma)=P\left(\frac{X-\mu}{\sigma}\le a\right) =P(Z\le a),\qquad\qquad(\ast)$$
where $Z$ has standard normal distribution.
For example, let $a=-1.5$.  Informally, if $X \le \mu-1.5\sigma$, we say that $X$ is at least $1.5$ standard deviation units down from the mean. So by $(\ast)$, we see that the probability that $X$ is at least $1.5$ standard deviation units down from the mean is $P(Z \le -1.5)$.  Similar considerations apply to $P(X \ge \mu-1.5\sigma)$.  The probability that $X$ is no more than $1.5$ standard deviation units down from the mean is $P(Z \ge -1.5)$. By the way, this is about  $0.9332$.   
Finally, we look at the particular question. First, a quick look at the geometry.  We have $P(X \ge 75)=0.7764$.  Since the normal is symmetric, we can see that $75$ is below the mean. Similarly, $83$ is below the mean. Since $0.7764$ is less than $0.9332$, we can see that $75$ is less than $1.5$ standard deviation units below the mean. Now it is time to calculate.
If you look in a table of the standard normal distribution, you will find that $0.7764$ is (approximately) the probability that a standard normal $Z$ is $\le 0.76$, and $0.7291$ is the probability that a standard normal is $\le 0.61$.  The same sort of information is also available through various programs, such as spreadsheets, statistics packages, and general purpose mathematics packages such as Maple or WolframAlpha.
By symmetry, this means that $P(Z \ge -0.76)=0.7764$, and $P(Z \ge -0.61)=0.7291$.
But our normal $X$ has probability $0.7764$ of being $\ge 75$.  That means that $75$ is $0.76$ standard deviation units down from the mean.  In symbols,
$$\mu-0.76\sigma=75.$$
Similarly, we obtain
$$\mu-0.61\sigma=83.$$
Now we have two linear equations in two unknowns, and we can readily solve them for $\mu$ and $\sigma$.  A natural start is to observe that $(\mu-0.61\sigma)-(\mu-0.76\sigma)=83-75$, so $\sigma=8/(0.15)$. Note that our calculation has adequate accuracy, but is not exact, since the table entries are not exact.
