Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Possible Duplicate:
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 $X$?

share|cite|improve this question

marked as duplicate by Sasha, JavaMan, t.b., Did, Rudy the Reindeer Dec 15 '11 at 19:06

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Duplicate: – tards Dec 15 '11 at 15:11
looks like homework. – NikBels Dec 15 '11 at 18:01
If you pay attention before you say the question is "duplicate", you would notice that in the previous question the probabilities were reversed and it was therefore impossible to calculate =) – franvergara66 Dec 16 '11 at 13:59

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.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.