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Let X be a normally distributed variable with unknown parameters μ and σ (sigma). If we know that P (X ≥ 75) = 0.7291 and P (X ≥ 83) = 0.7764. With the information given Is it possible to determine the values for μ and σ ?. It is possible these odds?, Personally I see no sense to this probabilities? because I believe that the resulting values ​​of these probabilities are simply not possible. Am I right?

P (X ≥ 75) = 0.7291 and P (X ≥ 83) = 0.7764
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A typo? It says "P (X ≥ 75) = 0.7291 and P(X ≥ 83) = 0.7764." Obviously if $X\ge 83$ then $X \ge 75$, so $\Pr(X\ge75)$ has to be at least a big as $\Pr(X\ge83)$. Did the two numbers, 0.7764 and 0.7764, get interchanged? –  Michael Hardy Nov 4 '11 at 4:14

2 Answers 2

up vote 3 down vote accepted

No, those probabilities are not possible. See calculations below:

Let $Y = (X-\mu)/\sigma$. Thus,

$P(X \ge 75) = P(Y \ge (75-\mu)/\sigma) = 1 - P(Y \le (75-\mu)/\sigma)$

Thus,

$P(Y \le (75-\mu)/\sigma) = 1-0.7291 = 0.2709$

Using WolframAlpha, we know that if the above is true it must be that

$(75-\mu)/\sigma = -0.61$

Similarly, for the other probability we get:

$(83-\mu)/\sigma = -0.76$

An inspection of the above two equations shows that the standard deviation $\sigma$ is negative which is impossible by definition.

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But if we had seen a larger number for $\Pr(X\ge75)$ than for $\Pr(X\ge83)$, then certainly it would have been possible to find $\mu$ and $\sigma$. –  Michael Hardy Nov 4 '11 at 4:15
    
@MichaelHardy Of course and my answer shows the mechanics of doing it. –  tards Nov 4 '11 at 4:17

You are right. These probabilities are impossible because no matter how $X$ is distributed, we must have $\mathbb{P}(X ≥ 75)\geq \mathbb{P}(X ≥ 83)$.

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