1. [SOLVED] Normal Distrubtion Question

Question:
$X$ has a normal distribution, and $P( X > 73.05 = 0.0289 )$. Give that the variance of the distribution is $18$, find the mean.

Attempt:

$= 1 - 0.0289$
$= 0.9711$
$= -1.897$

$\sigma^2 = 18$ , $\sigma = 3\sqrt2$

$P\left( \frac{X - \mu}{\sigma} > 73.05 \right) = -1.897$

Don't know how to find the mean...

2. Originally Posted by looi76
Question:
$X$ has a normal distribution, and $P( X > 73.05 = 0.0289 )$. Give that the variance of the distribution is $18$, find the mean.

Attempt:

$= 1 - 0.0289$
$= 0.9711$
$= -1.897$

$\sigma^2 = 18$ , $\sigma = 3\sqrt2$

$P\left( \frac{X - \mu}{\sigma} > 73.05 \right) = -1.897$

Don't know how to find the mean...

The z-score corresponding to a cumulative standard normal probability of $0.9711$ is $1.897$.

So:

$
\frac{73.05-\mu}{\sqrt{18}}=1.897
$

Now solve for $\mu$.

RonL

3. Originally Posted by looi76
Question:
$X$ has a normal distribution, and $P( X > 73.05 = 0.0289 )$. Give that the variance of the distribution is $18$, find the mean.

Attempt:

$= 1 - 0.0289$
$= 0.9711$
$= -1.897$

$\sigma^2 = 18$ , $\sigma = 3\sqrt2$

$P\left( \frac{X - \mu}{\sigma} > 73.05 \right) = -1.897$

Don't know how to find the mean...
A question of similar ilk that might be of interest: http://www.mathhelpforum.com/math-he...n-problem.html