# Thread: [SOLVED] Normal Distrubtion Question

1. ## [SOLVED] Normal Distrubtion Question

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

Attempt:

$\displaystyle = 1 - 0.0289$
$\displaystyle = 0.9711$
$\displaystyle = -1.897$

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

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

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

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

Attempt:

$\displaystyle = 1 - 0.0289$
$\displaystyle = 0.9711$
$\displaystyle = -1.897$

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

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

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

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

So:

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

Now solve for $\displaystyle \mu$.

RonL

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

Attempt:

$\displaystyle = 1 - 0.0289$
$\displaystyle = 0.9711$
$\displaystyle = -1.897$

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

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

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