# Thread: Normal distribution, finding a value question!

1. ## Normal distribution, finding a value question!

For a normal distribution, with a mean of $\displaystyle 10.5$ and a variance of $\displaystyle 16.7$, determine the value of $\displaystyle X_{0}$ so that $\displaystyle P(10.5 < x < X_{0}) = 0.4987$.

I am very lost on this question, any and all help is greatly appreciated!

Answer is supposedly $\displaystyle X_{0}=22.76$

2. ## Re: Normal distribution, finding a value question!

Apply the fact that $\displaystyle P(10.5 < x < X_{0}) = P(x < X_{0}) - P(x<10.5 )$

3. ## Re: Normal distribution, finding a value question!

Originally Posted by pickslides
Apply the fact that $\displaystyle P(10.5 < x < X_{0}) = P(x < X_{0}) - P(x<10.5 )$
$\displaystyle 0.4987 = P(x < X_{0}) - 0.5$

$\displaystyle 0.9987 = P(x < X_{0})$

Is that correct so far? Not sure what I do next

4. ## Re: Normal distribution, finding a value question!

Looks great

$\displaystyle 0.9987 = P(x < X_{0})$

$\displaystyle \displaystyle 0.9987 = P\left(z < \frac{X_{0}-10.5}{16.7}\right)$

Using the Normal Distribution Table

$\displaystyle \displaystyle 3.1 = \frac{X_{0}-10.5}{16.7}$