# Solving for the Random Variable X in a Z score

I know I am going to hate myself when I find out the method of doing this is trivial. Anyways the problem is this. Let X be normally distributed with mean of 120 and standard deviation of 8. Find the interval such that 50% of the data values fall within. I Know that you have to do $\phi(z_2) -\phi(z_1)$ where $z_2=\dfrac{X_1-120}{\sigma}$ and $z_1=\dfrac{X_2-120}{\sigma}$ I know that this will be equal to $0.5$ and I also know that $0.5$ corresponds to a $z$ score of $0$. So how do I manipulate this to solve for $X_1$ and $X_2$

• Can you not use $X\leq 120$ or $X\geq 120$? Mar 20, 2014 at 17:08
• What do you mean? I am looking for a specific interval that 50% of the data values will lie in
Mar 20, 2014 at 17:11
• There are many such intervals. Two simple ones are $[-\infty,120]$ and $[120,\infty]$. If you start at $a < 120$ instead of at $-\infty$, it'll have to go a bit past 120, to make up for the elements within $[-\infty,a)$ that you've now excluded. Are you maybe looking for an interval that lies symmetrically around 120, i.e. an interval of the form $[120-d,120+d]$?
– fgp
Mar 20, 2014 at 17:11
• yes that is exactly what I am looking for
Requiring that $\Pr[L \le X \le U] = 0.5$ does not uniquely specify the interval $[L,U]$:
Here, $X \sim {\rm Normal}(\mu = 120, \sigma = 8)$, and we can clearly see that there are infinitely many intervals that contain half of the total area under the density. If we assume you want the interval that is symmetric about the mean, i.e., $\Pr[120-\delta \le X \le 120+\delta] = 0.5$, then standardizing gives $$0.5 = \Pr\left[-\frac{\delta}{8} \le \frac{X - \mu}{\sigma} \le \frac{\delta}{8}\right],$$ and it follows that we wish to find $z^* = \delta/8$ such that $\Pr[Z \le z^*] = 0.25 + 0.5 = 0.75$, or $$z^* = 0.67449.$$ Therefore, $\delta = 5.39592$, and the desired symmetric interval about $\mu = 120$ is $[114.604,125.396]$.