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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 $X_2$

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    $\begingroup$ Can you not use $X\leq 120$ or $X\geq 120$? $\endgroup$ Mar 20, 2014 at 17:08
  • $\begingroup$ What do you mean? I am looking for a specific interval that 50% of the data values will lie in $\endgroup$
    – adam
    Mar 20, 2014 at 17:11
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    $\begingroup$ 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 $[120-d,120+d]$? $\endgroup$
    – fgp
    Mar 20, 2014 at 17:11
  • $\begingroup$ yes that is exactly what I am looking for $\endgroup$
    – adam
    Mar 20, 2014 at 17:12

2 Answers 2

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Requiring that $\Pr[L \le X \le U] = 0.5$ does not uniquely specify the interval $[L,U]$:

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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 $[114.604,125.396]$.

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Recall The Empirical Rule : 68% of your data falls within 1 standard deviation of the mean. Now if you want 50% of your data falls within some standard deviation of the mean. call this t. Solve for t. This means the total area of the graph between z = 0 and z = t is 0.25 which means P(0 < z < t) = 0.25. From this you can find t. I look up the table in Triola book and it gives t = 0.68. So (x - 120)/8 = 0.68 ==> x(2) = 125.44. Also solve for the other end value: (x - 120)/8 = -0.68 ==> x(1) = 114.56. So the interval is: [114.56, 125.44]

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