# Normal distribution - absolute value solved

The random variable Y is normally distribution with mean = 8 and S.D = 5. Show that, P(|X−8|<6.2) = 0.785

What I did: (6.2 + 8)/5 = 2.84. The value of 2.84 is 0.9977 in the table. Normally, I would just double the value I have got if it was an absolute value question but here I'm stumped.

• You want the area from 1.8 to 14.2. You can look up the one sided tails for each and subtract – player100 Sep 29 '18 at 19:03
• I would say: P(|X−8|<6.2) = P(-6.2 < X - 8 < 6.2) = P( 1.8 < X < 14.2) – georg Sep 29 '18 at 19:07
• @georg did the same thing, and subtracted the two, answer isn't 0.785. – 4956 Sep 29 '18 at 19:12
• @4956 - Excel: 0,892512303 - 0,107487697 = 0,785024606 – georg Sep 29 '18 at 19:40

$$Z=(X-8)/5$$ is a standard normal, so $$P(|X-8|<6.2) = P\left(|Z|<\frac{6.2}{5}\right) = \Phi\left(\frac{6.2}{5}\right)-\Phi\left(-\frac{6.2}{5}\right)\approx0.785$$ where $$\Phi$$ is the standard normal CDF.
• @4956 $Z=(X-8)/5.$ It went into $Z.$ – spaceisdarkgreen Sep 29 '18 at 22:25
• @4956 $|Z|<\frac{6.2}{5}$ means the region $-\frac{6.2}{5}< Z < \frac{6.2}{5}.$ There are many ways to express the area under the bell curve over this region in terms of the CDF. I chose one. I have edited in a different choice that might be more apparent. – spaceisdarkgreen Sep 30 '18 at 8:14