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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.

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  • $\begingroup$ You want the area from 1.8 to 14.2. You can look up the one sided tails for each and subtract $\endgroup$ – player100 Sep 29 '18 at 19:03
  • $\begingroup$ I would say: P(|X−8|<6.2) = P(-6.2 < X - 8 < 6.2) = P( 1.8 < X < 14.2) $\endgroup$ – georg Sep 29 '18 at 19:07
  • $\begingroup$ @georg did the same thing, and subtracted the two, answer isn't 0.785. $\endgroup$ – 4956 Sep 29 '18 at 19:12
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    $\begingroup$ @4956 - Excel: 0,892512303 - 0,107487697 = 0,785024606 $\endgroup$ – georg Sep 29 '18 at 19:40
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$Z=(X-8)/5$ 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.

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  • $\begingroup$ thank you. but where did 8 go? $\endgroup$ – 4956 Sep 29 '18 at 19:34
  • $\begingroup$ @4956 $Z=(X-8)/5.$ It went into $Z.$ $\endgroup$ – spaceisdarkgreen Sep 29 '18 at 22:25
  • $\begingroup$ I'm sorry, I'm asking a lot of questions - but why have we taken a -6.2/5. Why the negative sign? $\endgroup$ – 4956 Sep 30 '18 at 8:01
  • $\begingroup$ @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. $\endgroup$ – spaceisdarkgreen Sep 30 '18 at 8:14
  • $\begingroup$ My first choice used the fact that the CDF evaluated at the negative value is the tail area, so I subtracted off twice the tail area, leaving the middle. $\endgroup$ – spaceisdarkgreen Sep 30 '18 at 8:16

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