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.
$Z=(X-8)/5$ 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.