find a z0 such that Pz0

find a z0 such that P(-z0<z<z0)=.8262

Solution

Here z shows the standard normal distribution. It is symmetric. Area between -z0 and z0 is 0.8262. So total area outside to the -z0 and z0 is 1-0.8262=0.1738. That is area to the left of -z0 or area to the right of z0 is 0.1738/2 = 0.0869. From z-table z-score -1.36 has area 0.0869 to its left. So by symetry z-score 1.36 has area 0.0869 to its right. That is area between -z0 and z0 is 0.8262. So required z0 is 1.36.