Question

Find the value of z such that 0.68260.6826 of the area lies between ?z?z and z.

Answer 1

Here we want to find z such that P( -z < Z < z) = 0.6826

Let P( -z < Z < z) = P( Z < z) - P(Z < -z) = [1 - P(Z > z) ] - P( Z < -z) = 1 - P( Z > z) - P(Z < -z)

= 1- P(Z < -z) -P(Z < -z) { because of symmetrty of normal distribution we P(Z > z) =P(Z < -z)

= 1 - 2* P(Z < -z)

So that   1 - 2* P(Z < -z) = 0.6826

this implies 2*  P(Z < -z) = 0.3174

This implies   P(Z < -z) =  0.3174 / 2 = 0.1587

Let's used standard normal table:

Look the following image

From the above image the -z value corresponding to the probability 0.1587 is -1

That is - z = -1 this implies that z = 1