Assume that z-scores are normally distributed with a mean of 0 and a standard deviation of 1. If P(-e<z<e)=0.2128P(e<z<e)=0.2128, what is ee?

1 Answer
May 4, 2017

e=0.27e=0.27

Explanation:

we have

Z~N(0,1)Z~N(0,1)

The Normal Distribution tables give probabilities for

P(z < x)P(z<x) so we need to adjust what we need to what the tables give us

P( -e < z< e )=2P( 0 < z < e)P(e<z<e)=2P(0<z<e)

by the symmetrical property of the Normal.

:. P (0 < z < e) =0.1064

now to use the tables we have to add

P(-oo < z < 0)=0.5

P( z < e )=0.6064

Using the tables in reverse we find

e=0.27