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

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Answer 1 · 2017-05-04T16:30:01

#e=0.27#

we have

#Z~N(0,1)#

The Normal Distribution tables give probabilities for

#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)#

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#