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

$e=0.27$

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?