Assume that
z
-scores are normally distributed with a mean of 0 and a standard deviation of 1.
If
P
(
z
<
e
)
=
0.0118
\displaystyle {P}{\left({z}<{e}\right)}={0.0118}
P
(
z
<
e
)
=
0.0118
, find
e
.
e
=
\displaystyle {e}=
e
=
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