Assume that
z
-scores are normally distributed with a mean of 0 and a standard deviation of 1.
If
P
(
z
>
d
)
=
0.9933
\displaystyle {P}{\left({z}>{d}\right)}={0.9933}
P
(
z
>
d
)
=
0.9933
, find
d
.
d
=
\displaystyle {d}=
d
=
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