Solution:
Given that,
a ) P( Z z ) = 0.1020
Using standard normal table
z = - 1.270
b ) P(z Z 0 ) = 0.1772
P ( Z 0 ) - P ( Z z ) = 0.1772
Using standard normal table
P ( Z z ) = 0.5000 + 0.1772
P ( Z z ) = 0.6772
z = 0.460
c ) P ( Z > z ) = 0.9929
= 1 - P ( Z < z ) = 0.9929
Using standard normal table
P ( Z < z ) = 1 - 0.9929
P ( Z < z ) = 0.0071
z = - 2.452
d ) P( 0.40 Z z ) = 0.3368
P ( Z z ) - P ( Z 0.40 ) = 0.3368
Using standard normal table
P ( Z z ) = 0.6554 + 0.3368
P ( Z z ) = 0.9922
z = 2.418
Find the following z values for the standard normal variable Z. (You may find it useful...
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