Let Z be the standard normal random variable. Find z>0 so that the area between
ID: 3056231 • Letter: L
Question
Let Z be the standard normal random variable. Find z>0 so that the area between -z and +z is 0.99.
1.28
1.645
1.96
2.33
2.575
Let Z be the standard normal random variable. Find z so that the area to the left of z is 0.3156.
-0.80
-1.18
-0.48
1.28
None of the above
Let Z be the standard normal random variable. What is P(Z>-2.38)?
0.5239
0.7764
0.9162
0.9913
None of the above
Let Z be the standard normal random variable. What is P(Z>0.38)?
0.0087
0.0708
0.0838
0.1075
None of the above
Let Z be the standard normal random variable. What is P(1<Z<1.50)?
0.0640
0.1938
0.1579
0.0919
None of the above
Let Z be the standard normal random variable. What is P(Z<-0.38)?
0.0571
0.0918
0.1401
0.3520
None of the above
Let Z be the standard normal random variable. What is P(Z<1.38)?
0.5239
0.7764
0.9162
0.9913
None of the above
Explanation / Answer
a) P(-z < Z < z) = 0.99
or, P(Z < z) - P(Z < -z) = 0.99
or, P(Z < z) - (1 - P(Z < z)) = 0.99
or, 2*P(Z < z) - 1 = 0.99
or, P(Z < z) = 0.995
or, z = 2.575
Option-E) 2.575
b) P(Z < z) = 0.3156
or, z = -0.48
Option-C) -0.48
c) P(Z > -2.38) = 1 - P(Z < -2.38) = 1 - 0.0087 = 0.9913
Option-D) 0.9913
d) P(Z > 0.38) = 1 - P(Z < 0.38) = 1 - 0.6480 = 0.3520
Option-E) none of the above
e) P(1 < Z < 1.5) = P(Z < 1.5) - P(Z < 1) = 0.9332 - 0.8413 = 0.0919
Option-D) 0.0919
f) P(Z < -0.38) = 0.3520
Option-D) 0.3520
g) P(Z < 1.38) = 0.9162
Option-C) 0.9162
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