Answer
a. $0.4332$
b. $0.3944$
c. $0.0344$
d. $0.1029$
e. $0.2912$
f. $0.8284$
g. $0.0401$
h. $0.8997$
i. $0.0170$
j. $0.9131$
Work Step by Step
Use table E or other resources for probability and z-value relations,
we can find the area under the standard normal distribution for the following:
a. Between 0 and 1.50
$P(0\lt z\lt1.5)=P(z=1.5)-P(z=0)=0.9332-0.5000=0.4332$
b. Between 0 and -1.25
$P(-1/25\lt z\lt0)=P(z=0)-P(z=-1.25)=0.5000-0.1056=0.3944$
c. Between 1.56 and 1.96
$P(1.56\lt z\lt1.96)=P(z=1.96)-P(z=1.56)=0.9750-0.0406=0.0344$
d. Between -1.20 and -2.25
$P(-2.25\lt z\lt -1.20)=P(z=-1.20)-P(z=-2.25)=0.1151-0.0122=0.1029$
e. Between -0.06 and 0.73
$P(-0.06\lt z\lt 0.73)=P(z=0.73)-P(z=-0.06)=0.7673-0.4761=0.2912$
f. Between 1.10 and -1.80
$P(-1.80\lt z\lt1.10)=P(z=1.10)-P(z=-1.80)=0.8643-0.0359=0.8284$
g. To the right of z = 1.75
$P(z\gt1.75)=1-P(z=1.75)=1-0.9599=0.0401$
h. To the right of z=-1.28
$P(z\gt -1.28)=1-P(z=-1.28)=1-0.1003=0.8997$
i. To the left of z=-2.12
$P(z\lt -2.12)=P(z= -2.12)=0.0170$
j. To the left of z = 1.36
$P(z\lt 1.36)=P(z= 1.36)=0.9131$
