Question

Determine the area under the standard normal curve that lies between:

a. z= -.31 and z=1.61

b. z=-2.01 and z=.26

c. z=-1.20 and z=2.11

d. z=1.92 and z=2.43

Find the value of z_a

a. z_.56=

b. Z_.23=

c. Z_.81=

d. Z_.06=

Answer 1

Solution,

1) Using standard normal table,  

a) P( -0.31 < Z < 1.61)

= P( Z < 1.61) - P( Z < -0.31)

= 0.9463 - 0.3783

= 0.5680

b) P( -2.01 < Z < 0.26)

= P( Z < 0.26) - P( Z < -2.01)

= 0.6026 - 0.0222

= 0.5804

c) P( -1.20 < Z < 2.11)

= P( Z < 2.11) - P( Z < -1.20)

= 0.9826 - 0.1151

= 0.8675

d) P( 1.92 < Z < 2.43)

= P( Z < 2.43) - P( Z < 1.92)

= 0.9925 - 0.9726

= 0.0199

2) Using standard normal table,

a) P(Z > z) = 0.56

= 1 - P(Z < z) = 0.56  

= P(Z < z) = 1 - 0.56

= P(Z < z ) = 0.44

= P(Z < -0.15 ) = 0.44  

z = -0.15

b) P(Z > z) = 0.23

= 1 - P(Z < z) = 0.23

= P(Z < z) = 1 - 0.23

= P(Z < z ) = 0.77

= P(Z < 0.74 ) = 0.77

z = 0.74

c) P(Z > z) = 0.81

= 1 - P(Z < z) = 0.81  

= P(Z < z) = 1 - 0.81

= P(Z < z ) = 0.19

= P(Z < -0.88 ) = 0.19  

z = -0.88

d) P(Z > z) = 0.06

= 1 - P(Z < z) = 0.06

= P(Z < z) = 1 - 0.06

= P(Z < z ) = 0.94

= P(Z < 1.55 ) = 0.94  

z = 1.55