Answer:
a. P(0 < z < 3.00) = 0.4987
b. P(0 < z < 1.00) = 0.3414
c. P(0 < z < 2.00) = 0.4773
d. P(0 < z < 0.79) = 0.2852
e. P(-3.00 < z < 0) = 0.4987
f. P(-1.00 < z < 0) = 0.3414
g. P(-1.58 < z < 0) = 0.4429
h. P(-0.79 < z < 0) = 0.2852
Explanation:
Find the area under the standard normal probability distribution between the following pairs of z-scores.
a. z=0 and z=3.00
From the standard normal distribution tables,
P(Z< 0) = 0.5 and P (Z< 3.00) = 0.9987
Thus;
P(0 < z < 3.00) = 0.9987 - 0.5
P(0 < z < 3.00) = 0.4987
b. b. z=0 and z=1.00
From the standard normal distribution tables,
P(Z< 0) = 0.5 and P (Z< 1.00) = 0.8414
Thus;
P(0 < z < 1.00) = 0.8414 - 0.5
P(0 < z < 1.00) = 0.3414
c. z=0 and z=2.00
From the standard normal distribution tables,
P(Z< 0) = 0.5 and P (Z< 2.00) = 0.9773
Thus;
P(0 < z < 2.00) = 0.9773 - 0.5
P(0 < z < 2.00) = 0.4773
d. z=0 and z=0.79
From the standard normal distribution tables,
P(Z< 0) = 0.5 and P (Z< 0.79) = 0.7852
Thus;
P(0 < z < 0.79) = 0.7852- 0.5
P(0 < z < 0.79) = 0.2852
e. z=−3.00 and z=0
From the standard normal distribution tables,
P(Z< -3.00) = 0.0014 and P(Z< 0) = 0.5
Thus;
P(-3.00 < z < 0 ) = 0.5 - 0.0013
P(-3.00 < z < 0) = 0.4987
f. z=−1.00 and z=0
From the standard normal distribution tables,
P(Z< -1.00) = 0.1587 and P(Z< 0) = 0.5
Thus;
P(-1.00 < z < 0 ) = 0.5 - 0.1586
P(-1.00 < z < 0) = 0.3414
g. z=−1.58 and z=0
From the standard normal distribution tables,
P(Z< -1.58) = 0.0571 and P(Z< 0) = 0.5
Thus;
P(-1.58 < z < 0 ) = 0.5 - 0.0571
P(-1.58 < z < 0) = 0.4429
h. z=−0.79 and z=0
From the standard normal distribution tables,
P(Z< -0.79) = 0.2148 and P(Z< 0) = 0.5
Thus;
P(-0.79 < z < 0 ) = 0.5 - 0.2148
P(-0.79 < z < 0) = 0.2852