Find the area under the standard normal probability distribution between the following pairs of z-scores. a. z=0 and z=3.00 e. z=-3.00 and z=0 b. z=0 and z=1.00 f. z=-1.00 and z=0 c. z=0 and z=2.00 g. z=negative 1.19 and z=0 d. z=0 and z=0.61 h. z=-0.61 and z=0

a) 0.49865

b) 0.34134

c) 0.47725

d) 0.22907

e) 0.49865

f) 0.34134

g) 0.38298

h) 0.22907

Step-by-step explanation:

* Lets explain how to solve the problem

a) P(0 < z < 3)

- From the normal distribution table of z

∵ P(0 < z < 3) = 0.99865 - 0.50000 = 0.49865

∴ P(0 < z < 3) = 0.49865

b) P(0 < z < 1)

- From the normal distribution table of z

∵ P(0 < z < 1) = 0.84134 - 0.50000 = 0.34134

∴ P(0 < z < 1) = 0.34134

c) P(0 < z < 2)

- From the normal distribution table of z

∵ P(0 < z < 2) = 0.97725 - 0.50000 = 0.47725

∴ P(0 < z < 2) = 0.47725

d) P(0 < z < 0.61)

- From the normal distribution table of z

∵ P(0 < z < 0.61) = 0.72907 - 0.50000 = 0.22907

∴ P(0 < z < 0.61) = 0.22907

e) P(-3 < z < 0)

- From the normal distribution table of z

∵ P(-3 < z < 0) = 0.50000 - 0.00135 = 0.49865

∴ P(-3 < z < 0) = 0.49865

f) P(-1 < z < 0)

- From the normal distribution table of z

∵ P(-1 < z < 0) = 0.50000 - 0.15866 = 0.34134

∴ P(-1 < z < 0) = 0.34134

g) P(-1.19 < z < 0)

- From the normal distribution table of z

∵ P(-1.19 < z < 0) = 0.50000 - 0.11702 = 0.38298

∴ P(-1.19 < z < 0) = 0.38298

h) P(-0.61 < z < 0)

- From the normal distribution table of z

∵ P(-0.61 < z < 0) = 0.50000 - 0.27093 = 0.22907

∴ P(-0.61 < z < 0) = 0.22907

the answer is the option b

the range is all real numbers

step-by-step explanation:

we know that

the absolute value parent function is equal to

see the attached figure to better understand the problem

so

a. it is v-shaped > is true

b. the range is all real numbers > is not true

because the range is the > [0,∞)

all real numbers greater than or equal to zero

c. it goes through the origin > is true

d. the domain is all real numbers > is true

step-by-step explanation:

1. perpendicular

2. parallel

3. perpendicular