Let z denote a variable that has a standard normal distribution. determine the value z* to satisfy the conditions below. (round all answers to two decimal places.) (a) p(z < z*) = 0.0256 z* = (b) p(z < z*) = 0.0098 z* = (c) p(z < z*) = 0.0493 z* = (d) p(z > z*) = 0.02 z* = (e) p(z > z*) = 0.0098 z* = (f) p(z > z* or z < -z*) = 0.1974 z* =

a)

b)

c)

d)

e)

f)

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Solution to the problem

a)  

For this case we need to find a value on the standard normal distribution that accumulates 0.0256 of the area on the left tail and we can use the following excel code: "=NORM.INV(0.0256,0,1)" and we got:

b)

For this case we need to find a value on the standard normal distribution that accumulates 0.0098 of the area on the left tail and we can use the following excel code: "=NORM.INV(0.0098,0,1)" and we got:

c)

For this case we need to find a value on the standard normal distribution that accumulates 0.0493 of the area on the left tail and we can use the following excel code: "=NORM.INV(0.0493,0,1)" and we got:

d)

For this case we need to find a value on the standard normal distribution that accumulates 0.02 of the area on the right tail and we can use the following excel code: "=NORM.INV(1-0.02,0,1)" and we got:

e)  

For this case we need to find a value on the standard normal distribution that accumulates 0.0098 of the area on the right tail and we can use the following excel code: "=NORM.INV(1-0.0098,0,1)" and we got:

f)

For this case we need to find a value on the standard normal distribution that accumulates 0.1974/2=0.0987 of the area on the right tail and we can use the following excel code: "=NORM.INV(1-0.0987,0,1)" and we got:

the answer is a

step-by-step explanation: