Let z denote a variable that has a standard normal distribution. Determine the value z* to satisfy the following conditions. (Round your answers to two decimal places.) a. P(z < z*) = 0.0244
z* =

b. P(z < z*) = 0.0098
z* =

c. P(z < z*) = 0.0496
z* =

d. P(z > z*) = 0.0204
z* =

e. P(z > z*) = 0.0098
z* =

(f) P(z > z* or z < -z*) = 0.201
z* =

a) z* = -1.97

b) z* = -2.33

c) z* = -1.65

d) z* = 2.04

e) z* = 2.33

f) z* = -1.25.

Step-by-step explanation:

Z-score:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

a. P(z < z*) = 0.0244

We have to look at the ztable, and find z which has a pvalue of 0.0244. So it is z* = -1.97

b. P(z < z*) = 0.0098

We have to look at the ztable, and find z which has a pvalue of 0.0098. So it is z* = -2.33

c. P(z < z*) = 0.0496

We have to look at the ztable, and find z which has a pvalue of 0.0496. So it is z* = -1.65

d. P(z > z*) = 0.0204

We have to look at the ztable, and find z which has a pvalue of 1 - 0.0204 = 0.9796. So z* = 2.04

e. P(z > z*) = 0.0098

We have to look at the ztable, and find z which has a pvalue of 1 - 0.0098 = 0.9902. So z* = 2.33

(f) P(z > z* or z < -z*) = 0.201

This is z which has a pvalue of 0.201/2 = 0.1055. So it is z* = -1.25.

560

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check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.