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.

answer: 59

step-by-step explanation:

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)