Question

Given that z is a standard normal random variable, find z for each situation. (Round your answers to two decimal places.)

(a)The area to the left of z is 0.1841.
(b)The area between −z and z is 0.9534.
(c)The area between −z and z is 0.2206.
(d)The area to the left of z is 0.9948.
(e)The area to the right of z is 0.6915.

Answer 1

The z-scores for different situations are calculated using the z-table.

Step-by-step explanation:

The given problem involves finding the z-scores for different situations using the z-table.

(a) To find the z-score when the area to the left of z is 0.1841, we locate this value in the z-table and find the corresponding z-score to be approximately -0.88.

(b) To find the z-score when the area between -z and z is 0.9534, we locate this value in the z-table and find the corresponding z-score to be approximately 1.96.

(c) To find the z-score when the area between -z and z is 0.2206, we locate this value in the z-table and find the corresponding z-score to be approximately 0.76.

(d) To find the z-score when the area to the left of z is 0.9948, we locate this value in the z-table and find the corresponding z-score to be approximately 2.59.

(e) To find the z-score when the area to the right of z is 0.6915, we subtract this value from 1 to get 0.3085. We then locate 0.3085 in the z-table and find the corresponding z-score to be approximately -0.51.