Standard Normal Distribution. In Exercises 17–36, assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph and find the probability of the given scores. If using technology instead of Table A-2, round answers to four decimal places.
Between − 3.90 and 2.00
Step 1 of 1
In finding the area under standard normal curve, you can use the statistical tables at the back of your textbook, or use your calculator.
In using the tables, remember that areas are reported as P(Z < a)
Therefore, you need to use the following relationships
P(Z > a) = 1 - P(Z < a)
P(a < Z < b) = P(Z < b) - P(Z < a)
Applying this principles, we get
P(-3.9 < Z < 2.00) = P(Z < 2) - P(Z < -3.9)
= 0.9772 - 0.0001