6-1 The Standard Normal Distribution 253 z 5 21.00 0.1587 This area 0.1587 z 5 22.50 0.0062 minus this area 2 0.0062 22.50 21.00 0.1525 equals this area 5 0.1525 FIGURE 6-7 Finding the Area Between Two z Scores INTERPRETATION Using the correspondence between probability and area, we conclude that there is a probability of 0.1525 that a randomly selected subject has a bone density reading between -1.00 and -2.50. Another way to interpret this result is to state that 15.25% of people have osteopenia, with bone density readings between -1.00 and -2.50. YOUR TURN. Do Exercise 11 “Standard Normal Distribution.” Example 5 can be generalized as the following rule: The area corresponding to the region between two z scores can be found by finding the difference between the two areas found in Table A-2. Figure 6-8 illustrates this general rule. The shaded region B can be found by calculating the difference between two areas found from Table A-2. 0 z Left z Right A B Shaded area B 5 (areas A and B combined) — (area A) FIGURE 6-8 Finding the Area Between Two z Scores HINT Don’t try to memorize a rule or formula for this case. Focus on understanding by using a graph. Draw a graph, shade the desired area, and then get creative to think of a way to find the desired area by working with cumulative areas from the left. Probabilities such as those in the preceding examples can also be expressed with the following notation. Notation P1a 6 z 6 b2 denotes the probability that the z score is between a and b. P1z 7 a2 denotes the probability that the z score is greater than a. P1z 6 a2 denotes the probability that the z score is less than a. With this notation, P1-2.50 6 z 6 -1.002 = 0.1525, states in symbols that the probability of a z score falling between -2.50 and -1.00 is 0.1525 (as in Example 5).
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