252 CHAPTER 6 Normal Probability Distributions subtracting 0.1587 from 1. The result is 0.8413. Even though Table A-2 is designed only for cumulative areas from the left, we can use it to find cumulative areas from the right, as shown in Figure 6-6. 0.1587 z 5 –1.00 1. Use z 5 21.00 in Table A-2 to find this area. 2. Because the total area is 1, this area is 1 2 0.1587 5 0.8413 FIGURE 6-6 Finding the Area to the Right of z = −1 Statdisk INTERPRETATION Because of the correspondence between probability and area, we conclude that the probability of randomly selecting someone with a bone density reading above -1 is 0.8413 (which is the area to the right of z = -1.00). We could also say that 84.13% of people have bone density levels above -1.00. YOUR TURN. Do Exercise 10 “Standard Normal Distribution.” Example 4 illustrates a way that Table A-2 can be used indirectly to find a cumulative area from the right. The following example illustrates another way that we can find an area indirectly by using Table A-2. Bone Density Test: Finding the Area Between Two Values EXAMPLE 5 A bone density test reading between -1.00 and -2.50 indicates that the subject has osteopenia, which is some bone loss. Find the probability that a randomly selected subject has a reading between -1.00 and -2.50. SOLUTION We are again dealing with normally distributed values having a mean of 0 and a standard deviation of 1. The values between -1.00 and -2.50 correspond to the shaded region in the third graph included in Figure 6-7. Table A-2 cannot be used to find that area directly, but we can use it to find the following: ■ The area to the left of z = -1.00 is 0.1587. ■ The area to the left of z = -2.50 is 0.0062. ■ The area between z = -2.50 and z = -1.00 (the shaded area at the far right in Figure 6-7) is the difference between the areas found in the preceding two steps:
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