256 CHAPTER 6 Normal Probability Distributions INTERPRETATION For the population of bone density test scores, 2.5% of the scores are equal to or less than -1.96 and 2.5% of the scores are equal to or greater than 1.96. Another interpretation is that 95% of all bone density test scores are between -1.96 and 1.96. YOUR TURN. Do Exercise 39 “Finding Bone Density Scores.” Critical Values DEFINITION For the standard normal distribution, a critical value is a z score on the borderline separating those z scores that are significantly low or significantly high. Common critical values are z = -1.96 and z = 1.96, and they are obtained as shown in Example 7. In Example 7, values of z = -1.96 or lower are significantly low because only 2.5% of the population have scores at or below -1.96, and the values at or above z = 1.96 are significantly high because only 2.5% of the population have scores at or above 1.96. Only 5% of all bone density scores are either -1.96 or lower or 1.96 or higher. Critical values will become extremely important in subsequent chapters. The following notation is used for critical z values found by using the standard normal distribution. Notation The expression za denotes the z score with an area of a to its right. (a is the Greek letter alpha.) Finding the Critical Value zA EXAMPLE 8 Find the value of z0.025. (Let a = 0.025 in the expression za.) YOUR TURN. Do Exercise 41 “Critical Values.” SOLUTION The notation of z0.025 is used to represent the z score with an area of 0.025 to its right. Refer to Figure 6-10 and note that the value of z = 1.96 has an area of 0.025 to its right, so z0.025 = 1.96. Note that z0.025 corresponds to a cumulative left area of 0.975. CAUTION When finding a value of za for a particular value of a, note that a is the area to the right of za, but Table A-2 and some technologies give cumulative areas to the left of a given z score. To find the value of za, resolve that conflict by using the value of 1 - a. For example, to find z0.1, refer to the z score with an area of 0.9 to its left. Examples 3 through 7 in this section are based on the real application of the bone density test, with scores that are normally distributed with a mean of 0 and standard deviation of 1, so that these scores have a standard normal distribution. Apart from the bone density test scores, it is rare to find such convenient parameters, because typical normal distributions have means different from 0 and standard deviations different from 1. In the next section we present methods for working with such normal distributions. New Technology, New Data, New Insight Residents of New York City believed that taxi cabs became scarce around rush hour in the late afternoon. Their complaints could not be addressed, because there were no data to support that alleged shortage. However, GPS units were installed on cabs and officials could then track their locations. After analyzing the GPS data, it was found that 20% fewer cabs were in service between 4:00 PM and 5:00 PM than in the preceding hour. Subjective beliefs and anecdotal stories were now substantiated with objective data. Two factors were found to be responsible for the late afternoon cab shortage. First, the 12-hour shifts were scheduled to change at 5:00 PM so that drivers on both shifts would get an equal share at a rush hour. Second, rising rents in Manhattan forced many cab companies to house their cabs in Queens, so drivers had to start returning around 4:00 PM so that they could make it back in time and avoid fines for being late. In recent years, the shortage of cabs has been alleviated with the growth of companies such as Uber and Lyft.
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