260 CHAPTER 6 Normal Probability Distributions 46. About % of the area is between z = -2 and z = 2 (or within 2 standard deviations of the mean). 47. About % of the area is between z = -3 and z = 3 (or within 3 standard deviations of the mean). 48. About % of the area is between z = -3.5 and z = 3.5 (or within 3.5 standard deviations of the mean). 49. Significance For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of scores that are a. significantly high (or at least 2 standard deviations above the mean). b. significantly low (or at least 2 standard deviations below the mean). c. not significant (or less than 2 standard deviations away from the mean). 50. Distributions In a continuous uniform distribution, m = minimum + maximum 2 and s = range 212 a. Find the mean and standard deviation for the distribution of the waiting times represented in Figure 6-2, which accompanies Exercises 5–8. b. For a continuous uniform distribution with m = 0 and s = 1, the minimum is -23 and the maximum is 23. For this continuous uniform distribution, find the probability of randomly selecting a value between –1 and 1, and compare it to the value that would be obtained by incorrectly treating the distribution as a standard normal distribution. Does the distribution affect the results very much? 6-1 Beyond the Basics Key Concept Most of the preceding section dealt with the real-world application of bone density scores, which have a normal distribution with m = 0 and s = 1. However, it is rare to find other real applications of a standard normal distribution. We now extend the methods of the previous section so that we can work with any nonstandard normal distribution (with a mean different from 0 and>or a standard deviation different from 1). The key is a simple conversion (Formula 6-2) that allows us to “standardize” any normal distribution so that x values can be transformed to z scores; then the methods of the preceding section can be used. 6-2 Real Applications of Normal Distributions FORMULA 6-2 z = x - m s 1roundz scores to 2 decimal places2 Figure 6-11 illustrates the conversion from a nonstandard to a standard normal distribution. The area in any normal distribution bounded by some score x (as in Figure 6-11a) is the same as the area bounded by the corresponding z score in the standard normal distribution (as in Figure 6-11b).
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