2. Excursions in Modern Mathematics, 7e: 16.3 - 2Copyright © 2010 Pearson Education, Inc. 16 Mathematics of Normal Distributions 16.1Approximately Normal Distributions of Data 16.2Normal Curves and Normal Distributions 16.3Standardizing Normal Data 16.4The 68-95-99.7 Rule 16.5Normal Curves as Models of Real- Life Data Sets 16.6Distribution of Random Events 16.7Statistical Inference

3. Excursions in Modern Mathematics, 7e: 16.3 - 3Copyright © 2010 Pearson Education, Inc. We have seen that normal curves don’t all look alike, but this is only a matter of perception. In fact, all normal distributions tell the same underlying story but use slightly different dialects to do it. One way to understand the story of any given normal distribution is to rephrase it in a simple common language–a language that uses the mean  and the standard deviation  as its only vocabulary. This process is called standardizing the data. Standardizing the Data

4. Excursions in Modern Mathematics, 7e: 16.3 - 4Copyright © 2010 Pearson Education, Inc. To standardize a data value x, we measure how far x has strayed from the mean  using the standard deviation  as the unit of measurement. A standardized data value is often referred to as a z-value. The best way to illustrate the process of standardizing normal data is by means of a few examples. z-value

5. Excursions in Modern Mathematics, 7e: 16.3 - 5Copyright © 2010 Pearson Education, Inc. Let’s consider a normally distributed data set with mean  = 45 ft and standard deviation  = 10 ft. We will standardize several data values, starting with a couple of easy cases. Example 16.4Standardizing Normal Data

6. Excursions in Modern Mathematics, 7e: 16.3 - 6Copyright © 2010 Pearson Education, Inc. ■ x 1 = 55 ft is a data point located 10 ft above (A in the figure) the mean  = 45 ft. Example 16.4Standardizing Normal Data

7. Excursions in Modern Mathematics, 7e: 16.3 - 7Copyright © 2010 Pearson Education, Inc. ■ Coincidentally, 10 ft happens to be exactly one standard deviation. The fact that x 1 = 55 ft is located one standard deviation above the mean can be rephrased by saying that the standardized value of x 1 = 55 is z 1 = 1. Example 16.4Standardizing Normal Data

8. Excursions in Modern Mathematics, 7e: 16.3 - 8Copyright © 2010 Pearson Education, Inc. ■ x 2 = 35 ft is a data point located 10 ft (i.e., one standard deviation) below the mean (B in the figure). This means that the standardized value of x 2 = 35 is z 2 = –1. Example 16.4Standardizing Normal Data

9. Excursions in Modern Mathematics, 7e: 16.3 - 9Copyright © 2010 Pearson Education, Inc. ■ x 3 = 50 ft is a data point that is 5 ft (i.e., half a standard deviation) above the mean (C in the figure). This means that the standardized value of x 3 = 50 is z 3 = 0.5. Example 16.4Standardizing Normal Data

10. Excursions in Modern Mathematics, 7e: 16.3 - 10Copyright © 2010 Pearson Education, Inc. ■ x 4 = 21.58 is... uh, this is a slightly more complicated case. How do we handle this one? First, we find the signed distance between the data value and the mean by taking their difference (x 4 –  ). In this case we get 21.58 ft – 45 ft = –23.42 ft. (Notice that for data values smaller than the mean this difference will be negative.) Example 16.4Standardizing Normal Data

11. Excursions in Modern Mathematics, 7e: 16.3 - 11Copyright © 2010 Pearson Education, Inc. ■ If we divide this difference by  = 10 ft, we get the standardized value z 4 = –2.342. This tells us the data point x 4 is –2.342 standard deviations from the mean  = 45 ft (D in the figure). Example 16.4Standardizing Normal Data

12. Excursions in Modern Mathematics, 7e: 16.3 - 12Copyright © 2010 Pearson Education, Inc. In Example 16.4 we were somewhat fortunate in that the standard deviation was  = 10, an especially easy number to work with. It helped us get our feet wet. What do we do in more realistic situations, when the mean and standard deviation may not be such nice round numbers? Other than the fact that we may need a calculator to do the arithmetic, the basic idea we used in Example 16.4 remains the same. Standardizing Values

13. Excursions in Modern Mathematics, 7e: 16.3 - 13Copyright © 2010 Pearson Education, Inc. In a normal distribution with mean  and standard deviation , the standardized value of a data point x is z = (x –  )/ . STANDARDIZING RULE

14. Excursions in Modern Mathematics, 7e: 16.3 - 14Copyright © 2010 Pearson Education, Inc. This time we will consider a normally distributed data set with mean  = 63.18 lb and standard deviation  = 13.27 lb. What is the standardized value of x = 91.54 lb? This looks nasty, but with a calculator, it’s a piece of cake: z = (x –  )/  = (91.54 – 63.18)/13.27 = 28.36/13.27 ≈ 2.14 Example 16.5Standardizing Normal Data: Part 2

15. Excursions in Modern Mathematics, 7e: 16.3 - 15Copyright © 2010 Pearson Education, Inc. One important point to note is that while the original data is given in pounds, there are no units given for the z-value. The units for the z-value are standard deviations, and this is implicit in the very fact that it is a z-value. Example 16.5Standardizing Normal Data: Part 2

16. Excursions in Modern Mathematics, 7e: 16.3 - 16Copyright © 2010 Pearson Education, Inc. The process of standardizing data can also be reversed, and given a z-value we can go back and find the corresponding x-value. All we have to do is take the formula z = (x –  )/  and solve for x in terms of z. When we do this we get the equivalent formula x =  +  z. Given , , and a value for z, this formula allows us to “unstandardize” z and find the original data value x. Finding the Value of a Data Point

17. Excursions in Modern Mathematics, 7e: 16.3 - 17Copyright © 2010 Pearson Education, Inc. Consider a normal distribution with mean  = 235.7 m and standard deviation  = 41.58 m. What is the data value x that corresponds to the standardized z-value z = –3.45? Example 16.6“Unstandardizing” a z-Value

18. Excursions in Modern Mathematics, 7e: 16.3 - 18Copyright © 2010 Pearson Education, Inc. We first compute the value of –3.45 standard deviations: –3.45  = –3.45  41.58 m = –143.451 m. The negative sign indicates that the data point is to be located below the mean. Thus, x = 235.7 m – 143.451 m = 92.249 m. Example 16.6“Unstandardizing” a z-Value