Copyright © 2012 Pearson Education, Inc. All rights reserved Chapter 9 Statistics

Copyright © 2012 Pearson Education, Inc. All rights reserved 9.1 Frequency Distributions; Measures of Central Tendency

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9 - 4 Figure 1 © 2012 Pearson Education, Inc.. All rights reserved.

9 - 5 Figure 2 © 2012 Pearson Education, Inc.. All rights reserved.

9 - 6 © 2012 Pearson Education, Inc.. All rights reserved.

9 - 7 Your Turn 1 Find the mean of the following data: 12, 17, 21, 25, 27, 38, 49. Solution: © 2012 Pearson Education, Inc.. All rights reserved.

9 - 8 © 2012 Pearson Education, Inc.. All rights reserved.

9 - 9 © 2012 Pearson Education, Inc.. All rights reserved.

© 2012 Pearson Education, Inc.. All rights reserved.

© 2012 Pearson Education, Inc.. All rights reserved.

Your Turn 2 Find the mean of the following grouped frequency. IntervalMidpoint, x Frequency f Product x f Total = 27Total = 536 © 2012 Pearson Education, Inc.. All rights reserved.

Your Turn 2 continued Solution : A column for the midpoint of each interval has been added. The numbers in this column are found by adding the endpoints of each interval and dividing by 2. For the interval 0–6, the midpoint is (0 + 6)/2 = 3. The numbers in the product column on the right are found by multiplying each frequency by its corresponding midpoint. Finally, we divide the total of the product column by the total of the frequency column to get © 2012 Pearson Education, Inc.. All rights reserved.

© 2012 Pearson Education, Inc.. All rights reserved.

Your Turn 3 and Example 6 (c) Your Turn: Find the median of the data 12, 17, 21, 25, 27, 38, 49. Solution: The median is the middle number; in this case, 25. (Note that the numbers are already arranged in numerical order.) In this list, three numbers are smaller than 25 and three are larger. 12, 17, 21, 25, 27, 38, (c) Find the median for each list of numbers 47, 59, 32, 81, 74, 153. Solution: First arrange the numbers in numerical order, from smallest to largest 32, 47, 59, 74, 81, 153. There are six numbers here; the median is the mean of the two middle numbers. © 2012 Pearson Education, Inc.. All rights reserved.

© 2012 Pearson Education, Inc.. All rights reserved.

Figure 3 © 2012 Pearson Education, Inc.. All rights reserved.

Copyright © 2012 Pearson Education, Inc. All rights reserved 9.2 Measures of Variation

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© 2012 Pearson Education, Inc.. All rights reserved.

© 2012 Pearson Education, Inc.. All rights reserved.

© 2012 Pearson Education, Inc.. All rights reserved.

Your Turn 1 © 2012 Pearson Education, Inc.. All rights reserved. Find the range, variance, and standard deviation for the list of numbers: 7, 11, 16, 17, 19, 35. Solution: The highest number here is 35; the lowest is 7. The range is the difference between these numbers, or 35 − 7 = 28. The mean of the numbers is Continued Number xSquare of the Number x Total = 2301

Your Turn 1 continued The standard deviation s is © 2012 Pearson Education, Inc.. All rights reserved.

Figure 4 © 2012 Pearson Education, Inc.. All rights reserved.

Figure 5 © 2012 Pearson Education, Inc.. All rights reserved.

© 2012 Pearson Education, Inc.. All rights reserved.

© 2012 Pearson Education, Inc.. All rights reserved.

Your Turn 2 Find the standard deviation for the grouped frequency distribution. Intervalxx2x2 ffx Total = 27Total = 12,528 © 2012 Pearson Education, Inc.. All rights reserved.

Your Turn 2 continued © 2012 Pearson Education, Inc.. All rights reserved.

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© 2012 Pearson Education, Inc.. All rights reserved.

Copyright © 2012 Pearson Education, Inc. All rights reserved 9.3 The Normal Distribution

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Figure 6 © 2012 Pearson Education, Inc.. All rights reserved.

Figure 7 © 2012 Pearson Education, Inc.. All rights reserved.

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Figure 9 © 2012 Pearson Education, Inc.. All rights reserved.

Figure 10 © 2012 Pearson Education, Inc.. All rights reserved.

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Figure 11 © 2012 Pearson Education, Inc.. All rights reserved.

Figure 12 © 2012 Pearson Education, Inc.. All rights reserved.

Figure 13 © 2012 Pearson Education, Inc.. All rights reserved.

Figure 14 © 2012 Pearson Education, Inc.. All rights reserved.

Figure 15 © 2012 Pearson Education, Inc.. All rights reserved.

Figure 16 © 2012 Pearson Education, Inc.. All rights reserved.

Your Turn 2 Find a value of z satisfying the following conditions. (a) 2.5% of the area is to the left of z. (b) 20.9% of the area is to the right of z. Solution: (a) Use the table backwards. Look in the body of the table for an area of , and find the corresponding value of z using the left column and the top column of the table. You should find that z = −1.96. (b) If 20.9% of the area is to the right, 79.1% is to the left. Find the value of z corresponding to an area of The closest value is z = © 2012 Pearson Education, Inc.. All rights reserved.

Your Turn 3 Find the z-score for x = 20 if a normal distribution has a mean 35 and standard deviation 20. Solution: © 2012 Pearson Education, Inc.. All rights reserved.

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© 2012 Pearson Education, Inc.. All rights reserved.

Figure 17 © 2012 Pearson Education, Inc.. All rights reserved.

Figure 18 © 2012 Pearson Education, Inc.. All rights reserved.

Figure 19 © 2012 Pearson Education, Inc.. All rights reserved.

Your Turn 4 Dixie Office Supplies finds that its sales force drives an average of 1200 miles per month per person, with a standard deviation of 150 miles. Assume that the number of miles driven by a salesperson is closely approximated by a normal distribution. Find the probability that a sales person drives between 1275 and 1425 miles. Solution: We will find the z-score for both x 1 = 1275 and x 2 =1425. Continued © 2012 Pearson Education, Inc.. All rights reserved.

Your Turn 4 continued From the table, z 1 = leads to an area of , while z 2 = corresponds to A total of − = or %, of the drivers travel between 1275 and 1425 miles per month. The probability that a driver travels between 1275 miles and 1425 miles per month is © 2012 Pearson Education, Inc.. All rights reserved.

Figure 20 © 2012 Pearson Education, Inc.. All rights reserved.

Copyright © 2012 Pearson Education, Inc. All rights reserved 9.4 Normal Approximation to the Binomial Distribution

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Your Turn 1 Suppose a die is rolled 12 times. Find the mean and standard deviation of the number of sixes rolled. Solution: Using n =12 and p = 1/ 6, the mean is © 2012 Pearson Education, Inc.. All rights reserved.

Figure 21 © 2012 Pearson Education, Inc.. All rights reserved.

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Figure 24 © 2012 Pearson Education, Inc.. All rights reserved.