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

Copyright © 2012 Pearson Education, Inc. All rights reserved

Chapter 9

Statistics


9.1

Frequency Distributions; Measures of Central

Tendency

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Figure 1

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

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Figure 2


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Your Turn 1Find the mean of the following data: 12, 17, 21, 25, 27, 38, 49.

Solution:


1 2The mean of the numbers , ,...... is

.

nn x x x

xx

n

12 17 21 25 27 38 497

x

1897

1 2Let 12, 17, and so on. Here 7, since there are 7 numbers.x x n

27

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Your Turn 2

Find the mean of the following grouped frequency.

Interval Midpoint, x

Frequency f

Product x f

0-6 3 2 6

7-13 10 4 40

14-20 17 7 119

21-27 24 10 240

28-34 31 3 93

35-41 38 1 38

Total = 27 Total = 536


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Your Turn 2 continuedSolution : 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


53627

xfx

n 19.85.

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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, 49.

6 (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.


59 74 133 1Median 66

2 2 2

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Figure 3



9.2

Measures of Variation

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9 - 21


9 - 22


9 - 23


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Your Turn 1


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

7 11 16 17 19 35 10517.5.

6 6

Number x Square of the Number x2

7 49

11 121

16 256

17 269

19 361

35 1225

Total = 2301

2 22Variance

1x nx

sn

22301 6(17.5)6 1

463.5

5 92.7.

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Your Turn 1 continued

The standard deviation s is


2 2

1x nx

sn

92.7s

9.628.

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Figure 4


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Figure 5


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Your Turn 2

Find the standard deviation for the grouped frequency distribution.

Interval x x2 f fx2

0-6 3 9 2 18

7-13 10 100 4 400

14-20 17 289 7 2023

21-27 24 576 10 5760

28-34 31 961 3 2883

35-41 38 1444 1 1444

Total = 27

Total =12,528


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Your Turn 2 continued


2 2

Standard deviation 1

fx nxs

n

2

Mean was calc12,528 27(19.85)

27 1ulated on slide 12.

8.52.

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9.3

The Normal Distribution

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Figure 6


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Figure 7


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Figure 8


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Figure 9


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Figure 10


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Figure 11


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Figure 12


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Figure 13


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Figure 14


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Figure 15


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Figure 16


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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 0.0025, 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 0.7910. The closest

value is z = 0.81.


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Your Turn 3Find the z-score for x = 20 if a normal distribution has a mean

35 and standard deviation 20.

Solution:


Here is the mean and is the standard deviation. x

z

20 3520

0.75

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Figure 17


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Figure 18


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Figure 19


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Your Turn 4Dixie 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 x1 = 1275 and x2=1425.

Continued


1

1

For 1275,

1275 1200

150 0.500

x

z

2

2

For 1425,

1425 1200

150 1.500

x

z

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Your Turn 4 continued From the table, z1 = 0.500 leads to an area of 0.6915, while

z2 = 1.500 corresponds to 0.9332.

A total of 0.9332 − 0.6915 = 0.2417 or 24.17 %, 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 0.2417.


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Figure 20



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


112 2.

6np

The standard deviation is

1 1(1 ) 12 1 1.291.

6 6np p

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Figure 21


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Figure 22


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Figure 23


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Figure 24


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

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