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G. Measures of Dispersion and Asymmetry.

1. Range

Downing & Clark, problem 7 above (Use data to find IQR). Review solutions and terms on page 41 (36 in 3 rd ed.) of Downing &

Clark.

2. The Variance and Standard Deviation of Ungrouped Data.

Text exercises 3.1b, 3.2b, 3.6, 3.37, 3.24 [3.1b, 3.2b, 3.7, 3.37, 3.23] (3.1b, 3.2b, 3.7, 3.23, 3.33)

3. The Variance and Standard Deviation of Grouped Data.

Text exercises 3.28, 3.30 (3.68, 3.70) (work 3.30 in thousands), Downing & Clark pg 42 or 37, problems 6,7 (Find sample standard deviation – hint: run problem 6 in hundreds) (Note that you can use the Excel or Minitab techniques in the graded assignment to compute and sum the fx

and

2 fx

columns in problems 6 and 7. ), Problems G1, G2. Graded Assignment 1

4. Skewness and Kurtosis.

Find the standard deviation, coefficient of variation and measures of skewness in Text problem 3.1, 3.2. Problems G3A, G4 (See

251wrksht).

5. Review a. Grouped Data. b. Ungrouped Data.

Section 3 is in this document.

---------------------------------------------------------------------------------------------------------------------------.

Exercise 3.28 (Formerly 3.68): The following data is given. Find a) the mean and b) the variance.

Class Intervals f

0 - under 10

10 - under 20

20 - under 30

30 - under 40

10

20

40

20

40 - under 50

Total

10

100

Solution: Of course, it is unnecessary to do everything in the table below, but you should know how to do the problem using both computational and definitional formulas. class f x fx fx

2

 x

 x

0-10 10 5 50 250 -20

10-20 20 15 300 4500 -10

20-30 40 25 1000 25000 0

30-40 20 35 700 24500 10

40-50 10 45 450 20250 20

100 2500 74500 f

 x

 x

-200

-200

0

200

200

0

So a) s

2 x

 f

 n

 n fx

 fx n

2

1 n x

100

2500

100

2

,

25 fx

100

2500 ,

 fx and, using the computational formula,

74600

100

1

 

2

2 

74500

12000

99

,

121 .

21212 . f

 x

 x

2

4000

2000

0

2000

4000

12000

1

251solnG2 1/31/08 (Open this document in 'Page Layout' view!)

Or using the definitional formula s

2 s

121 .

21212

11 .

0096 and C

 f

 x

 x

2 n

1 s x

 11 .

0096

25

12000

99

0 .

4404 .

121 .

21212 . So

Exercise 3.30 (Formerly 3.70): The following sample data is given. f

M is frequency in March, f

A

in

April. Find a) the means and b) the standard deviations. c) Have these statistics changed substantially between the months?

Class Intervals f

M f

A

$0 - under $2000

$2000 - under $4000

$4000 - under $5000

$5000 - under $8000

$8000 - under $10000

6

13

17

10

4

10

14

13

10

0

$10000 - under $12000

Total

0

50

3

50

Solution: Note that the intervals are not even. Only computational formulas are used below. Numbers are in thousands. class f

M x

0- 2 6 1 f x f x

2 f x f x f x

2

M M

A

A A

6.0 6.00 10 1.0 10.0 10.00

2- 4 13 3

4- 5 17 4.5 76.5 344.25 13 4.5 58.5 263.25

5- 8 10 6.5 65.0 422.50 10 6.5 65.0 422.50

8-10 4 9

10-12 0 11

50

So a)

 f

 n x

M

 n f

M

50 x

,

222

50 f

M

.

5 x

39.0 117.00 14 3.0 42.0 126.00

36.0 324.00 0 9.0 0.0 0.00

0.0 0.00 3 11.0 33.0 363.00

222.5 1213.75 50

222

4 .

45

.

5 x

A

,

 f

M x

 n f

A

2 x

1213

208

50

.

75

.

5

,

4 .

17 f

A x

208.5 1184.75

208 .

5 and

 f

A x

2 

1184 .

75 . b ) Using the computational formula, s s

2

M

2

A

 f f

M

A n n x x

2

2

1

1 n x

2

M n x

2

A 

1213

11184

.

75

.

75

50

50

50

50

1

1

4 .

45

2

223 .

625

49

4 .

56378 . So s

M

4 .

56378

4 .

17

2

315 .

305

6 .

43480 . So s

A

4 .

56378

49

Though the mean has fallen somewhat (6%), there are more items in both the largest and smallest

2 .

1363 .

2 .

5367 categories, making the standard deviation larger (by 19%).

2

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Downing and Clark, pg. 37, Application 6: Find sample standard deviation of the data below– hint: run problem 6 in hundreds) . The given data is group number , (group), frequency: 1 , ( 0-1000), 12; 2 , (1000-

2000), 15; 3 , (2000-3000), 19; 4 , (3000-4000), 22; 5 , (4000-5000), 30; 6 , (5000-6000), 56; 7 , (6000-7000),

48; 8 , (7000-8000), 40; 9 , (8000-9000), 30; 10 , (9000-10000), 16; 11 , (10000-11000), 4; 12 , (11000-

12000), 2.

Solution: To use the computational method, construct the following table. Note that we have converted the data into hundreds so that the group (2000-3000) is treated as (20-30). This means that the hundreds, the fx

2

column is in

 

2 

ten thousands and the fx

3

column is in

 

3  fx column is in millions.

Row group f x fx fx

2 fx

3

1 0-1000 12 5 60 300 1500

2 1000-2000 15 15 225 3375 50625

3 2000-3000 19 25 475 11875 296875

4 3000-4000 22 35 770 26950 943250

5 4000-5000 30 45 1350 60750 2733750

6 5000-6000 56 55 3080 169400 9317000

7 6000-7000 48 65 3120 202800 13182000

8 7000-8000 40 75 3000 225000 16875000

9 8000-9000 30 85 2550 216750 18423750

10 9000-10000 16 95 1520 144400 13718000

11 10000-11000 4 105 420 44100 4630500

12 11000-12000 2 115 230 26450 30417500

294 16800 1132150

So

 f

 n

294 ,

 fx

16800 ,

 fx

2 

1132150 , x

 n fx

16800

294

57 .

142857 (in hundreds

– so actually 5714.2857) and, using the computational formula to find the sample variance , s

2 

 fx n

2

1 n x

2

1132150

294

294

57

1

.

142857

2

172150

293

587 .

54266 (in

 

2 

ten thousands). So s

587 .

54266

24 .

2393 (in hundreds – so actually 2423.93) and C

 s

 24 .

2393

57 .

142857

0 .

4242 . x

Downing and Clark, pg. 37, Application 7: Find sample standard deviation of the data given in the first three columns below.

Row group f x fx fx

2 fx

3

1 0-10 122 5 610 3050 15250

2 10-20 180 15 2700 40500 607500

3 20-30 256 25 6400 160000 4000000

4 30-40 350 35 12250 428750 15006250

5 40-50 311 45 13995 629775 28339876

6 50-60 278 55 15290 840950 46252248

7 60-70 250 65 16250 1056250 68656248

8 70-80 211 75 15825 1186875 89015624

9 80-90 180 85 15300 1300500 110542496

10 90-100 175 95 16625 1579375 150040624

11 100-110 143 105 15015 1576575 165540368

12 110-120 120 115 13800 1587000 182504992

13 120-130 106 125 13250 1656250 207031248

14 130-140 99 135 13365 1804275 243577120

15 140-150 97 145 14065 2039425 295716640

16 150-160 75 155 11625 1801875 279290624

Total 2953 196365 17691424 1886137088

So

 f

 n

2953 ,

 fx

196365 and

 fx

2 

17691424 . This means x

 n fx

196365

2953

66 .

496783 and, using the computational formula to find the sample variance ,

251solnG2 1/31/08 (Open this document in 'Page Layout' view!)

 fx

2  n x

2 n

1

2953

66 .

496783

2953

1

2 s

2 

17691424

4633783 .

2

1569 .

7098 . So

2952 s

1569 .

7098

39 .

6196 .

Problem G1: If the mean return for an industry is 10% with a standard deviation of 6%, out of 100 firms how many do you expect to have returns above 22%?

Solution: Use Chebyshef's Rule. For any number k greater than 1, at least 1

1 k

2

 of the points will fall within k standard deviations of the mean (i. e. in the interval

  k

) and no more than

1

of the points k

2 will fall beyond k standard deviations of the mean . Note that 22% is above the mean by a number of standard deviations. Since x

  

22

10

12 and

12

12

6

2 , k

2 .

If the proportion in the tails is no

1 more than k

2

, we can compute

1 k

2 above 22%. This would be 25 firms.

1

2

2

1

4

and conclude that at most one quarter of the points are

Problem G2: If the mean is 5 and the standard deviation is 2, find an interval that must contain at least the central two-thirds of the observations.

Solution: Again use Chebyshef's Rule. If we want a central interval with no less than 1/3 of the observations, then the tails must contain at most 1/3. So we can say that

1 k

2 that

 k

5

1 .

732

3

 

1 .

732 . We know that

5

3 .

464

 

5 and

 

2 , so that our interval is

, or 1.536 to 8.464.

1

3

. So k

2

  k

3 , which means

4

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