section 3-2

Chapter 3Data Description

Section 3-2

Measures of Central Tendency


Section 3-2

Exercise #3

61, 11, 1, 3, 2, 30, 18, 3, 7

Find (a) the mean.

Find (b) the median.

Find (c) the mode.

Find (d) the midrange.

The data above are the number of burglaries reported for a specific year for nine western Pennsylvania universities.

X =

xn

=


136

9 = 15.1

Find (a) the mean.

61, 11, 1, 3, 2, 30, 18, 3, 7

Find (b) the median.


MD: 1, 2, 3, 3, 7, 11, 18, 30, 61

MD: 7

61, 11, 1, 3, 2, 30, 18, 3, 7

Find (c) the mode.

Mode =


3

61, 11, 1, 3, 2, 30, 18, 3, 7

Find (d) the midrange.


MR =

1 + 61

2 = 31

61, 11, 1, 3, 2, 30, 18, 3, 7

MD: 1, 2, 3, 3, 7, 11, 18, 30, 61

Which measure of average might be thebest in this case? Explain your answer.

X =

xn

= 1369

= 15.1

MD = 7

Mode = 3 MR = 31

61, 11, 1, 3, 2, 30, 18, 3, 7

Which measure of average might be thebest in this case? Explain your answer.

The median is probably the best measure of average because 61 is an extremely large data value and makes the meanartificially high.

61, 11, 1, 3, 2, 30, 18, 3, 7


Section 3-2

Exercise #5

Find (a) the mean, (b) the median, (c) the mode, and (d) the midrange.

A researcher claims that each year, there is an average of 300 victims of identity theft in major cities. Twelve wererandomly selected, and the number of victims of identity theft in each city is shown.

574 229 663 372 1,202 88

117 239 465 136 189 75

(a) X =

Xn


X = 4349, n = 12

X =

4349

12 362

574 229 663 372 1,202 88

117 239 465 136 189 75

(b) Arrange data in increasing order:


75 88 117 136 189 229 239 372 465 574 663 1202

median

574 229 663 372 1,202 88

117 239 465 136 189 75


MD =

229 + 239

2 = 234

574 229 663 372 1,202 88

117 239 465 136 189 75

(b) Arrange data in increasing order:

(c) No data are repeated.


No mode.

574 229 663 372 1,202 88

117 239 465 136 189 75

(d) Arrange data in increasing order.


75 88 117 136 189 229 239 372 465 574 663 1202

574 229 663 372 1,202 88

117 239 465 136 189 75


MR =

75 + 1202

2 = 638.5

574 229 663 372 1,202 88

117 239 465 136 189 75

(d) Arrange data in increasing order.


X = 362, MD = 234, no mode, MR = 638.5

Can you conclude that the researcher was correct?

574 229 663 372 1,202 88

117 239 465 136 189 75

It seems that the average number of identity thefts is higher than 300.

Section 3-2

Exercise #17


Eighty randomly selected light bulbs were tested to determine their lifetimes (in hours). This frequency distribution was obtained.

5107.5-118.5

1496.5-107.5

1885.5-96.5

2574.5-85.5

1263.5-74.5

652.5-63.5

FrequencyClass Boundaries Find the (a) mean and (b) modal class.

107.5 – 118.5

96.5 – 107.5

85.5 – 96.5

74.5 – 85.5

63.5 – 74.5

52.5 – 63.5

Boundaries f • Xm Xm f

8281269

348658

20002580

16381891

142814102

5655113

680780

• =

f XmX n

modal class: 74.5 – 85.5

= 680780 = 85.1

Eighty randomly selected light bulbs were tested to determine their lifetimes (in hours). This frequency distribution was obtained. Find the (a) mean and (b) modal class.

Section 3-2

Exercise #19


Find the (a) mean and (b) modal class.

The cost per load (in cents) of 35 laundry detergents tested by a consumer organization is shown.

262-68

055-61

148-54

641-47

534-40

1227-33

720-26

213-19

FrequencyClass Limits

62 – 68

55 – 61

48 – 54

41 – 47

34 – 40

27 – 33

20 – 26

13 – 19

Class

54.5 – 61.5

61.5 – 68.5

47.5 – 54.5

40.5 – 47.5

33.5 – 40.5

26.5 – 33.5

19.5 – 26.5

12.5 – 19.5

f • XmfXmBoundarieslimits

32216

161723

3601230

185537

264644

51151

0058

130265

35 1183

X =

f •Xmn

modal class: 26.5 – 33.5

= 33.8 = 118335

Find the (a) mean and (b) modal class.

The cost per load (in cents) of 35 laundry detergents tested by a consumer organization is shown.


Section 3-3

Exercise #7

Find the range.

The number of incidents where policies were needed for a sample of ten schools in Allegheny County is 7, 37, 3, 8, 48, 11, 6, 0, 10, 3. Assume the data represent samples.

range = 48 – 0 = 48

22

2 – /

= – 1

X X ns

n

Use the shortcut formula for the unbiased estimator to compute the variance and standard deviation.

X = 133 n = 10 X 2 = 4061


22

4061 – 133 / 10 =

10 – 1s 254.7

s 254.7 16

Use the shortcut formula for the unbiased estimator to compute the variance and standard deviation.


range = 48 s2 = 254.7 s = 16

The number of incidents where policies were needed for a sample of ten schools in Allegheny County is 7, 37, 3, 8, 48, 11, 6, 0, 10, 3.

Is the data consistent or does it vary? Explain your answer.

By any of these measures, it can be said that the data can vary.


Section 3-3

Exercise #21

2539-602

1475-539

0411-474

2347-410

0283-346

5219-282

0155-218

291-154

1327-90

fNumber

The data shows the number of murders in 25 selected cities.

Find the variance and standard deviation.

2

1

0

2

0

5

0

2

13

1141

506.5

0

757

0

1252.5

0

245

760.5

650,940.5

256,542.25

0

286,524.5

0

313,751.25

0

30,012.5

44,489.25

570.5

506.5

442.5

378.5

314.5

250.5

186.5

122.5

58.5

539-602

475-538

411-474

347-410

283-346

219-282

155-218

91-154

27-90

Class f • Xm2

f • Xm Xmf

f • Xm2 = 1,582,260.25 f • Xm = 4662.5



2

• 2• – 2 =

– 1

f Xf X

nsn

f • Xm2 = 1,582,260.25 f • Xm = 4662.5



24662.51 582 260 25 –

25= 24

, , .

= 29,696

s = 29,696 172.3

s2

Section 3-3

Exercise #33


The mean of a distribution is 20 and the standard deviation is 2. Answer each. Use Chebyshev’s theorem.

a. At least what percentage of the values will fall between 10 and 30?

b. At least what percentage of the values will fall between 12 and 28?

a. Subtract the mean from the larger value: 30 – 20 = 10

Divide by the standard deviation to get k: 10

2 = 5

b. Subtract the mean from the larger value: 28 – 20 = 8. Divide by the standard

deviation to get k: 8

2 = 4

1–

1

52 = 0.96 or 96%

0.9375 or 93.75% 1–

1

42 =


Section 3-3

Exercise #41

The average U.S. yearly per capita consumption of citrus fruits is 26.8 pounds. Suppose that the distribution of fruit amounts consumed is bell-shaped with a standard deviation equal to 4.2 pounds.

What percentage of Americans would you expect to consume more than 31 pounds of citrus fruit per year?

= 4.2 = 26.8

By the Empirical Rule, 68% of consumption is within 1 standard deviation of the mean. Then 1/2 of 32%, or 16%, of consumption would be more than 31 pounds of citrus fruit per year.

22.6 3126.8

34%34% 16%16%


Section 3-4

Exercise #13

Which of the following exam scores has a better relative position?

a. A score of on an ex 42 = 39 aam with nd = 4X s

b. A score of on an ex76 = 71

am with and = 3X s

z =

42 – 39

4 =

z =

76 – 71

3 =

3

4

5

3

Section 3-4

Exercise #22


Find the percentile ranks of each weight in the data set. The weights are in pounds.

Data: 78, 82, 86, 88, 92, 97

Percentile =

number of values below + 0.5

total number of values 100%

Data: 78, 82, 86, 88, 92, 97

For 78,

For 82,

For 86,

0 + 0.56

100% =

1 + 0.56

100% =

2 + 0.5

6 100% =

8th

percentile

25th

percentile

42nd

percentile

Percentile =

number of values below + 0.5

total number of values 100%

Data: 78, 82, 86, 88, 92, 97

For 88,

For 92,

For 97,

3 + 0.5

6 100% =

4 + 0.5

6 100% =

5 + 0.5

6 100% =

58th

percentile

75th

percentile

92nd

percentile


Section 3-4

Exercise #23

What value corresponds to the 30th percentile?

Find the percentile ranks of each weight in the data set. The weights are in pounds.

78, 82, 86, 88, 92, 97

c =

6(30)

100 = 1.8 or 2

Therefore, the answer is the2nd in the series, or 82.


Section 3-5

Exercise #1

Minimum:

Q1:

Median:

Q3:

Maximum:

Interquartile Range:

Data arranged in order:6, 8, 12, 19, 27, 32, 54

Identify the five number summary and find the interquartile range.

8, 12, 32, 6, 27, 19, 54

32

19

8

6

54

32 – 8 = 24


Section 3-5

Exercise #9

Use the boxplot to identify the maximum value, minimum value, median, first quartile, third quartile, and interquartile range.

Minimum:

Q1:

Median:

Q3:

Maximum:

Interquartile Range:

55

65

70

90

95

90 – 65 = 25

10095908580757065605550

Section 3-5

Exercise #15


9.8 8.0 13.9 4.4 3.9 21.715.9 3.2 11.7 24.8 34.1 17.6

These data are the number of inches of snow reported in randomly selected cities for September 1 through January 10. Construct a boxplot and comment on the skewness of the data.

0 3530252015105

Data arranged in order :

Minimum:

Q1:

MD:

Q3:

Maximum:3.2 34.1

11.7 + 13.9

2 = 12.8

4.4 + 8.0

2 = 6.2

17.6 + 21.7

2 = 19.65

9.8 8.0 13.9 4.4 3.9 21.715.9 3.2 11.7 24.8 34.1 17.6

Comment on the skewness of the data.

35302520151050

3.2 34.112.8 19.656.2

The distribution is positively skewed.


Section 3-5

Exercise #16

These data represent the volumes in cubic yards of the largest dams in the United States and in South America.

Construct a boxplot of the data for each region and compare the distributions.

50,000 52,435 62,850 66,500 77,700 78,008 92,000125,628

United States

46,563 56,242102,014105,944274,026311,539

South America

For USA:

50,000 52,435 62,850 66,500 77,700 78,008 92,000125,628

United States

Min = 50,000 Max = 125,628

MD = 72,100 Q1 = 57,642.5 Q3 = 85,004

125,62872,100

50 100 150 200 250 300 350

For South America:

Min = 46,563 Max = 311,539

MD = 103,979 Q1 = 56,242

Q3 = 274,026

46,563 56,242102,014105,944274,026311,539

South America

50 100 150 200 250 300

311,539103,979

125,62872,100

50 100 150 200 250 300

311,539103,979

South America

USA

Compare the distributions:

section 3-2

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