section 3-2
DESCRIPTION
Chapter 3 Data Description. Section 3-2. Measures of Central Tendency. Chapter 3 Data Description. Section 3-2. Exercise #3. 1, 2, 3, 3, 7, 11, 18, 30, 61. MD:. MD: 7. MD: 1, 2, 3, 3, 7, 11, 18, 30, 61. Chapter 3 Data Description. Section 3-2. Exercise #5. - PowerPoint PPT PresentationTRANSCRIPT
Chapter 3Data Description
Section 3-2
Measures of Central Tendency
Chapter 3Data Description
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
=
The data above are the number of burglaries reported for a specific year for nine western Pennsylvania universities.
136
9 = 15.1
Find (a) the mean.
61, 11, 1, 3, 2, 30, 18, 3, 7
Find (b) the median.
The data above are the number of burglaries reported for a specific year for nine western Pennsylvania universities.
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 =
The data above are the number of burglaries reported for a specific year for nine western Pennsylvania universities.
3
61, 11, 1, 3, 2, 30, 18, 3, 7
Find (d) the midrange.
The data above are the number of burglaries reported for a specific year for nine western Pennsylvania universities.
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
Chapter 3Data Description
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
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.
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:
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.
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
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.
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.
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.
No mode.
574 229 663 372 1,202 88
117 239 465 136 189 75
(d) Arrange data in increasing order.
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.
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
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.
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.
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.
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
Chapter 3Data Description
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
Chapter 3Data Description
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.
Chapter 3Data Description
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
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.
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.
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 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.
Chapter 3Data Description
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
The data shows the number of murders in 25 selected cities.
Find the variance and standard deviation.
2
• 2• – 2 =
– 1
f Xf X
nsn
f • Xm2 = 1,582,260.25 f • Xm = 4662.5
The data shows the number of murders in 25 selected cities.
Find the variance and standard deviation.
24662.51 582 260 25 –
25= 24
, , .
= 29,696
s = 29,696 172.3
s2
Section 3-3
Exercise #33
Chapter 3Data Description
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 =
Chapter 3Data Description
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%
Chapter 3Data Description
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
Chapter 3Data Description
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
Chapter 3Data Description
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.
Chapter 3Data Description
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
Chapter 3Data Description
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
Chapter 3Data Description
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.
Chapter 3Data Description
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: