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Probability and Statistics Solution Manual | Reference Guide Chapter 2 | Representing Data Numerically
Outliers
Pages 64–65
1.
IQR = Q3 − Q1 = 38 − 34 = 4
2.
IQR = Q3 − Q1 = 15.5 − 8 = 7.5
3.
Q1: left side of box; 124
Q3: right side of box; 136
IQR = Q3 − Q1 = 136 − 124 = 12
Q1: left side of box; 1.3
4.
8.
Order the data from least to greatest: {0.05, 0.34,
0.55, 0.86, 0.99, 1.23, 1.54, 1.54, 1.65, 1.98}
Median: 0.05, 0.34, 0.55, 0.86, 0.99 | 1.23, 1.54, 1.54,
1.65, 1.98; the median is 1.11.
Q1: 0.05, 0.34, 0.55, 0.86, 0.99 | 1.23, 1.54, 1.54, 1.65,
1.98; Q1 is 0.55.
Q3: 0.05, 0.34, 0.55, 0.86, 0.99 | 1.23, 1.54, 1.54, 1.65,
1.98; Q3 is 1.54.
IQR = Q3 − Q1 = 1.54 − 0.55 = 0.99
Q3: right side of box; 2.15
IQR = Q3 − Q1 = 2.15 − 1.3 = 0.85
9.
Order the data from least to greatest: {8, 10, 10, 11,
20, 20, 21, 27, 28, 29}
Order the data from least to greatest: {8, 10, 10, 11,
20, 20, 21, 27, 28, 29}
Median: 8, 10, 10, 11, 20 | 20, 21, 27, 28, 29; the
median is 20.
Median: 8, 10, 10, 11, 20 | 20, 21, 27, 28, 29;
the median is 20.
Q1: 8, 10, 10, 11, 20 | 20, 21, 27, 28, 29; Q1 is 10.
Q1: 8, 10, 10, 11, 20 | 20, 21, 27, 28, 29; Q1 is 10.
Q3: 8, 10, 10, 11, 20 | 20, 21, 27, 28, 29; Q3 is 27.
Q3: 8, 10, 10, 11, 20 | 20, 21, 27, 28, 29; Q3 is 27.
IQR = Q3 − Q1 = 27 − 10 = 17
IQR = Q3 − Q1 = 27 − 10 = 17
6.
Order the data from least to greatest: {90, 110, 125,
140, 185, 225, 300, 305, 305, 330, 585}
Lower fence = Q1 − 1.5(IQR) = 10 − 1.5(17)
= −15.5
Upper fence = Q3 + 1.5(IQR) = 27 + 1.5(17) = 52.5
Median: 90, 110, 125, 140, 185, 225, 300, 305, 305,
330, 585; the median is 225.
There are no data values less than the lower fence or
greater than the upper fence. So there are no outliers.
Q1: 90, 110, 125, 140, 185, 225, 300, 305, 305, 330,
585; Q1 is 125.
10.
Order the data from least to greatest: {90, 110, 125,
140, 185, 225, 300, 305, 305, 330, 585}
Q3: 90, 110, 125, 140, 185, 225, 300, 305, 305, 330,
585; Q3 is 305.
Median: 90, 110, 125, 140, 185, 225, 300, 305, 305,
330, 585; the median is 225.
IQR = Q3 − Q1 = 305 − 125 = 180
Order the data from least to greatest: {0.1, 2.5, 2.6,
2.9, 2.9, 3.1, 3.1, 3.3, 3.8, 3.9, 4.0}
Q1: 90, 110, 125, 140, 185, 225, 300, 305, 305, 330,
585; Q1 is 125.
Median: 0.1, 2.5, 2.6, 2.9, 2.9, 3.1, 3.1, 3.3, 3.8, 3.9,
4.0; the median is 3.1.
Q3: 90, 110, 125, 140, 185, 225, 300, 305, 305, 330,
585; Q3 is 305.
IQR = Q3 − Q1 = 305 − 125 = 180
Q1: 0.1, 2.5, 2.6, 2.9, 2.9, 3.1, 3.1, 3.3, 3.8, 3.9, 4.0;
Q1 is 2.6.
Lower fence = Q1 − 1.5(IQR) = 125 − 1.5(180)
= −145
Q3: 0.1, 2.5, 2.6, 2.9, 2.9, 3.1, 3.1, 3.3, 3.8, 3.9, 4.0;
Q3 is 3.8.
Upper fence = Q3 + 1.5(IQR) = 305 + 1.5(180) = 575
IQR = Q3 − Q1 = 3.8 − 2.6 = 1.2
The data value 585 is greater than the upper fence.
So 585 is an outlier
5.
7.
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Probability and Statistics Solution Manual | Reference Guide Chapter 2 | Representing Data Numerically
11.
Order the data from least to greatest: {0.1, 2.5, 2.6, 2.9, 2.9, 3.1, 3.1, 3.3, 3.8, 3.9, 4.0}
Median: 0.1, 2.5, 2.6, 2.9, 2.9, 3.1, 3.1, 3.3, 3.8, 3.9, 4.0; the median is 3.1.
Q1: 0.1, 2.5, 2.6, 2.9, 2.9, 3.1, 3.1, 3.3, 3.8, 3.9, 4.0; Q1 is 2.6.
Q3: 0.1, 2.5, 2.6, 2.9, 2.9, 3.1, 3.1, 3.3, 3.8, 3.9, 4.0; Q3 is 3.8.
IQR = Q3 − Q1 = 3.8 − 2.6 = 1.2
Lower fence = Q1 − 1.5(IQR) = 2.6 − 1.5(1.2) = 0.8
Upper fence = Q3 + 1.5(IQR) = 3.8 + 1.5(1.2) = 5.6
The data value 0.1 is less than the lower fence. So 0.1 is an outlier.
12.
Order the data from least to greatest: {0.05, 0.34, 0.55, 0.86, 0.99, 1.23, 1.54, 1.54, 1.65, 1.98}
Median: 0.05, 0.34, 0.55, 0.86, 0.99 | 1.23, 1.54, 1.54, 1.65, 1.98; the median is 1.11.
Q1: 0.05, 0.34, 0.55, 0.86, 0.99 | 1.23, 1.54, 1.54, 1.65, 1.98; Q1 is 0.55.
Q3: 0.05, 0.34, 0.55, 0.86, 0.99 | 1.23, 1.54, 1.54, 1.65, 1.98; Q3 is 1.54.
IQR = Q3 − Q1 = 1.54 − 0.55 = 0.99
Lower fence = Q1 − 1.5(IQR) = 0.55 − 1.5(0.99) = −0.935
Upper fence = Q3 + 1.5(IQR) = 1.54 + 1.5(0.99) = 3.025
There are no data values less than the lower fence or greater than the upper fence.
So there are no outliers.
13.
Order the data from least to greatest: {11, 14, 14, 15, 18, 19, 21, 26, 27, 30, 46}
Find five-number summary.
Minimum is the least value, which is 11.
Maximum is the greatest value, which is 46.
Median: 11, 14, 14, 15, 18, 19, 21, 26, 27, 30, 46; the median is 19.
Q1: 11, 14, 14, 15, 18, 19, 21, 26, 27, 30, 46; Q1 is 14.
Q3: 11, 14, 14, 15, 18, 19, 21, 26, 27, 30, 46; Q3 is 27.
minimum = 11, Q1 = 14, median = 19, Q3 = 27, maximum = 46
Draw a box with Q1 and Q3 as sides. Draw a segment from the minimum value
to Q1 and Q3 to the maximum value.
11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46
14.
{11, 14, 14, 15, 18, 19, 21, 26, 27, 30, 46}
Q1: 11, 14, 14, 15, 18, 19, 21, 26, 27, 30, 46; Q1 is 14.
Q3: 11, 14, 14, 15, 18, 19, 21, 26, 27, 30, 46; Q3 is 27.
IQR = Q3 − Q1 = 27 − 14 = 13
15.
{11, 14, 14, 15, 18, 19, 21, 26, 27, 30, 46}
Q1: 11, 14, 14, 15, 18, 19, 21, 26, 27, 30, 46; Q1 is 14.
Q3: 11, 14, 14, 15, 18, 19, 21, 26, 27, 30, 46; 1.54; Q3 is 27.
IQR = Q3 − Q1 = 27 − 14 = 13
Lower fence = Q1 − 1.5(IQR) = 14 − 1.5(13) = −5.5
Upper fence = Q3 + 1.5(IQR) = 27 + 1.5(13) = 46.5
There are no data values less than the lower fence or greater than the upper fence. So 46 is not an outlier.
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Copying or distributing without K12’s written consent is prohibited.
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Probability and Statistics Solution Manual | Reference Guide Chapter 2 | Representing Data Numerically
16.
Order the data from least to greatest: {0, 1, 2, 2, 2, 3, 3, 3, 3, 5, 6}
Median: 0, 1, 2, 2, 2, 3, 3, 3, 3, 5, 6; the median is 3.
Q1: 0, 1, 2, 2, 2, 3, 3, 3, 3, 5, 6; Q1 is 2.
Q3: 0, 1, 2, 2, 2, 3, 3, 3, 3, 5, 6; Q3 is 3.
IQR = Q3 − Q1 = 3 − 2 = 1
17.
Order the data from least to greatest: {0, 1, 2, 2, 2, 3, 3, 3, 3, 5, 6}
Median: 0, 1, 2, 2, 2, 3, 3, 3, 3, 5, 6; the median is 3.
Q1: 0, 1, 2, 2, 2, 3, 3, 3, 3, 5, 6; Q1 is 2.
Q3: 0, 1, 2, 2, 2, 3, 3, 3, 3, 5, 6; Q3 is 3.
IQR = Q3 − Q1 = 3 − 2 = 1
Lower fence = Q1 − 1.5(IQR) = 2 − 1.5(1) = 0.5
Upper fence = Q3 + 1.5(IQR) = 3 + 1.5(1) = 4.5
The data value 0 is less than the lower fence, and the data values 5 and 6 are
greater than the upper fence. So 0, 5, and 6 are outliers.
18.
There are 19 data values in the data set. The median is the 10th value, which is $5.95.
Q1 is the middle value of the 9 data values to the left of the median; Q1 is $4.99.
Q3 is the middle value of the 9 data values to the right of the median; Q3 is $6.99.
IQR = Q3 − Q1 = 6.99 − 4.99 = 2; $2
19.
There are 19 data values in the data set. The median is the 10th value, which is $5.95.
Q1 is the middle value of the 9 data values to the left of the median; Q1 is $4.99.
Q3 is the middle value of the 9 data values to the right of the median; Q3 is $6.99.
IQR = Q3 − Q1 = 6.99 − 4.99 = 2; the IQR is $2.
Lower fence = Q1 − 1.5(IQR) = 4.99 − 1.5(2) = 1.99
Upper fence = Q3 + 1.5(IQR) = 6.99 + 1.5(2) = 9.99
The data value $0.95 is less than the lower fence. So $0.95 is an outlier.
20.
The minimum value is the least value, which is $0.99. The maximum value is the
greatest value, which is $7.99. There are 19 data values in the data set. The median
is the 10th value, which is $5.95. Q1 is the middle value of the 9 data values to the
left of the median; Q1 is $4.99. Q3 is the middle value of the 9 data values to the
right of the median; Q3 is $6.99.
IQR = Q3 − Q1 = 6.99 − 4.99 = 2; the IQR is $2.
Lower fence = Q1 − 1.5(IQR) = 4.99 − 1.5(2) = 1.99
Upper fence = Q3 + 1.5(IQR) = 6.99 + 1.5(2) = 9.99
The data value 0.95 is less than the lower fence. So 0.95 is an outlier and is marked
with a dot on the box plot. The next greatest value in the data set, 4.89, becomes the
end of the left whisker. Draw a box with Q1 and Q3 as sides. Draw a segment from 4.89
to Q1 and Q3 to the maximum value 7.99.
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
6.5
7
7.5
8
8.5
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Probability and Statistics Solution Manual | Reference Guide Chapter 2 | Representing Data Numerically
21.
The highest price is a data value that is not greater
than the upper fence for the data set.
23.
Order the data from least to greatest with n as the
least value: {n, 39, 42, 44, 47, 48, 51, 59, 65, 67, 72}
There are 19 data values in the data set. The median
is the 10th value, which is $5.95. Q1 is the middle
value of the 9 data values to the left of the median;
Q1 is $4.99. Q3 is the middle value of the 9 data
values to the right of the median; Q3 is $6.99.
Median: n, 39, 42, 44, 47, 48, 51, 59, 65, 67, 72;
the median is 48.
Q1: n, 39, 42, 44, 47, 48, 51, 59, 65, 67, 72; Q1 is 42.
Q3: n, 39, 42, 44, 47, 48, 51, 59, 65, 67, 72; Q3 is 65.
IQR = Q3 − Q1 = 65 − 42 = 23
n is the greatest whole number that is less than the
lower fence.
Lower fence = Q1 − 1.5(IQR) = 42 − 1.5(23) = 7.5
n=7
IQR = Q3 − Q1 = 6.99 − 4.99 = 2; the IQR is $2.
Upper fence = Q3 + 1.5(IQR) = 6.99 + 1.5(2) = 9.99
So $9.99 is the highest price a CD could sell for and
not be an outlier.
22.
Order the data from least to greatest with n as the
greatest value: {39, 42, 44, 47, 48, 51, 59, 65, 67, 72, n}
Median: 39, 42, 44, 47, 48, 51, 59, 65, 67, 72, n; the
median is 51.
Q1: 39, 42, 44, 47, 48, 51, 59, 65, 67, 72, n; Q1 is 44.
Q3: 39, 42, 44, 47, 48, 51, 59, 65, 67, 72, n; Q3 is 67.
IQR = Q3 − Q1 = 67 − 44 = 23
n is the least whole number that is greater than the
upper fence.
Upper fence = Q3 + 1.5(IQR) = 67 + 1.5(23) = 101.5
n = 102
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Copying or distributing without K12’s written consent is prohibited.
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