Question

the following data give the margin of victory for a football championship over 15 years
15 4 4 10 11 31 3 25 8 4 3 4 6 14 8

a. find the mean and median margin of victory
b. identify the outlier in the data set. if the outlier is eliminated, what are the new mean and median?

a. find the mean and median of the weights

the mean is
(round to the nearest tenth as needed.)

Explanation:

Step1: Sum the data values

First, we list the data points: 15, 4, 4, 10, 11, 31, 3, 25, 8, 4, 3, 4, 6, 14, 8. Let's sum them up.
\[ \begin{align*} &15 + 4 + 4 + 10 + 11 + 31 + 3 + 25 + 8 + 4 + 3 + 4 + 6 + 14 + 8\\ =&15+(4+4)+10+11+31+3+25+8+(4+3+4)+6+14+8\\ =&15 + 8+10+11+31+3+25+8+11+6+14+8\\ =&(15+8)+10+11+31+3+25+8+11+6+14+8\\ =&23+10+11+31+3+25+8+11+6+14+8\\ =&(23+10)+11+31+3+25+8+11+6+14+8\\ =&33+11+31+3+25+8+11+6+14+8\\ =&(33+11)+31+3+25+8+11+6+14+8\\ =&44+31+3+25+8+11+6+14+8\\ =&(44+31)+3+25+8+11+6+14+8\\ =&75+3+25+8+11+6+14+8\\ =&(75+3)+25+8+11+6+14+8\\ =&78+25+8+11+6+14+8\\ =&(78+25)+8+11+6+14+8\\ =&103+8+11+6+14+8\\ =&(103+8)+11+6+14+8\\ =&111+11+6+14+8\\ =&(111+11)+6+14+8\\ =&122+6+14+8\\ =&(122+6)+14+8\\ =&128+14+8\\ =&(128+14)+8\\ =&142+8\\ =&150 \end{align*} \]
Wait, that can't be right. Wait, let's count the number of data points. Let's list them again: 15, 4, 4, 10, 11, 31, 3, 25, 8, 4, 3, 4, 6, 14, 8. Let's count: 1 (15), 2 (4), 3 (4), 4 (10), 5 (11), 6 (31), 7 (3), 8 (25), 9 (8), 10 (4), 11 (3), 12 (4), 13 (6), 14 (14), 15 (8). So 15 data points. Let's sum again carefully:

15 + 4 = 19; 19 + 4 = 23; 23 + 10 = 33; 33 + 11 = 44; 44 + 31 = 75; 75 + 3 = 78; 78 + 25 = 103; 103 + 8 = 111; 111 + 4 = 115; 115 + 3 = 118; 118 + 4 = 122; 122 + 6 = 128; 128 + 14 = 142; 142 + 8 = 150. Wait, that's correct? Wait, 15 numbers. So sum is 150? Wait, no, 15 + 4 + 4 + 10 + 11 + 31 + 3 + 25 + 8 + 4 + 3 + 4 + 6 + 14 + 8. Let's add the larger numbers: 15 + 10 + 11 + 31 + 25 + 14 = 15+10=25; 25+11=36; 36+31=67; 67+25=92; 92+14=106. Then the smaller numbers: 4 + 4 + 3 + 8 + 4 + 3 + 4 + 6 + 8. 4+4=8; 8+3=11; 11+8=19; 19+4=23; 23+3=26; 26+4=30; 30+6=36; 36+8=44. Then total sum: 106 + 44 = 150. Yes, that's correct.

Step2: Calculate the mean

The mean is the sum divided by the number of data points. The number of data points \( n = 15 \). So mean \( \bar{x}=\frac{\text{Sum}}{n}=\frac{150}{15} = 10 \). Wait, but let's check again. Wait, 31 is a data point. Wait, 15 + 4 + 4 + 10 + 11 + 31 + 3 + 25 + 8 + 4 + 3 + 4 + 6 + 14 + 8. Let's add 31 and 25: 56; 15 + 10 + 11 + 14 = 50; 4 + 4 + 4 + 4 + 3 + 3 + 8 + 8 + 6 = 4*4=16; 3*2=6; 8*2=16; 6. So 16+6+16+6=44. Then 56 + 50 = 106; 106 + 44 = 150. So sum is 150. Number of data points is 15. So mean is 150/15 = 10.

Step3: Find the median

To find the median, we first order the data in ascending order. Let's sort the data: 3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31. Wait, wait, let's list all data points: 3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31. Wait, how many data points? 15. So the median is the middle value, which is the 8th value (since (15 + 1)/2 = 8th term). Let's count: 1:3, 2:3, 3:4, 4:4, 5:4, 6:4, 7:6, 8:8, 9:8, 10:10, 11:11, 12:14, 13:15, 14:25, 15:31. Wait, that's not right. Wait, original data: 15, 4, 4, 10, 11, 31, 3, 25, 8, 4, 3, 4, 6, 14, 8. Let's sort them properly:

Start with the smallest: 3, 3 (from 3 and 3), then 4, 4, 4, 4 (four 4s), then 6, then 8, 8 (two 8s), then 10, 11, 14, 15, 25, 31. Wait, let's list all in order:

3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31. Wait, that's 15 numbers. The 8th number is the median. Let's count:

1:3

2:3

3:4

4:4

5:4

6:4

7:6

8:8

Yes, the 8th term is 8. Wait, but earlier when we calculated the mean, we got 10, but let's check the data again. Wait, 31 is a data point. Wait, maybe I made a mistake in the sum. Wait, 31 + 25 = 56; 15 + 10 + 11 + 14 = 50; 4 + 4 + 4 + 4 + 3 + 3 + 8 + 8 + 6 = 4*4=16, 3*2=6, 8*2=16, 6. So 16+6=22, 22+16=38, 38+6=44. Then 56 + 50 = 106, 106 + 44 = 150. So sum is 150. Number of data points is 15. So mean is 10. But the median: when sorted, the data is [3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31]. Wait, no, wait, 8 appears twice? Wait, original data: 8, 8? Let's check the original data: 15, 4, 4, 10, 11, 31, 3, 25, 8, 4, 3, 4, 6, 14, 8. So the 9th term is 8, 15th term is 8? Wait, no, let's list all 15 terms:

1:3

2:3

3:4

4:4

5:4

6:4

7:6

8:8

9:8

10:10

11:11

12:14

13:15

14:25

15:31

Ah, I see, I missed a term earlier. So the sorted data is: 3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31. Wait, no, that's 15 terms? Let's count: 1 (3), 2 (3), 3 (4), 4 (4), 5 (4), 6 (4), 7 (6), 8 (8), 9 (8), 10 (10), 11 (11), 12 (14), 13 (15), 14 (25), 15 (31). Yes, 15 terms. So the median is the 8th term? Wait, (15 + 1)/2 = 8, so the 8th term. The 8th term is 8? Wait, no, the 7th term is 6, 8th is 8, 9th is 8, 10th is 10. Wait, no, let's index from 1 to 15:

1: 3

2: 3

3: 4

4: 4

5: 4

6: 4

7: 6

8: 8

9: 8

10: 10

11: 11

12: 14

13: 15

14: 25

15: 31

Yes, so the 8th term is 8. So median is 8.

Wait, but this contradicts the mean of 10. Wait, maybe I made a mistake in the sum. Let's recalculate the sum:

Data points: 3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31.

Let's sum them:

3 + 3 = 6

6 + 4 + 4 + 4 + 4 = 6 + 16 = 22

22 + 6 = 28

28 + 8 + 8 = 28 + 16 = 44

44 + 10 = 54

54 + 11 = 65

65 + 14 = 79

79 + 15 = 94

94 + 25 = 119

119 + 31 = 150. Yes, sum is 150. Number of terms is 15. So mean is 150/15 = 10. Median is the 8th term, which is 8.

Wait, but let's check part (b) about the outlier. The outlier is likely 31, since it's much larger than the other values. Let's check the interquartile range (IQR) to confirm. First, find Q1 and Q3. The data is sorted: [3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31]. The median is the 8th term (8), so the lower half is the first 7 terms: [3, 3, 4, 4, 4, 4, 6[LLM SSE On Failure]

Answer:

Step1: Sum the data values

First, we list the data points: 15, 4, 4, 10, 11, 31, 3, 25, 8, 4, 3, 4, 6, 14, 8. Let's sum them up.

$$
LATEXBLOCK0
$$

Wait, that can't be right. Wait, let's count the number of data points. Let's list them again: 15, 4, 4, 10, 11, 31, 3, 25, 8, 4, 3, 4, 6, 14, 8. Let's count: 1 (15), 2 (4), 3 (4), 4 (10), 5 (11), 6 (31), 7 (3), 8 (25), 9 (8), 10 (4), 11 (3), 12 (4), 13 (6), 14 (14), 15 (8). So 15 data points. Let's sum again carefully:

15 + 4 = 19; 19 + 4 = 23; 23 + 10 = 33; 33 + 11 = 44; 44 + 31 = 75; 75 + 3 = 78; 78 + 25 = 103; 103 + 8 = 111; 111 + 4 = 115; 115 + 3 = 118; 118 + 4 = 122; 122 + 6 = 128; 128 + 14 = 142; 142 + 8 = 150. Wait, that's correct? Wait, 15 numbers. So sum is 150? Wait, no, 15 + 4 + 4 + 10 + 11 + 31 + 3 + 25 + 8 + 4 + 3 + 4 + 6 + 14 + 8. Let's add the larger numbers: 15 + 10 + 11 + 31 + 25 + 14 = 15+10=25; 25+11=36; 36+31=67; 67+25=92; 92+14=106. Then the smaller numbers: 4 + 4 + 3 + 8 + 4 + 3 + 4 + 6 + 8. 4+4=8; 8+3=11; 11+8=19; 19+4=23; 23+3=26; 26+4=30; 30+6=36; 36+8=44. Then total sum: 106 + 44 = 150. Yes, that's correct.

Step2: Calculate the mean

The mean is the sum divided by the number of data points. The number of data points \( n = 15 \). So mean \( \bar{x}=\frac{\text{Sum}}{n}=\frac{150}{15} = 10 \). Wait, but let's check again. Wait, 31 is a data point. Wait, 15 + 4 + 4 + 10 + 11 + 31 + 3 + 25 + 8 + 4 + 3 + 4 + 6 + 14 + 8. Let's add 31 and 25: 56; 15 + 10 + 11 + 14 = 50; 4 + 4 + 4 + 4 + 3 + 3 + 8 + 8 + 6 = 44=16; 32=6; 8*2=16; 6. So 16+6+16+6=44. Then 56 + 50 = 106; 106 + 44 = 150. So sum is 150. Number of data points is 15. So mean is 150/15 = 10.

Step3: Find the median

To find the median, we first order the data in ascending order. Let's sort the data: 3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31. Wait, wait, let's list all data points: 3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31. Wait, how many data points? 15. So the median is the middle value, which is the 8th value (since (15 + 1)/2 = 8th term). Let's count: 1:3, 2:3, 3:4, 4:4, 5:4, 6:4, 7:6, 8:8, 9:8, 10:10, 11:11, 12:14, 13:15, 14:25, 15:31. Wait, that's not right. Wait, original data: 15, 4, 4, 10, 11, 31, 3, 25, 8, 4, 3, 4, 6, 14, 8. Let's sort them properly:

Start with the smallest: 3, 3 (from 3 and 3), then 4, 4, 4, 4 (four 4s), then 6, then 8, 8 (two 8s), then 10, 11, 14, 15, 25, 31. Wait, let's list all in order:

3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31. Wait, that's 15 numbers. The 8th number is the median. Let's count:

1:3

2:3

3:4

4:4

5:4

6:4

7:6

8:8

Yes, the 8th term is 8. Wait, but earlier when we calculated the mean, we got 10, but let's check the data again. Wait, 31 is a data point. Wait, maybe I made a mistake in the sum. Wait, 31 + 25 = 56; 15 + 10 + 11 + 14 = 50; 4 + 4 + 4 + 4 + 3 + 3 + 8 + 8 + 6 = 44=16, 32=6, 8*2=16, 6. So 16+6=22, 22+16=38, 38+6=44. Then 56 + 50 = 106, 106 + 44 = 150. So sum is 150. Number of data points is 15. So mean is 10. But the median: when sorted, the data is [3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31]. Wait, no, wait, 8 appears twice? Wait, original data: 8, 8? Let's check the original data: 15, 4, 4, 10, 11, 31, 3, 25, 8, 4, 3, 4, 6, 14, 8. So the 9th term is 8, 15th term is 8? Wait, no, let's list all 15 terms:

1:3

2:3

3:4

4:4

5:4

6:4

7:6

8:8

9:8

10:10

11:11

12:14

13:15

14:25

15:31

Ah, I see, I missed a term earlier. So the sorted data is: 3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31. Wait, no, that's 15 terms? Let's count: 1 (3), 2 (3), 3 (4), 4 (4), 5 (4), 6 (4), 7 (6), 8 (8), 9 (8), 10 (10), 11 (11), 12 (14), 13 (15), 14 (25), 15 (31). Yes, 15 terms. So the median is the 8th term? Wait, (15 + 1)/2 = 8, so the 8th term. The 8th term is 8? Wait, no, the 7th term is 6, 8th is 8, 9th is 8, 10th is 10. Wait, no, let's index from 1 to 15:

1: 3

2: 3

3: 4

4: 4

5: 4

6: 4

7: 6

8: 8

9: 8

10: 10

11: 11

12: 14

13: 15

14: 25

15: 31

Yes, so the 8th term is 8. So median is 8.

Wait, but this contradicts the mean of 10. Wait, maybe I made a mistake in the sum. Let's recalculate the sum:

Data points: 3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31.

Let's sum them:

3 + 3 = 6

6 + 4 + 4 + 4 + 4 = 6 + 16 = 22

22 + 6 = 28

28 + 8 + 8 = 28 + 16 = 44

44 + 10 = 54

54 + 11 = 65

65 + 14 = 79

79 + 15 = 94

94 + 25 = 119

119 + 31 = 150. Yes, sum is 150. Number of terms is 15. So mean is 150/15 = 10. Median is the 8th term, which is 8.

Wait, but let's check part (b) about the outlier. The outlier is likely 31, since it's much larger than the other values. Let's check the interquartile range (IQR) to confirm. First, find Q1 and Q3. The data is sorted: [3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31]. The median is the 8th term (8), so the lower half is the first 7 terms: [3, 3, 4, 4, 4, 4, 6[LLM SSE On Failure]