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 sum all the given data points: \(15 + 4 + 4 + 10 + 11 + 31 + 3 + 25 + 8 + 4 + 3 + 4 + 6 + 14 + 8\). Let's calculate that:
\(15+4 = 19\); \(19+4 = 23\); \(23+10 = 33\); \(33+11 = 44\); \(44+31 = 75\); \(75+3 = 78\); \(78+25 = 103\); \(103+8 = 111\); \(111+4 = 115\); \(115+3 = 118\); \(118+4 = 122\); \(122+6 = 128\); \(128+14 = 142\); \(142+8 = 150\). So the sum is \(150\).

Step2: Calculate the mean

The mean is the sum divided by the number of data points. There are \(15\) data points. So the mean is \(\frac{150}{15}=10.0\)? Wait, wait, no, wait. Wait, let's re - check the data. Wait, the data is: 15, 4, 4, 10, 11, 31, 3, 25, 8, 4, 3, 4, 6, 14, 8. Let's count the number of terms: 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 terms. Now sum again:

\(3 + 3+4 + 4+4 + 4+6 + 8+8 + 10+11 + 14+15 + 25+31\)

Let's group the smaller numbers: \(3+3 = 6\); \(4\times4=16\); \(6\); \(8\times2 = 16\); \(10\); \(11\); \(14\); \(15\); \(25\); \(31\).

Now sum these groups: \(6+16 = 22\); \(22 + 6=28\); \(28+16 = 44\); \(44+10 = 54\); \(54+11 = 65\); \(65+14 = 79\); \(79+15 = 94\); \(94+25 = 119\); \(119+31 = 150\). So sum is \(150\). Then mean is \(\frac{150}{15}=10.0\)? Wait, but let's check again. Wait, maybe I made a mistake in the sum. Wait, 15 + 4 is 19, +4 is 23, +10 is 33, +11 is 44, +31 is 75, +3 is 78, +25 is 103, +8 is 111, +4 is 115, +3 is 118, +4 is 122, +6 is 128, +14 is 142, +8 is 150. Yes, sum is 150. Number of data points is 15. So mean is \(150\div15 = 10.0\). Wait, but let's check the median.

Step3: Find the median

To find the median, we first order the data from least to greatest. Let's order the data: \(3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31\). Wait, wait, when we order the data:

The data points are: 3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31. Wait, let's count the number of terms. Wait, original data has 15 terms. Let's list them in order:

3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31. Wait, let's check the 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. The median is the middle term, which is the 8th term (since \((15 + 1)\div2=8\)). The 8th term in the ordered list is 8. Wait, wait, no: when we order the data:

Wait, let's do it step by step. Let's list all data points:

3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31. Wait, the 8th term is 8? Wait, no, let's index them:

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, the 8th term is 8. So the median is 8. Wait, but earlier when we calculated the mean, we got 10.0. Wait, but let's check the sum again. Wait, 3+3 = 6, 4*4 = 16, 6, 8*2 = 16, 10, 11, 14, 15, 25, 31. So 6+16 = 22, +6 = 28, +16 = 44, +10 = 54, +11 = 65, +14 = 79, +15 = 94, +25 = 119, +31 = 150. Yes, sum is 150. Number of data points is 15. So mean is 10.0.

Wait, but maybe I made a mistake in the data. Wait, the problem says "margin of victory for a football championship over 15 years" and the data is 15, 4, 4, 10, 11, 31, 3, 25, 8, 4, 3, 4, 6, 14, 8. Let's count the number of numbers: 15 (1), 4 (2), 4 (3), 10 (4), 11 (5), 31 (6), 3 (7), 25 (8), 8 (9), 4 (10), 3 (11), 4 (12), 6 (13), 14 (14), 8 (15). So 15 numbers. So sum is 150, mean is 10.0, median 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. If we remove 31, the new data set has 14 numbers. Let's recalculate the mean and median for part (b) later, but for part (a), the mean is 10.0 and median is 8.

Wait, but wait, when I calculated the sum, is that correct? Let's add again:

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. Yes, sum is 150. Number of data points: 15. So mean is \(150\div15 = 10.0\).

Answer:

The mean is \(\boldsymbol{10.0}\) (rounded to the nearest tenth) and the median is \(\boldsymbol{8}\). (For part (a) of the question, the mean is 10.0)