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 10.0
(round to the nearest tenth as needed.)

the median is 8
(type an integer or a decimal.)

b which, if any, of the margins of victory would be considered the outlier? select the correct choice below and, if necessary, fill in the answer box to complete your choice

a. the outlier is
(type an integer or a decimal.)

b. none of the margins would be considered an outlier

Explanation:

Step1: Identify the data set

The data set for margin of victory over 15 years is: 15, 4, 4, 10, 11, 31, 3, 25, 8, 4, 3, 4, 6, 14, 8 (assuming the last number is 8, maybe a typo earlier). Let's list them in order: 3, 3, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31. Wait, wait, original data: 15, 4, 4, 10, 11, 31, 3, 25, 8, 4, 3, 4, 6, 14, 8. Let's sort them: 3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31. Wait, that's 15 numbers? Wait 3,3,4,4,4,4,6,8,8,10,11,14,15,25,31 – no, wait, let's count: 3 (1), 3 (2), 4 (3), 4 (4), 4 (5), 4 (6), 6 (7), 8 (8), 8 (9), 10 (10), 11 (11), 14 (12), 15 (13), 25 (14), 31 (15). Yes, 15 numbers.

Step2: Find the mean (part a)

Mean is sum of all data divided by number of data points (n=15). Let's calculate the sum: 3 + 3 + 4 + 4 + 4 + 4 + 6 + 8 + 8 + 10 + 11 + 14 + 15 + 25 + 31. Let's add step by step:

3+3=6; 6+4=10; 10+4=14; 14+4=18; 18+4=22; 22+6=28; 28+8=36; 36+8=44; 44+10=54; 54+11=65; 65+14=79; 79+15=94; 94+25=119; 119+31=150. Wait, sum is 150? Then mean is 150 / 15 = 10.0. That matches the given mean.

Step3: Find the median (part a)

Median is the middle value when data is sorted. For n=15 (odd), the median is the (15+1)/2 = 8th term. Sorted data: 3, 3, 4, 4, 4, 4, 6, 8, 8, 10, 11, 14, 15, 25, 31. The 8th term is 8. So median is 8. That matches the given median.

Step4: Identify the outlier (part b)

An outlier is a value that is much larger or smaller than the rest. Let's check the data. Most values are between 3 and 15, except 25 and 31. Wait, but let's use the interquartile range (IQR) method. First, find Q1 and Q3.

For n=15, the data is divided into lower half (first 7 numbers: 3,3,4,4,4,4,6) and upper half (last 7 numbers: 10,11,14,15,25,31 – wait no, wait n=15, so the median is the 8th term (8). So lower half is first 7 terms (positions 1-7): 3,3,4,4,4,4,6. Q1 is the median of lower half: (7+1)/2 = 4th term. Lower half sorted: 3,3,4,4,4,4,6. 4th term is 4. Upper half is positions 9-15: 8,10,11,14,15,25,31? Wait no, wait after median (8th term:8), the upper half is terms 9-15: 8,10,11,14,15,25,31? Wait no, sorted data: 3,3,4,4,4,4,6,8,8,10,11,14,15,25,31. So terms 1-7: 3,3,4,4,4,4,6 (Q1 is 4th term:4). Terms 9-15: 8,10,11,14,15,25,31? Wait no, term 8 is 8, term 9 is 8, term 10 is 10, term 11 is 11, term 12 is 14, term 13 is 15, term 14 is 25, term 15 is 31. Wait, upper half is terms 9-15: 8,10,11,14,15,25,31. The median of upper half (Q3) is the 4th term of upper half: 14 (since upper half has 7 terms, (7+1)/2=4th term: 8,10,11,14,15,25,31 – 4th term is 14).

IQR = Q3 - Q1 = 14 - 4 = 10.

Lower fence = Q1 - 1.5*IQR = 4 - 15 = -11 (no data below this).

Upper fence = Q3 + 1.5*IQR = 14 + 15 = 29.

So any data above 29 is an outlier. 31 is above 29 (25 is below 29). So 31 is the outlier? Wait, but let's check the sum again. Wait original data: 15,4,4,10,11,31,3,25,8,4,3,4,6,14,8. Wait maybe I made a mistake in sorting. Wait the original data as given: 15, 4, 4, 10, 11, 31, 3, 25, 8, 4, 3, 4, 6, 14, 8. Let's list all numbers: 15,4,4,10,11,31,3,25,8,4,3,4,6,14,8. Let's count: 15 numbers. Let's sort them correctly: 3,3,4,4,4,4,6,8,8,10,11,14,15,25,31. Wait 3 (1), 3 (2), 4 (3), 4 (4), 4 (5), 4 (6), 6 (7), 8 (8), 8 (9), 10 (10), 11 (11), 14 (12), 15 (13), 25 (14), 31 (15). Yes. So 25 is 14th term, 31 is 15th. Upper fence is 29, so 31 is above 29, so 31 is the outlier. Wait but 25 is 25, which is below 29 (25 < 29), so 25 is not an outlier. 31 is above 29, so 31 is the outlier.

Wait but let's check the sum again. Wait when we calculated the sum earlier, we got 150. Let's recalculate: 3+3=6; +4=10; +4=14; +4=18; +4=22; +6=28; +8=36; +8=44; +10=54; +11=65; +14=79; +15=94; +25=119; +31=150. Yes. So mean is 150/15=10.0.

Now, if we eliminate the outlier (31), the new data set has 14 numbers. Let's list the data without 31: 3,3,4,4,4,4,6,8,8,10,11,14,15,25.

Step5: New mean (after eliminating outlier)

Sum of new data: 150 - 31 = 119. Number of data points: 14. So mean = 119 / 14 = 8.5.

Step6: New median (after eliminating outlier)

For n=14 (even), median is the average of the 7th and 8th terms. Sorted data without 31: 3,3,4,4,4,4,6,8,8,10,11,14,15,25. 7th term:6; 8th term:8. Median = (6 + 8)/2 = 7? Wait no, wait sorted data: 3,3,4,4,4,4,6,8,8,10,11,14,15,25. Wait positions 1-14: 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. Wait 7th term is 6, 8th term is 8. So median is (6 + 8)/2 = 7? Wait but maybe I made a mistake. Wait original median was 8 (with 31). If we remove 31, the data is 14 numbers. Let's check the sorted data again: 3,3,4,4,4,4,6,8,8,10,11,14,15,25. Yes, 14 numbers. The median is the average of the 7th and 8th terms. 7th term:6, 8th term:8. So (6+8)/2=7. But wait, maybe the outlier is 31? Wait but let's check the original data again. Wait the user's data has "31" – maybe that's the outlier. Alternatively, maybe 25? Wait no, 25 is below 29. Wait 31 is above 29, so 31 is the outlier.

But let's check the options. Option A: The outlier is [number], Option B: None. So since 31 is an outlier (above upper fence), the outlier is 31.

Now, if we eliminate 31, new mean: sum was 150, minus 31 is 119, divided by 14: 119/14=8.5. New median: sorted data without 31: 3,3,4,4,4,4,6,8,8,10,11,14,15,25. The 7th term is 6, 8th term is 8, so median is (6+8)/2=7.

But the user's question is part b: identify the outlier. So the outlier is 31, so option A: The outlier is 31.

Answer:

a) The mean is 10.0, the median is 8.
b) A. The outlier is 31. If the outlier (31) is eliminated, the new mean is 8.5 and the new median is 7.

(Note: For part b, the outlier identification: using IQR, 31 is above the upper fence (29), so it's the outlier. After removing 31, mean = 119/14 = 8.5, median = (6 + 8)/2 = 7.)