Question

mean absolute deviation example 2: find the mean, median, and mad of the number of runs allowed by tim lincecum: 0,0,0,2,4,4,5,6,6,8. mean = 35/10 = 3.5, median = 4, mad = (3.5+3.5+3.5+1.5+0.5+0.5+1.5+2.5+2.5+4.5)/10... multiple-choice question: what is the average of the differences he just wrote down? options: 3.5, 2.5, 2.4

Explanation:

Step1: Identify the data set

The data set of runs allowed is: \(0, 0, 0, 2, 4, 4, 5, 6, 6, 8\) (assuming the handwritten "4,4" is correct, so we have 10 data points as the mean calculation used \(25/10 = 2.5\)? Wait, no, earlier mean was \(25/10 = 2.5\)? Wait, let's recalculate the mean. Wait, the sum of \(0 + 0 + 0 + 2 + 4 + 4 + 5 + 6 + 6 + 8\): \(0\times3 = 0\), \(2 + 4 + 4 + 5 + 6 + 6 + 8 = 2 + 8 + 10 + 14 = 34\)? Wait, no, maybe the data is \(0,0,0,2,4,5,6,6,8\) and one more? Wait, the mean was calculated as \(25/10 = 2.5\)? Wait, \(25\) divided by \(10\) is \(2.5\), so sum is \(25\). Let's check: \(0 + 0 + 0 + 2 + 4 + 4 + 5 + 6 + 6 + 8\) – wait, \(0\times3 = 0\), \(2 + 4 + 4 = 10\), \(5 + 6 + 6 + 8 = 25\)? No, \(10 + 25 = 35\). Wait, maybe the data is \(0,0,0,2,4,5,6,6,8\) and a typo? Wait, the MAD calculation: the differences from the mean (let's assume mean is \(2.5\) as per the first calculation? Wait, no, the mean written is \(25/10 = 2.5\)? Wait, \(25\) divided by \(10\) is \(2.5\). Then the differences (absolute deviations) from the mean:

For each data point:

• \(|0 - 2.5| = 2.5\) (three times)
• \(|2 - 2.5| = 0.5\)
• \(|4 - 2.5| = 1.5\) (two times)
• \(|5 - 2.5| = 2.5\)
• \(|6 - 2.5| = 3.5\) (two times)
• \(|8 - 2.5| = 5.5\)

Wait, but the handwritten MAD calculation has numbers like \(3.5, 3.5, 3.5, 1.5, 0.5, 0.5, 0.5, 1.5, 1.5, 2.5\)? Wait, maybe the mean is \(4\)? No, the median is \(4\). Wait, the median is \(4\) (since the data is ordered: \(0,0,0,2,4,4,5,6,6,8\), the middle two are \(4\) and \(4\), so median is \(4\)). But the mean was calculated as \(25/10 = 2.5\)? Wait, maybe the data is \(0,0,0,2,4,5,6,6,8\) with a missing \(0\)? No, the count is 10. Wait, maybe the data is \(0,0,0,2,4,4,5,6,6,8\) – sum is \(0+0+0+2+4+4+5+6+6+8 = 35\), so mean is \(35/10 = 3.5\) (as written: \(35/10 = 3.5\)). Ah, that's correct. So mean is \(3.5\).

Now, the absolute deviations (differences from mean, absolute value):

• For \(0\): \(|0 - 3.5| = 3.5\) (three times)
• For \(2\): \(|2 - 3.5| = 1.5\)
• For \(4\): \(|4 - 3.5| = 0.5\) (two times)
• For \(5\): \(|5 - 3.5| = 1.5\)
• For \(6\): \(|6 - 3.5| = 2.5\) (two times)
• For \(8\): \(|8 - 3.5| = 4.5\)

Wait, but the handwritten MAD calculation has \(3.5, 3.5, 3.5, 1.5, 0.5, 0.5, 1.5, 2.5, 2.5, 4.5\)? Wait, no, the handwritten numbers are "3.5,3.5,3.5,1.5,0.5,0.5,1.5,2.5,2.5,4.5" – let's sum these: \(3.5\times3 = 10.5\), \(1.5\times2 = 3\), \(0.5\times2 = 1\), \(2.5\times2 = 5\), \(4.5\). Total: \(10.5 + 3 + 1 + 5 + 4.5 = 24\). Then MAD is \(24/10 = 2.4\)? Wait, no, the question is "What is the average of the differences he just wrote down?" So the differences (absolute deviations) are the numbers he wrote: \(3.5, 3.5, 3.5, 1.5, 0.5, 0.5, 1.5, 2.5, 2.5, 4.5\). Let's sum them:

\(3.5 + 3.5 + 3.5 + 1.5 + 0.5 + 0.5 + 1.5 + 2.5 + 2.5 + 4.5\)

Calculate step by step:

• \(3.5\times3 = 10.5\)
• \(1.5\times2 = 3\)
• \(0.5\times2 = 1\)
• \(2.5\times2 = 5\)
• \(4.5\)

Now sum all: \(10.5 + 3 = 13.5\); \(13.5 + 1 = 14.5\); \(14.5 + 5 = 19.5\); \(19.5 + 4.5 = 24\)

Now, average (mean) of these differences: \(24 \div 10 = 2.4\)

Wait, but let's check the options: 3.5, 2.5, 2.4. So the average is \(24/10 = 2.4\).

Step1: Sum the differences

The differences are: \(3.5, 3.5, 3.5, 1.5, 0.5, 0.5, 1.5, 2.5, 2.5, 4.5\)

Sum = \(3.5 \times 3 + 1.5 \times 2 + 0.5 \times 2 + 2.5 \times 2 + 4.5\)

= \(10.5 + 3 + 1 + 5 + 4.5\)

= \(24\)

Step2: Calculate the average

Average = \(\frac{\text{Sum}}{\text{Number of data points}}\) = \(\frac{24}{10}\) = \(2.4\)

Answer:

2.4 (corresponding to the option with 2.4)