Try This Revisited
Let's return to the Try This problem presented at the beginning of this lesson:
Vishal commutes to work each day. He finds that his commute time varies dramatically from day to day depending on the time of day that he leaves, traffic conditions, weather, etc. Vishal's morning commute times, in minutes, are shown for a 10-day period.
- \(38\)
- \(32\)
- \(48\)
- \(33\)
- \(34\)
- \(61\)
- \(37\)
- \(35\)
- \(59\)
- \(43\)
Vishal has calculated that his mean commute time is \(42\) minutes, but he wonders by how much, on average, his commute times differ from his mean commute time. Determine such a measure for Vishal.
Solution
Let \(x\) represent Vishal's commute time. We begin by verifying Vishal's mean commute time.
Recall that the mean of a set of numbers, \(\bar{x}\), is calculated by adding all of the values and dividing by the number of data points:
\(\begin{align*} \bar{x} &= \dfrac{38 +32 +48 +33 +34+ 61 +37 +35 +59 +43}{10} \\ &= \dfrac{420}{10} \\ &= 42 \end{align*}\)
Vishal would like to know how long, on average, his commute times deviate (or vary) from the mean. The table shows the deviation of each data point from the mean.
| \(x\) |
\(x-\bar{x}\) |
| \(38\) |
\(-4\) |
| \(32\) |
\(-10\) |
| \(48\) |
\(6\) |
| \(33\) |
\(-9\) |
| \(34\) |
\(-8\) |
| \(61\) |
\(19\) |
| \(37\) |
\(-5\) |
| \(35\) |
\(-7\) |
| \(59\) |
\(17\) |
| \(43\) |
\(1\) |
Since the table shows how far each data point deviates from the mean (\(x-\bar{x}\) is the difference between each data point and the mean), it seems reasonable to determine the average of these numbers: Vishal would like to know, on average, how much his commute times differ from the mean commute time.
However, if you determine the sum of all of the values of \(x-\bar{x}\), you will find that the sum is \(0\), so that the mean of these deviations will also be \(0\). It turns out that this occurs for any data set!
The problem is that the positive and negative values of \(x-\bar{x}\) cancel out when they are added together. However, Vishal is only interested in how much his commute times vary from the mean commute time, and not whether the times are above or below his mean commute time. Thus, he is actually interested in the mean of the absolute value of the deviations:
| \(x\) |
\(x-\bar{x}\) |
\(|x-\bar{x}|\) |
| \(38\) |
\(-4\) |
\(4\) |
| \(32\) |
\(-10\) |
\(10\) |
| \(48\) |
\(6\) |
\(6\) |
| \(33\) |
\(-9\) |
\(9\) |
| \(34\) |
\(-8\) |
\(8\) |
| \(61\) |
\(19\) |
\(19\) |
| \(37\) |
\(-5\) |
\(5\) |
| \(35\) |
\(-7\) |
\(7\) |
| \(59\) |
\(17\) |
\(17\) |
| \(43\) |
\(1\) |
\(1\) |
The average of the absolute value of the deviations from the mean is:
\(\dfrac{4+10+6+9+8+19+5+7+17+1}{10}=8.6\)
Thus, Vishal's commute times have an average absolute deviation from the mean of \(8.6\) minutes. This measure is known as the mean deviation.
Did You Know?
The mean deviation, as calculated in the Try This Revisited, is one measure of spread of a data set, a measure that quantifies if the data is tightly clustered around the mean or widely spread around the mean. A more common measure of spread is called the standard deviation. It is based on \((x-\bar{x})^2\) rather than \(|x-\bar{x}|\), both of which create a list of non-negative numbers.