Section 6.9: Measures of Variation
- Page ID
- 216509
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- Compute the variance of a data set
- Compute the standard deviation of a data set
- Interpret standard deviation
In the previous section, when we collect data, whether it’s test scores, daily temperatures, or the ages of people in a group, we often start by looking at measures of center, like the mean or median. These tell us what a “typical” value looks like. But center alone never tells the whole story. Imagine two classes that both have an average test score of 80:
- In one class, almost everyone scored between 78 and 82.
- In the other, half the class scored above 95 and the other half below 65.
Same average (mean and median are both 80), totally different situations. This is where measures of variation become essential.
The main three measures of variation are called the range, variance, and standard deviation.
Note: Some statistical textbooks may include interquartile range also as a measure of variation.
What Are Measures of Variation?
Measures of variation (also called measures of spread or dispersion) describe how spread out or clustered data values are. They help answer questions like:
- Are most values close to the center?
- Are the values widely scattered?
- Are there extreme values far from the rest?
Understanding variation helps us interpret data more accurately and avoid misleading conclusions.
Why Variation Matters
Even if two datasets share the same mean or median, their variation can differ dramatically. Variation tells us:
- Consistency (e.g., are students' scores steady or all over the place?)
- Risk (e.g., does investment return vary a lot or stay predictable?)
- Reliability (e.g., is a machine producing parts with uniform size?)
In everyday life, low variation often means predictability, and high variation may mean uncertainty or diversity.
The range is the simplest measure of variation. It is calculated by the formula:
Range = Maximum value − Minimum value
It shows the total span of the data but is very sensitive to outliers.
Note: Range was discussed earlier in Section 6.?. Outliers will be discussed further in Section 6.11.
Find the range of the data set: 3, 3, 5, 7, 8, 8, 11, 12, 17
✅ Solution:
Make sure the data set is in ascending order, which it is. Now, use the formula:
Range = Maximum value − Minimum value
Range = 17 − 3
Range = 14
Thus, the range is 14.
Find the range of the data set: 6, 8, 9, 17, 12, 15, 11
✅ Solution:
Make sure the data set is in ascending order, which it is not. Placing the data in ascending order, we have 6, 8, 9, 11, 12, 15, 17. Now, use the formula:
Range = Maximum value − Minimum value
Range = 17 − 6
Range = 11
Thus, the range is 11.
The standard deviation of a data set is a measure of how spread out the values are. It tells you, on average, how far the data values tend to be from the mean. A small standard deviation means the values are close to the mean (low variation). A large standard deviation means the values are more spread out (high variation).
How to find a standard deviation of a sample set:
- Step 1 — List the data. (Ascending order is not necessary).
- Step 2 — Find the mean.
- Step 3 — Subtract the mean from each value in the data set.
- Step 4 — Square each difference.
- Step 5 — Add up all of the squared differences.
- Step 6 — Divide by \(n-1\). (Note: This value is called the variance).
- Step 7 — Take the square root.
The mathematical formula for the steps above is the following:
For a sample data set, the sample standard deviation, \(s\), is defined as:
\[s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}\nonumber\]
where: \(x_i\) = each individual value, \(\bar{x}\) = sample mean, and \(n\) is the sample size.
A great example to understand standard deviation and the role it takes on is looking at gas prices at different gas stations. When comparing gas prices across different gas stations in a city or region, it’s common to find that each station has its own average price (mean). Some stations consistently charge a little less, while others consistently charge a little more. However, even though these mean prices differ, the standard deviations for the stations are often quite similar. This happens because the patterns of pricing tend to move together. A gas station that is usually one of the cheapest will remain one of the cheapest even when prices fluctuate. Likewise, a station that is typically more expensive will continue to be among the higher‑priced stations. In other words, the whole market tends to shift up or down at the same time due to factors like wholesale prices, taxes, and supply conditions. As a result, while the overall level of prices (the mean) may differ from station to station, the amount of variation they experience (the standard deviation) tends to be similar. Prices at each station rise and fall by roughly comparable amounts over time, even if they start from different “baseline” levels.
So although one gas station may average 20 cents cheaper than another, both stations often fluctuate by roughly the same amount day to day or week to week. This shared pattern of movement explains why their standard deviations are similar even when their means are not.
Although standard deviation will become especially important later when we study the normal distribution, our focus in this section is simply on how to calculate it. For now, we treat standard deviation as a numerical summary that tells us how spread out the values in a data set are around the mean. It gives us a single number that reflects the typical distance between each data point and the average. By learning to compute standard deviation now, we build the foundation needed for understanding more advanced ideas, such as how the normal distribution uses standard deviation to describe predictable patterns in real‑world data. At this stage, however, our goal is only to become comfortable with the calculation process and to recognize standard deviation as a basic measure of variation.
Suppose a professor records the number of hours ten students studied for an exam: 3, 4, 5, 6, 6, 6, 7, 11. Round your answer to the nearest hundredth.
✅ Solution:
- Step 1 — List the data. (Ascending order is not necessary): 3, 4, 5, 6, 6, 6, 7, 11
- Step 2 — Find the mean: \(\bar{x} = \frac{\sum x_i}{8} = \frac{x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8}{8}=\frac{3 + 4 + 5 + 6 + 6 + 6 + 7 + 11}{8}=\frac{48}{8}=6\)
- Step 3 — Subtract the mean from each value in the data set: Here, it is easier to do this in a table.
- Step 4 — Square each difference: Here, this is also easier to do this in a table.
| Step 1: Data | Step 3: Subtract the mean | Step 4: Square the difference |
|---|---|---|
| 3 | 3 − 6 = −3 | (−3)2 = 9 |
| 4 | 4 − 6 = −2 | (−2)2 = 4 |
| 5 | 5 − 6 = −1 | (−1)2 = 1 |
| 6 | 6 − 6 = 0 | (0)2 = 0 |
| 6 | 6 − 6 = 0 | (0)2 = 0 |
| 6 | 6 − 6 = 0 | (0)2 = 0 |
| 7 | 7 − 6 = 1 | (1)2 = 1 |
| 11 | 11 − 6 = 5 | (5)2 = 25 |
- Step 5 — Add up all of the squared differences: 9 + 4 + 1 + 0 + 0 + 0 + 1 + 25 = 40
- Step 6 — Divide by n − 1. (Note: This result here is called the variance): \(\frac{40}{8 − 1}=\frac{40}{7}=5.7142857142857\)
- Step 7 — Take the square root: \(\sqrt{5.7142857142857}=2.3904572186688\approx2.39\)
So, the standard deviation is 2.39.
✅ Alternative Solution:
Let's use the mathematical formula for standard deviation of a sample, \(s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}\). If we focus only on the numerator, \((x_i - \bar{x})^2\), we can create the same three column table from above, yet labeling the headers which are the actual steps 1, 3, and 4 in the table. Those steps mathematically are \(x_i\), \((x_i - \bar{x})\), and \((x_i - \bar{x})^2\), respectively.
| \(x_i\) | \((x_i - \bar{x})\) | \((x_i - \bar{x})^2\) |
|---|---|---|
| 3 | 3 − 6 = −3 | (−3)2 = 9 |
| 4 | 4 − 6 = −2 | (−2)2 = 4 |
| 5 | 5 − 6 = −1 | (−1)2 = 1 |
| 6 | 6 − 6 = 0 | (0)2 = 0 |
| 6 | 6 − 6 = 0 | (0)2 = 0 |
| 6 | 6 − 6 = 0 | (0)2 = 0 |
| 7 | 7 − 6 = 1 | (1)2 = 1 |
| 11 | 11 − 6 = 5 | (5)2 = 25 |
| Sum = 40 |
\[s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}\nonumber\]
\[s = \sqrt{\frac{40}{8-1}}\nonumber\]
\[s = \sqrt{\frac{40}{7}}\nonumber\]
\[s = \sqrt{5.7142857142857}\nonumber\]
\[s = 2.3904572186688\nonumber\]
\[s\approx2.39\nonumber\]
So, the standard deviation is 2.39.
Find the standard deviation of the following data set: 5, 7, 10, 5, 9, 9, 8, 2, 7, 8. Find the standard deviation of the data set. Round your answer to the nearest hundredth.
✅ Solution:
- Step 1 — List the data. (Ascending order is not necessary): 5, 7, 10, 5, 9, 9, 8, 2, 7, 8
- Step 2 — Find the mean: \(\bar{x} = \frac{\sum x_i}{10} = \frac{x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10}}{8}=\frac{5 + 7 + 10 + 5 + 9 + 9 + 8 + 2 + 7 + 8}{10}=\frac{70}{10}=7\)
- Step 3 — Subtract the mean from each value in the data set: Here, it is easier to do this in a table.
- Step 4 — Square each difference: Here, this is also easier to do this in a table.
| Step 1: Data | Step 3: Subtract the mean | Step 4: Square the difference |
|---|---|---|
| 5 | 5 − 7 = −2 | (−2)2 = 4 |
| 7 | 7 − 7 = 0 | (0)2 = 0 |
| 10 | 10 − 7 = 3 | (3)2 = 9 |
| 5 | 5 − 7 = −2 | (−2)2 = 4 |
| 9 | 9 − 7 = 2 | (2)2 = 4 |
| 9 | 9 − 7 = 2 | (2)2 = 4 |
| 8 | 8 − 7 = 1 | (1)2 = 1 |
| 2 | 2 − 7 = −5 | (−5)2 = 25 |
| 7 | 7 − 7 = 0 | (0)2 = 0 |
| 8 | 8 − 7 = 1 | (1)2 =1 |
- Step 5 — Add up all of the squared differences: 4 + 0 + 9 + 4 + 4 + 4 + 1 + 25 + 0 + 1 = 52
- Step 6 — Divide by n − 1. (Note: The result here is called the variance): \(\frac{52}{10 − 1}=\frac{52}{9}=5.777777777777778\)
- Step 7 — Take the square root: \(\sqrt{5.77777777}=2.403700850309326\approx2.40\)
So, the standard deviation is 2.40.
✅ Alternative Solution:
Let's use the mathematical formula for standard deviation of a sample, \(s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}\). If we focus only on the numerator, \((x_i - \bar{x})^2\), we can create the same three column table from above, yet labeling the headers which are the actual steps 1, 3, and 4 in the table. Those steps mathematically are \(x_i\), \((x_i - \bar{x})\), and \((x_i - \bar{x})^2\), respectively.
| \(x_i\) | \((x_i - \bar{x})\) | \((x_i - \bar{x})^2\) |
|---|---|---|
| 5 | 5 − 7 = −2 | (−2)2 = 4 |
| 7 | 7 − 7 = 0 | (0)2 = 0 |
| 10 | 10 − 7 = 3 | (3)2 = 9 |
| 5 | 5 − 7 = −2 | (−2)2 = 4 |
| 9 | 9 − 7 = 2 | (2)2 = 4 |
| 9 | 9 − 7 = 2 | (2)2 = 4 |
| 8 | 8 − 7 = 1 | (1)2 = 1 |
| 2 | 2 − 7 = −5 | (−5)2 = 25 |
| 7 | 7 − 7 = 0 | (0)2 = 0 |
| 8 | 8 − 7 = 1 | (1)2 =1 |
| Sum = 52 |
\[s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}\nonumber\]
\[s = \sqrt{\frac{70}{10-1}}\nonumber\]
\[s = \sqrt{\frac{70}{9}}\nonumber\]
\[s = \sqrt{5.777777777777778}\nonumber\]
\[s = 2.403700850309326\nonumber\]
\[s\approx2.39\nonumber\]
So, the standard deviation is 2.40.
Car battery warranties typically offer 1–4 years of free replacement, often followed by a pro-rated period up to 84 months (7 years), with coverage based on purchase date and type. One brand of battery logged the last 15 batteries that came back defective under their pro-rated 84-month warranty. Here is the data on the time for each battery lasted:
51, 37, 60, 62, 55, 71, 42, 65, 68, 74, 40, 64, 68, 59, 22
Find the standard deviation of the data set. Round your answer to the nearest hundredth.
✅ Solution:
- Step 1 — List the data. (Ascending order is not necessary): 51, 37, 60, 62, 55, 71, 42, 65, 68, 74, 40, 64, 68, 59, 22
- Step 2 — Find the mean: \(\bar{x} = \frac{\sum x_i}{15} = \frac{x_1 + x_2 + x_3 + \cdots + x_{15}}{15}=\frac{51 + 37 + 60 + 62 + 55 + 71 + 42 + 65 + 68 + 74 + 40 + 64 + 68 + 59 + 22}{15}=\frac{838}{15}=55.86666666666667\approx55.87\)
- Step 3 — Subtract the mean from each value in the data set: Here, it is easier to do this in a table.
- Step 4 — Square each difference: Here, this is also easier to do this in a table.
| Step 1: Data | Step 3: Subtract the mean | Step 4: Square the difference |
|---|---|---|
| 51 | 51 − 55.87 = −4.87 | (−4.87)2 = 23.7169 |
| 37 | 37 − 55.87 = −18.87 | (−18.87)2 = 356.0769 |
| 60 | 60 − 55.87 = −4.13 | (−4.13)2 = 17.0569 |
| 62 | 62 − 55.87 = 6.13 | (6.13)2 = 37.5769 |
| 55 | 55 − 55.87 = −0.87 | (−0.87)2 = 0.7569 |
| 71 | 71 − 55.87 = 15.13 | (15.13)2 = 228.9169 |
| 42 | 42 − 55.87 = −13.87 | (−13.87)2 = 192.3769 |
| 65 | 65 − 55.87 = 9.13 | (9.13)2 = 83.3569 |
| 68 | 68 − 55.87 = 12.13 | (12.13)2 = 147.1369 |
| 74 | 74 − 55.87 = 18.13 | (18.13)2 = 328.6969 |
| 40 | 40 − 55.87 = −15.87 | (−15.87)2 = 251.8569 |
| 64 | 64 − 55.87 = 8.13 | (8.13)2 = 66.0969 |
| 68 | 68 − 55.87 = 12.13 | (12.13)2 = 147.1369 |
| 59 | 59 − 55.87 = 3.13 | (3.13)2 = 9.7969 |
| 22 | 22 − 55.87 = −33.87 | (−33.87)2 = 1,147.1769 |
- Step 5 — Add up all of the squared differences: The sum of the last column is 3,037.7335
- Step 6 — Divide by n − 1. (Note: This result here is called the variance): \(\frac{3,037.7335}{15 − 1}=\frac{3,037.7335}{14}=216.98096428571\)
- Step 7 — Take the square root: \(\sqrt{216.98096428571}=14.730273734242\approx14.73\)
So, the standard deviation is 14.73.
✅ Alternative Solution:
Let's use the mathematical formula for standard deviation of a sample, \(s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}\). If we focus only on the numerator, \((x_i - \bar{x})^2\), we can create the same three column table from above, yet labeling the headers which are the actual steps 1, 3, and 4 in the table. Those steps mathematically are \(x_i\), \((x_i - \bar{x})\), and \((x_i - \bar{x})^2\), respectively.
| \(x_i\) | \((x_i - \bar{x})\) | \((x_i - \bar{x})^2\) |
|---|---|---|
| 51 | 51 − 55.87 = −4.87 | (−4.87)2 = 23.7169 |
| 37 | 37 − 55.87 = −18.87 | (−18.87)2 = 356.0769 |
| 60 | 60 − 55.87 = −4.13 | (−4.13)2 = 17.0569 |
| 62 | 62 − 55.87 = 6.13 | (6.13)2 = 37.5769 |
| 55 | 55 − 55.87 = −0.87 | (−0.87)2 = 0.7569 |
| 71 | 71 − 55.87 = 15.13 | (15.13)2 = 228.9169 |
| 42 | 42 − 55.87 = −13.87 | (−13.87)2 = 192.3769 |
| 65 | 65 − 55.87 = 9.13 | (9.13)2 = 83.3569 |
| 68 | 68 − 55.87 = 12.13 | (12.13)2 = 147.1369 |
| 74 | 74 − 55.87 = 18.13 | (18.13)2 = 328.6969 |
| 40 | 40 − 55.87 = −15.87 | (−15.87)2 = 251.8569 |
| 64 | 64 − 55.87 = 8.13 | (8.13)2 = 66.0969 |
| 68 | 68 − 55.87 = 12.13 | (12.13)2 = 147.1369 |
| 59 | 59 − 55.87 = 3.13 | (3.13)2 = 9.7969 |
| 22 | 22 − 55.87 = −33.87 | (−33.87)2 = 1,147.1769 |
| Sum = 3,037.7335 |
\[s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}\nonumber\]
\[s = \sqrt{\frac{3,037.7335}{15-1}}\nonumber\]
\[s = \sqrt{\frac{3,037.7335}{14}}\nonumber\]
\[s = \sqrt{216.98096428571}\nonumber\]
\[s = 14.730273734242\nonumber\]
\[s\approx14.73\nonumber\]
So, the standard deviation is 14.73.
Variance of a sample data set, like standard deviation, measures how far values tend to be from the mean on average. However, variance measures the average squared distance from the mean. Because the variance uses squared distances, this makes it sensitive to large deviations. The sample variance is simply the square of the sample standard deviation, or
\[s^2=\frac{\sum (x - \bar{x})^2}{n - 1}\nonumber\]
Note: This is the same formula for sample standard deviation without the square root.
Refer to Example #6.9.3. Suppose a professor records the number of hours ten students studied for an exam: 3, 4, 5, 6, 6, 6, 7, 11. Find the variances of the data set. Round your answer to the nearest hundredth.
✅ Solution:
So, to find the variance of a data set, just perform the first six steps for standard deviation. Do not perform Step 7. Do not take the square root. When you are finished with Step 6, that value is the variance.
- Step 1 — List the data. (Ascending order is not necessary): 3, 4, 5, 6, 6, 6, 7, 11
- Step 2 — Find the mean: \(\bar{x} = \frac{\sum x_i}{8} = \frac{x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8}{8}=\frac{3 + 4 + 5 + 6 + 6 + 6 + 7 + 11}{8}=\frac{48}{8}=6\)
- Step 3 — Subtract the mean from each value in the data set: Here, it is easier to do this in a table.
- Step 4 — Square each difference: Here, this is also easier to do this in a table.
| Step 1: Data | Step 3: Subtract the mean | Step 4: Square the difference |
|---|---|---|
| 3 | 3 − 6 = −3 | (−3)2 = 9 |
| 4 | 4 − 6 = −2 | (−2)2 = 4 |
| 5 | 5 − 6 = −1 | (−1)2 = 1 |
| 6 | 6 − 6 = 0 | (0)2 = 0 |
| 6 | 6 − 6 = 0 | (0)2 = 0 |
| 6 | 6 − 6 = 0 | (0)2 = 0 |
| 7 | 7 − 6 = 1 | (1)2 = 1 |
| 11 | 11 − 6 = 5 | (5)2 = 25 |
- Step 5 — Add up all of the squared differences: 9 + 4 + 1 + 0 + 0 + 0 + 1 + 25 = 40
- Step 6 — Divide by n − 1. \(\frac{40}{8 − 1}=\frac{40}{7}=5.7142857142857\approx5.71\)
So, the variance is 5.71.
✅ Alternative Solution:
Let's use the mathematical formula for standard deviation of a sample, \(s^2 = {\frac{\sum(x_i - \bar{x})^2}{n-1}}\). If we focus only on the numerator, \((x_i - \bar{x})^2\), we can create the same three column table from above, yet labeling the headers which are the actual steps 1, 3, and 4 in the table. Those steps mathematically are \(x_i\), \((x_i - \bar{x})\), and \((x_i - \bar{x})^2\), respectively.
| \(x_i\) | \((x_i - \bar{x})\) | \((x_i - \bar{x})^2\) |
|---|---|---|
| 3 | 3 − 6 = −3 | (−3)2 = 9 |
| 4 | 4 − 6 = −2 | (−2)2 = 4 |
| 5 | 5 − 6 = −1 | (−1)2 = 1 |
| 6 | 6 − 6 = 0 | (0)2 = 0 |
| 6 | 6 − 6 = 0 | (0)2 = 0 |
| 6 | 6 − 6 = 0 | (0)2 = 0 |
| 7 | 7 − 6 = 1 | (1)2 = 1 |
| 11 | 11 − 6 = 5 | (5)2 = 25 |
| Sum = 40 |
\[s^2 = {\frac{\sum(x_i - \bar{x})^2}{n-1}}\nonumber\]
\[s^2 = {\frac{40}{8-1}}\nonumber\]
\[s^2 = {\frac{40}{7}}\nonumber\]
\[s^2 = {5.7142857142857}\nonumber\]
\[s^2\approx5.71\nonumber\]
So, the variance is 5.71.
If the variance of a sample data set is known to be 100, then what is the standard deviation?
✅ Solution:
Since the variance is given, the standard deviation is the square root of the variance. So, \(\sqrt{100}=10\).
The standard deviation is 10.
If the standard deviation of a sample data set is known to be 8, then what is the variance?
✅ Solution:
Since the standard deviation is given, the variance is the square of the standard deviation. So, \(8^2=64\).
The variance is 8.
If the variance of a sample data set is known to be 7, then what is the standard deviation?
✅ Solution:
Since the variance is given, the standard deviation is the square root of the variance. So, \(\sqrt7=2.6457513110646\approx2.65\).
The standard deviation is 2.65.
If the standard deviation of a sample data set is known to be 1.2, then what is the variance?
✅ Solution:
Since the standard deviation is given, the variance is the square of the standard deviation. So, \(1.2^2=1.44\).
The variance is 1.44.
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Find the range of the data set: 2, 5, 5, 5, 6, 8, 12, 15
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Find the range of the data set: 5, 4, 3, 11, 12, 13
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Find the standard deviation and variance of the following sample data set: 7, 6, 8, 5, 6, 7, 8, 9
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Ten students flip a coin 100 times and record the number of times it lands on heads. The results for the number of heads is 44, 49, 52, 62, 53, 48, 54, 49, 46, 51. Find the standard deviation and variance of the data set. Round your answer to the nearest hundredth.
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If the variance of a sample data set is known to be 4, then what is the standard deviation?
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If the standard deviation of a sample data set is known to be 9, then what is the variance?
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PLACE THIS IN HOMEWORK: Winter Olympic Games.
| Year | Location | No. of Gold Medals |
|---|---|---|
| 1924 | Chamonix, France | 2 |
| 1928 | St. Moritz, Switzerland | 2 |
| 1932 | Lake Placid, New York, USA | 4 |
| 1936 | Garmisch-Partenkirchen, Germany | 0 |
| 1948 | St. Moritz, Switzerland | 4 |
| 1952 | Oslo, Norway | 6 |
| 1956 | Cortina d'Ampezzo, Italy | 3 |
| 1960 | Squaw Valley, California, USA | 4 |
| 1964 | Innsbruck, Austria | 2 |
| 1968 | Grenoble, France | 5 |
| 1972 | Sapporo, Jappan | 2 |
| 1976 | Innsbruck, Austria | 3 |
| 1980 | Lake Placid, New York USA | 4 |
| 1984 | Sarajevo, Yugoslavia | 4 |
| 1988 | Calgary, Alberta, Canada | 1 |
| 1992 | Albertville, France | 4 |
| 1994 | Lillehammer, Norway | 5 |
| 1998 | Nagano, Japan | 3 |
| 2002 | Salt Lake City, Utah, USA | 13 |
| 2006 | Turin, Italy | 9 |
| 2010 | Vancouver, British Columbia, Canada | 15 |
| 2014 | Sochi, Russia | 9 |
| 2018 | Pyeongchang, South Korea | 8 |
| 2022 | Beijing, China | 9 |
Find the mean, median, mode, and midrange of the magnitude ratings.
- Answers
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- Range = 13.
- Mean = 10.
- Median = 6.5, Mode = 5, 6, 7 & 8.
- Median = 19, Mode = None.
- Mean = 7.15 , Median = 7.15, Mode = 7., Midrange = 7.15.