Chapter 2: Descriptive Statistics
2.7 Measures of the Spread of the Data
Learning Objectives
By the end of this section, the student should be able to:
- Recognize, describe, and calculate the measures of the spread of data: variance, standard deviation, and range.
- Use the mean and the standard deviation to calculate z-scores and the empirical rule.
An important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean; in other data sets, the data values are more widely spread out from the mean. The most common measure of variation, or spread, is the standard deviation. The standard deviation is a number that measures how far data values are from their mean.
The Standard Deviation
- provides a numerical measure of the overall amount of variation in a data set, and
- can be used to determine whether a particular data value is close to or far from the mean.
The standard deviation provides a measure of the overall variation in a data set.
The standard deviation is always positive or zero. The standard deviation is small when the data are all concentrated close to the mean, exhibiting little variation or spread. The standard deviation is larger when the data values are more spread out from the mean, exhibiting more variation.
Suppose that we are studying the amount of time customers wait in line at the checkout at supermarket A and supermarket B. the average wait time at both supermarkets is five minutes. At supermarket A, the standard deviation for the wait time is two minutes; at supermarket B the standard deviation for the wait time is four minutes.
Because supermarket B has a higher standard deviation, we know that there is more variation in the wait times at supermarket B. Overall, wait times at supermarket B are more spread out from the average; wait times at supermarket A are more concentrated near the average.
The standard deviation can be used to determine whether a data value is close to or far from the mean.
Suppose that Rosa and Binh both shop at supermarket A. Rosa waits at the checkout counter for seven minutes and Binh waits for one minute. At supermarket A, the mean waiting time is five minutes and the standard deviation is two minutes. The standard deviation can be used to determine whether a data value is close to or far from the mean.
Rosa waits for seven minutes:
- Seven is two minutes longer than the average of five; two minutes is equal to one standard deviation.
- Rosa's wait time of seven minutes is two minutes longer than the average of five minutes.
- Rosa's wait time of seven minutes is one standard deviation above the average of five minutes.
Binh waits for one minute.
- One is four minutes less than the average of five; four minutes is equal to two standard deviations.
- Binh's wait time of one minute is four minutes less than the average of five minutes.
- Binh's wait time of one minute is two standard deviations below the average of five minutes.
- A data value that is two standard deviations from the average is just on the borderline for what many statisticians would consider to be far from the average. Considering data to be far from the mean if it is more than two standard deviations away is more of an approximate "rule of thumb" than a rigid rule. In general, the shape of the distribution of the data affects how much of the data is further away than two standard deviations. (You will learn more about this in later chapters.)
A number line may help you understand standard deviation. If we were to put five and seven on a number line, seven is to the right of five. We say, then, that seven is one standard deviation to the right of five because [latex]5 + (1)(2) = 7[/latex].
If one were also part of the data set, then one is two standard deviations to the left of five because [latex]5 + (–2)(2) = 1[/latex].
- In general, a value = mean + (#ofSTDEV)(standard deviation)
- where #ofSTDEVs = the number of standard deviations
- #ofSTDEV does not need to be an integer
- One is two standard deviations less than the mean of five because: [latex]1 = 5 + (–2)(2)[/latex].
The equation value = mean + (# of STDEVs)(standard deviation) can be expressed for a sample and for a population.
- sample: [latex]x = \overline{x} + (\text{\# of STDEV}) \cdot (s)[/latex]
- Population: [latex]x = \mu + (\text{\# of STDEV}) \cdot (\sigma)[/latex]
The lower case letter [latex]s[/latex] represents the sample standard deviation and the Greek letter [latex]\sigma[/latex] (sigma, lower case) represents the population standard deviation.
The symbol [latex]\overline{x}[/latex] is the sample mean and the Greek symbol [latex]\mu[/latex] is the population mean.
Calculating the Standard Deviation
If [latex]x[/latex] is a number, then the difference "x – mean" is called its deviation. In a data set, there are as many deviations as there are items in the data set. The deviations are used to calculate the standard deviation. If the numbers belong to a population, in symbols a deviation is [latex]x – \mu[/latex]. For sample data, in symbols a deviation is [latex]x - \overline{x}[/latex].
The procedure to calculate the standard deviation depends on whether the numbers are the entire population or are data from a sample. The calculations are similar, but not identical. Therefore the symbol used to represent the standard deviation depends on whether it is calculated from a population or a sample. The lower case letter [latex]s[/latex] represents the sample standard deviation and the Greek letter [latex]\sigma[/latex] (sigma, lower case) represents the population standard deviation. If the sample has the same characteristics as the population, then [latex]s[/latex] should be a good estimate of [latex]\sigma[/latex].
To calculate the standard deviation, we need to calculate the variance first. The variance is the average of the squares of the deviations (the [latex]x - \overline{x}[/latex] values for a sample, or the [latex]x – \mu[/latex] values for a population). The symbol [latex]{\sigma}^{2}[/latex] represents the population variance; the population standard deviation [latex]\sigma[/latex] is the square root of the population variance. The symbol [latex]s^2[/latex] represents the sample variance; the sample standard deviation [latex]s[/latex] is the square root of the sample variance. You can think of the standard deviation as a special average of the deviations.
If the numbers come from a census of the entire population and not a sample, when we calculate the average of the squared deviations to find the variance, we divide by [latex]N[/latex], the number of items in the population. If the data are from a sample rather than a population, when we calculate the average of the squared deviations, we divide by [latex]n – 1[/latex], one less than the number of items in the sample.
Formulas for the Sample Standard Deviation
- [latex]s=\sqrt{\frac{\Sigma {\left(x-\overline{x}\right)}^{2}}{n-1}}[/latex] or [latex]s=\sqrt{\frac{\Sigma f{\left(x-\overline{x}\right)}^{2}}{n-1}}[/latex]
- For the sample standard deviation, the denominator is [latex]n - 1[/latex], that is the sample size MINUS 1.
Formulas for the Population Standard Deviation
- [latex]\sigma=\sqrt{\frac{\Sigma{\left(x-\mu\right)}^{2}}{N}}[/latex] or [latex]\sigma = \sqrt{\frac{\Sigma f{\left(x–\mu\right)}^{2}}{N}}[/latex]
- For the population standard deviation, the denominator is [latex]N[/latex], the number of items in the population.
In these formulas, [latex]f[/latex] represents the frequency with which a value appears. For example, if a value appears once, [latex]f[/latex] is one. If a value appears three times in the data set or population, [latex]f[/latex] is three.
Note
It is important to concentrate on two things with the standard deviation:
- Whether or not you need to find [latex]s[/latex], the sample standard deviation, or [latex]\sigma[/latex], the population standard deviation.
- Correctly using and interpreting the information that the standard deviation gives us.
Sampling Variability of a Statistic
The statistic of a sampling distribution was discussed in Descriptive Statistics: Measuring the Center of the Data. How much the statistic varies from one sample to another is known as the sampling variability of a statistic. You typically measure the sampling variability of a statistic by its standard error. The standard error of the mean is an example of a standard error. It is a special standard deviation and is known as the standard deviation of the sampling distribution of the mean. You will cover the standard error of the mean in the chapter The Central Limit Theorem (not now). The notation for the standard error of the mean is [latex]\frac{\sigma }{\sqrt{n}}[/latex] where [latex]\sigma[/latex] is the standard deviation of the population and n is the size of the sample.
Example
In a fifth-grade class, the teacher was interested in the average age and the sample standard deviation of the ages of her students. The following data are the ages for a SAMPLE of [latex]n = 20[/latex] fifth grade students. The ages are rounded to the nearest half year:
9; 9.5; 9.5; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5
[latex]\overline{x}=\frac{9 + 9.5(2) + 10(4) + 10.5(4) + 11(6) + 11.5(3)}{20}=10.525[/latex]
**Note, the method shown above is using the frequencies to calculate the mean. Adding all the values individually is also an option.
The average age is 10.53 years, rounded to two places.
The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating [latex]s[/latex].
| Data | Freq. | Deviations | Deviations2 | (Freq.)(Deviations2) |
|---|---|---|---|---|
| x | f | (x – [latex]\overline{x}[/latex]) | (x – [latex]\overline{x}[/latex])2 | (f)(x – [latex]\overline{x}[/latex])2 |
| 9 | 1 | 9 – 10.525 = –1.525 | (–1.525)2 = 2.325625 | 1 × 2.325625 = 2.325625 |
| 9.5 | 2 | 9.5 – 10.525 = –1.025 | (–1.025)2 = 1.050625 | 2 × 1.050625 = 2.101250 |
| 10 | 4 | 10 – 10.525 = –0.525 | (–0.525)2 = 0.275625 | 4 × 0.275625 = 1.1025 |
| 10.5 | 4 | 10.5 – 10.525 = –0.025 | (–0.025)2 = 0.000625 | 4 × 0.000625 = 0.0025 |
| 11 | 6 | 11 – 10.525 = 0.475 | (0.475)2 = 0.225625 | 6 × 0.225625 = 1.35375 |
| 11.5 | 3 | 11.5 – 10.525 = 0.975 | (0.975)2 = 0.950625 | 3 × 0.950625 = 2.851875 |
| The total is 9.7375 |
The sample variance, [latex]s^2[/latex], is equal to the sum of the last column (9.7375) divided by the total number of data values minus one [latex](20 – 1)[/latex]:
[latex]{s}^{2}=\frac{9.7375}{20-1}=0.5125[/latex]
The sample standard deviation s is equal to the square root of the sample variance:
[latex]s=\sqrt{0.5125}=0.715891,[/latex] which is rounded to two decimal places, [latex]s=0.72[/latex].
Now, recall that value = mean + (#ofSTDEVs)(standard deviation).
- For a sample: [latex]x = \overline{x} + \text{(\#ofSTDEVs)(s)}[/latex]
- For a population: [latex]x= \mu + \text{(\#ofSTDEVs)(} \sigma \text{)}[/latex]
- Find the value that is one standard deviation above the mean. Find ([latex]\overline{x} + 1s[/latex]).
- Find the value that is two standard deviations below the mean. Find ([latex]\overline{x} -2s[/latex]).
- Find the values that are 1.5 standard deviations from (below and above) the mean.
Solution
- [latex](\overline{x} + 1s) = 10.53 + (1)(0.72) = 11.25[/latex]
- [latex](\overline{x} – 2s) = 10.53 – (2)(0.72) = 9.09[/latex]
- One standard deviation below the mean is [latex](\overline{x} – 1.5s) = 10.53 – (1.5)(0.72) = 9.45[/latex]. One standard deviation below the mean is [latex](\overline{x}+ 1.5s) = 10.53 + (1.5)(0.72) = 11.61[/latex].
Your Turn!
On a baseball team, the ages of each of the players are as follows:
21; 21; 22; 23; 24; 24; 25; 25; 28; 29; 29; 31; 32; 33; 33; 34; 35; 36; 36; 36; 36; 38; 38; 38; 40
Find the mean and standard deviation. Then find the value that is two standard deviations above the mean.
Solution
Mean: [latex]\overline{x} = 30.68[/latex]
Standard Deviation: [latex]s = 6.09[/latex]
Two standard deviations above the mean: [latex]( \overline{x} + 2s) = 30.68+ (2)(6.09) = 42.86[/latex]
Explanation of the standard deviation calculation shown in the table
The deviations show how spread out the data are about the mean. The data value 11.5 is farther from the mean than is the data value 11 which is indicated by the deviations 0.97 and 0.47. A positive deviation occurs when the data value is greater than the mean, whereas a negative deviation occurs when the data value is less than the mean. The deviation is –1.525 for the data value nine. If you add the deviations, the sum is always zero. (For the fifth grade teacher example, there are [latex]n = 20[/latex] deviations.) So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you make them positive numbers, and the sum will also be positive. The variance, then, is the "average" squared deviation.
The variance is a squared measure and does not have the same units as the data. Taking the square root solves the problem. The standard deviation measures the spread in the same units as the data.
Notice that instead of dividing by [latex]n = 20[/latex], the calculation is divided by [latex]n – 1 = 20 – 1 = 19[/latex] because the data is a sample. For the sample variance, we divide by the sample size minus one [latex](n – 1)[/latex]. Why not divide by [latex]n[/latex]? The answer has to do with the population variance. The sample variance is an estimate of the population variance. Based on the theoretical mathematics that lies behind these calculations, dividing by [latex](n – 1)[/latex] gives a better estimate of the population variance.
Note
Your concentration should be on what the standard deviation tells us about the data. The standard deviation is a number which measures how far the data are spread from the mean.
The standard deviation, [latex]s[/latex] or [latex]\sigma[/latex], is either zero or larger than zero. When the standard deviation is zero, there is no spread; that is, all the data values are equal to each other. The standard deviation is small when the data are all concentrated close to the mean, and is larger when the data values show more variation from the mean. When the standard deviation is a lot larger than zero, the data values are very spread out about the mean; outliers can make [latex]s[/latex] or [latex]\sigma[/latex] very large.
The standard deviation, when first presented, can seem unclear. By graphing your data, you can get a better "feel" for the deviations and the standard deviation. You will find that in symmetrical distributions, the standard deviation can be very helpful but in skewed distributions, the standard deviation may not be much help. The reason is that the two sides of a skewed distribution have different spreads. In a skewed distribution, it is better to look at the first quartile, the median, the third quartile, the smallest value, and the largest value. Because numbers can be confusing, always graph your data. Display your data in a histogram or a box plot.
Example
Use the following data (first exam scores) from Susan Dean's spring pre-calculus class:
33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100
- Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places.
- Calculate the following to one decimal place:
- The sample mean
- The sample standard deviation
- The median
- The first quartile
- The third quartile
- IQR
- Construct a box plot and a histogram on the same set of axes. Make comments about the box plot, the histogram, and the chart.
Solution
- The chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places is shown below.
- The solutions are below:
- The sample mean = 73.5
- The sample standard deviation = 17.9
- The median = 73
- The first quartile = 61
- The third quartile = 90
- IQR = 90 – 61 = 29
- The x-axis goes from 32.5 to 100.5; y-axis goes from –2.4 to 15 for the histogram. The number of intervals is five, so the width of an interval is (100.5 – 32.5) divided by five, is equal to 13.6. Endpoints of the intervals are as follows: the starting point is 32.5, 32.5 + 13.6 = 46.1, 46.1 + 13.6 = 59.7, 59.7 + 13.6 = 73.3, 73.3 + 13.6 = 86.9, 86.9 + 13.6 = 100.5 = the ending value. No data values fall on an interval boundary.
| Data | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 33 | 1 | 0.032 | 0.032 |
| 42 | 1 | 0.032 | 0.064 |
| 49 | 2 | 0.065 | 0.129 |
| 53 | 1 | 0.032 | 0.161 |
| 55 | 2 | 0.065 | 0.226 |
| 61 | 1 | 0.032 | 0.258 |
| 63 | 1 | 0.032 | 0.29 |
| 67 | 1 | 0.032 | 0.322 |
| 68 | 2 | 0.065 | 0.387 |
| 69 | 2 | 0.065 | 0.452 |
| 72 | 1 | 0.032 | 0.484 |
| 73 | 1 | 0.032 | 0.516 |
| 74 | 1 | 0.032 | 0.548 |
| 78 | 1 | 0.032 | 0.580 |
| 80 | 1 | 0.032 | 0.612 |
| 83 | 1 | 0.032 | 0.644 |
| 88 | 3 | 0.097 | 0.741 |
| 90 | 1 | 0.032 | 0.773 |
| 92 | 1 | 0.032 | 0.805 |
| 94 | 4 | 0.129 | 0.934 |
| 96 | 1 | 0.032 | 0.966 |
| 100 | 1 | 0.032 | 0.998 |
The long left whisker in the box plot is reflected in the left side of the histogram. The spread of the exam scores in the lower 50% is greater ([latex]73 – 33 = 40[/latex]) than the spread in the upper 50% ([latex]100 – 73 = 27[/latex]). The histogram, box plot, and chart all reflect this. There are a substantial number of A and B grades (80s, 90s, and 100). The histogram clearly shows this. The box plot shows us that the middle 50% of the exam scores (IQR = 29) are Ds, Cs, and Bs. The box plot also shows us that the lower 25% of the exam scores are Ds and Fs.
Your Turn!
The following data show the different types of pet food stores in the area carry.
6; 6; 6; 6; 7; 7; 7; 7; 7; 8; 9; 9; 9; 9; 10; 10; 10; 10; 10; 11; 11; 11; 11; 12; 12; 12; 12; 12; 12;
Calculate the sample mean and the sample standard deviation to one decimal place.
Solution
[latex]\overline{x}=9.3[/latex]
[latex]s = 2.2[/latex]
Standard Deviation of Grouped Frequency Tables
Recall that for grouped data we do not know individual data values, so we cannot describe the typical value of the data with precision. In other words, we cannot find the exact mean, median, or mode. We can, however, determine the best estimate of the measures of center by finding the mean of the grouped data with the formula:
[latex]\text{Mean of Frequency Table}=\frac{\sum fm}{\sum f}[/latex] where [latex]f =[/latex] the frequency of the interval and [latex]m =[/latex] the midpoint of the interval.
Just as we could not find the exact mean, neither can we find the exact standard deviation. Remember that standard deviation describes numerically the expected deviation a data value has from the mean. In simple English, the standard deviation allows us to compare how “unusual” individual data is compared to the mean.
Example
Use the table below to discuss what the standard deviation tells us about the data for the first class.
| Class | Frequency, f | Midpoint, m | m2 | [latex]\overline{x}[/latex] | fm2 | Standard Deviation |
|---|---|---|---|---|---|---|
| 0–2 | 1 | 1 | 1 | 7.58 | 1 | 3.5 |
| 3–5 | 6 | 4 | 16 | 7.58 | 96 | 3.5 |
| 6–8 | 10 | 7 | 49 | 7.58 | 490 | 3.5 |
| 9–11 | 7 | 10 | 100 | 7.58 | 700 | 3.5 |
| 12–14 | 0 | 13 | 169 | 7.58 | 0 | 3.5 |
| 15–17 | 2 | 16 | 256 | 7.58 | 512 | 3.5 |
Solution
For this data set, we have the mean, [latex]\overline{x} = 7.8[/latex] and the standard deviation, [latex]s_x = 3.5[/latex]. This means that a randomly selected data value would be expected to be 3.5 units from the mean. If we look at the first class, we see that the class midpoint is equal to one. This is almost two full standard deviations from the mean since [latex]7.58 – 3.5 – 3.5 = 0.58[/latex].
Your Turn!
Find the standard deviation for the data in the table below.
| Class | Frequency, f | Midpoint, m | fm | [latex]\overline{x}[/latex] | [latex]m-\overline{x}[/latex] | [latex](m-\overline{x})^{2}[/latex] | [latex]f(m-\overline{x})^{2}[/latex] |
|---|---|---|---|---|---|---|---|
| 0–2 | 1 | 1 | 1 | 7.58 | -6.58 | 43.2964 | 43.2964 |
| 3–5 | 6 | 4 | 24 | 7.58 | -3.58 | 12.8164 | 76.8984 |
| 6–8 | 10 | 7 | 70 | 7.58 | -0.58 | 0.3364 | 3.364 |
| 9–11 | 7 | 10 | 70 | 7.58 | 2.42 | 5.8564 | 40.9948 |
| 12–14 | 0 | 13 | 0 | 7.58 | 5.42 | 29.3764 | 0 |
| 15–17 | 2 | 16 | 32 | 7.58 | 8.42 | 70.8964 | 141.7928 |
| SUM [latex]\Sigma[\latex] | 26 | 197 | 43.2964 |
Solution
The values in the second, third, and fourth columns of the table are used to calculate the mean of the grouped frequency table, the value in the fifth column. [latex]\overline{x}=\frac{\sum fm}{\sum f}=\frac{197}{26}\approx7.58[/latex]
After calculating [latex]\overline{x}[/latex], find the difference, [latex]m-\overline{x}[/latex], for each midpoint, m. Next, square each difference. In the final column, calculate the product of the frequency and the squared difference for each class. The table makes it easy to use the formula for calculating the standard deviation of a grouped frequency table: [latex]{s}_{x}=\sqrt{\frac{f{\left(m-\overline{x}\right)}^{2}}{n-1}}=\sqrt{\frac{43.2964}{26-1}\approx3.50[/latex]
While the formula for calculating the standard deviation is not complicated, the calculations are tedious. It is important you are careful, so you don't make mistakes.
Comparing Values from Different Data Sets
The standard deviation is useful when comparing data values that come from different data sets. If the data sets have different means and standard deviations, then comparing the data values directly can be misleading.
- For each data value, calculate how many standard deviations away from its mean the value is.
- Use the formula: [latex]\text{value} = \text{mean} + (\text{\#ofSTDEVs})(\text{standard deviation})[/latex]; solve for #ofSTDEVs.
- [latex]\text{\#ofSTDEVs} = \frac{\text{value – mean}}{\text{standard deviation}}[/latex]
- Compare the results of this calculation.
latex]\text{\#ofSTDEVs}[/latex] is often called a "z-score"; we can use the symbol z. In symbols, the formulas become:
| Sample | [latex]x= \overline{x} + zs[/latex] | [latex]z=\frac{x\text{ }-\text{ }\overline{x}}{s}[/latex] |
|---|---|---|
| Population | [latex]x = \mu + z \sigma[/latex] | [latex]z=\frac{x\text{ }-\text{ }\mu }{\sigma }[/latex] |
Example
Two students, John and Ali, from different high schools, wanted to find out who had the highest GPA when compared to his school. Which student had the highest GPA when compared to his school?
| Student | GPA | School Mean GPA | School Standard Deviation |
|---|---|---|---|
| John | 2.85 | 3.0 | 0.7 |
| Ali | 77 | 80 | 10 |
Solution
For each student, determine how many standard deviations (#ofSTDEVs) his GPA is away from the average, for his school. Pay careful attention to signs when comparing and interpreting the answer.
[latex]z= \text{\#ofSTDEVs} = \frac{\text{value }–\text{mean}}{\text{standard deviation}}=\frac{x-\mu }{\sigma }[/latex]
For John, [latex]z = \text{\#ofSTDEVs}=\frac{2.85–3.0}{0.7}=–0.21[/latex]
For Ali, [latex]z= \text{\#ofSTDEVs} = \frac{77-80}{10}=-0.3[/latex]
John has the better GPA when compared to his school because his GPA is 0.21 standard deviations below his school's mean while Ali's GPA is 0.3 standard deviations below his school's mean.
John's z-score of –0.21 is higher than Ali's z-score of –0.3. For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school.
Your Turn!
Two swimmers, Angie and Beth, from different teams, wanted to find out who had the fastest time for the 50 meter freestyle when compared to her team. Which swimmer had the fastest time when compared to her team?
| Swimmer | Time (seconds) | Team Mean Time | Team Standard Deviation |
|---|---|---|---|
| Angie | 26.2 | 27.2 | 0.8 |
| Beth | 27.3 | 30.1 | 1.4 |
Solution
For Angie: [latex]z=\frac{\text{26}\text{.2 – 27}\text{.2}}{\text{0}\text{.8}}=-1.25[/latex]
For Beth: [latex]z=\frac{\text{27}\text{.3}–\text{30}\text{.1}}{1.\text{4}}=-2[/latex]
The following lists give a few facts that provide a little more insight into what the standard deviation tells us about the distribution of the data.
- At least 75% of the data is within two standard deviations of the mean.
- At least 89% of the data is within three standard deviations of the mean.
- At least 95% of the data is within 4.5 standard deviations of the mean.
- This is known as Chebyshev's Rule.
- Approximately 68% of the data is within one standard deviation of the mean.
- Approximately 95% of the data is within two standard deviations of the mean.
- More than 99% of the data is within three standard deviations of the mean.
- This is known as the Empirical Rule.
- It is important to note that this rule only applies when the shape of the distribution of the data is bell-shaped and symmetric. We will learn more about this when studying the "Normal" or "Gaussian" probability distribution in later chapters.
Coefficient of Variation
Another useful way to compare distributions besides simple comparisons of means or standard deviations is to adjust for differences in the scale of the data being measured. Quite simply, a large variation in data with a large mean is different than the same variation in data with a small mean. To adjust for the scale of the underlying data the Coefficient of Variation (CV) has been developed. Mathematically: [latex]\text{CV}=\frac{s}{\overline{x}}*100[/latex] conditioned upon [latex]\overline{x} ≠ 0[/latex],where [latex]𝑠[/latex] is the standard deviation of the data and [latex]\overline{x}[/latex] is the mean.
We can see that this measures the variability of the underlying data as a percentage of the mean value; the center weight of the data set. This measure is useful in comparing risk where an adjustment is warranted because of differences in scale of two data sets. In effect, the scale is changed to common scale, percentage differences, and allows direct comparison of the two or more magnitudes of variation of different data sets.
Videos
Below are helpful videos for the content covered in this Section. Videos are provided from YouTube.
- How to Calculate the Range, Midrange, and Coefficient of Range
- How to Calculate the Standard Deviation
- Standard Deviation and Coefficient of Variation
- How to Calculate Variance
- Standard Deviation of Grouped Data
- Mean Absolute Deviation
- How to Calculate the Average Deviation
- How to Find the Range of a Data Set
- Describing Quantitative Data: Range and Standard Deviation
- Ex: Mode, Mid-range, Range, Standard Deviation Using Range Rule of Thumb
- Ex: Interpret the Mean and Standard Deviation of Two Data Sets
- Ex: Calculate the Sample Standard Deviation
Section 2.7 Review
The standard deviation can help you calculate the spread of data. There are different equations to use if you are calculating the standard deviation of a sample or of a population.
- The Standard Deviation allows us to compare individual data or classes to the data set mean numerically.
- [latex]s=\sqrt{\frac{{\sum }^{\text{}}{\left(x-\overline{x}\right)}^{2}}{n-1}}[/latex] or [latex]s=\sqrt{\frac{{\sum }^{\text{}}f{\left(x-\overline{x}\right)}^{2}}{n-1}}[/latex] is the formula for calculating the standard deviation of a sample. To calculate the standard deviation of a population, we would use the population mean, [latex]\mu[/latex], and the formula [latex]\sigma = \sqrt{\frac{{\sum }^{\text{}}{\left(x-\mu \right)}^{2}}{N}}[/latex] or [latex]\sigma = \sqrt{\frac{{\sum }^{\text{}}f{\left(x-\mu \right)}^{2}}{N}}[/latex].
Formula Review
Sample Standard Deviation
- [latex]s=\sqrt{\frac{\Sigma {\left(x-\overline{x}\right)}^{2}}{n-1}}[/latex] or [latex]s=\sqrt{\frac{\Sigma f{\left(x-\overline{x}\right)}^{2}}{n-1}}[/latex]
- For the sample standard deviation, the denominator is [latex]n - 1[/latex], that is the sample size MINUS 1.
Population Standard Deviation
- [latex]\sigma=\sqrt{\frac{\Sigma{\left(x-\mu\right)}^{2}}{N}}[/latex] or [latex]\sigma = \sqrt{\frac{\Sigma f{\left(x–\mu\right)}^{2}}{N}}[/latex]
- For the population standard deviation, the denominator is [latex]N[/latex], the number of items in the population.
Section 2.7 Practice
The following data are the distances between 20 retail stores and a large distribution center. The distances are in miles.
29; 37; 38; 40; 58; 67; 68; 69; 76; 86; 87; 95; 96; 96; 99; 106; 112; 127; 145; 150
Find the value that is one standard deviation below the mean.
Two baseball players, Fredo and Karl, on different teams wanted to find out who had the higher batting average when compared to his team. Which baseball player had the higher batting average when compared to his team?
| Baseball Player | Batting Average | Team Batting Average | Team Standard Deviation |
|---|---|---|---|
| Fredo | 0.158 | 0.166 | 0.012 |
| Karl | 0.177 | 0.189 | 0.015 |
Solution
- For Fredo: [latex]z=\frac{0.158\text{ – }0.166}{0.012}=-0.67[/latex]
- For Karl: [latex]z=\frac{0.177\text{ – }0.189}{0.015}=-0.8[/latex]
- Fredo’s z-score of –0.67 is higher than Karl’s z-score of –0.8. For batting average, higher values are better, so Fredo has a better batting average compared to his team.
Find the standard deviation for the following frequency tables using the formula.
-
Table 9: Grouped Frequency Table for Grades Grade Frequency 49.5–59.5 2 59.5–69.5 3 69.5–79.5 8 79.5–89.5 12 89.5–99.5 5 -
Table 10: Grouped Frequency Table for Daily Low Temperatures Daily Low Temperature Frequency 49.5–59.5 53 59.5–69.5 32 69.5–79.5 15 79.5–89.5 1 89.5–99.5 0 -
Table 11: Grouped Frequency Table for Points Per Game Points per Game Frequency 49.5–59.5 14 59.5–69.5 32 69.5–79.5 15 79.5–89.5 23 89.5–99.5 2
Solution
- [latex]{s}_{x}=\sqrt{\frac{\sum f{m}^{2}}{n}-{\overline{x}}^{2}}=\sqrt{\frac{193157.45}{30}-{79.5}^{2}}=10.88[/latex]
- [latex]{s}_{x}=\sqrt{\frac{\sum f{m}^{2}}{n}-{\overline{x}}^{2}}=\sqrt{\frac{380945.3}{101}-{60.94}^{2}}=7.62[/latex]
- [latex]{s}_{x}=\sqrt{\frac{\sum f{m}^{2}}{n}-{\overline{x}}^{2}}=\sqrt{\frac{440051.5}{86}-{70.66}^{2}}=11.14[/latex]
Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows:
| # of movies | Frequency |
|---|---|
| 0 | 5 |
| 1 | 9 |
| 2 | 6 |
| 3 | 4 |
| 4 | 1 |
- Find the sample mean [latex]\overline{x}[/latex].
- Find the approximate sample standard deviation, s.
Solution
- 1.48
- 1.12
One hundred teachers attended a seminar on mathematical problem solving. The attitudes of a representative sample of 12 of the teachers were measured before and after the seminar. A positive number for change in attitude indicates that a teacher's attitude toward math became more positive. The 12 change scores are as follows:
3; 8; –1; 2; 0; 5; –3; 1; –1; 6; 5; –2
- What is the mean change score?
- What is the standard deviation for this population?
- What is the median change score?
- Find the change score that is 2.2 standard deviations below the mean.
Refer to the graphs below to determine which of the following are true and which are false. Explain your solution to each part in complete sentences.
Long Image Description: This shows three graphs. The first is a histogram with a mode of 3 and fairly symmetrical distribution between 1 (minimum value) and 5 (maximum value). The second graph is a histogram with peaks at 1 (minimum value) and 5 (maximum value) with 3 having the lowest frequency. The third graph is a box plot. The first whisker extends from 0 to 1. The box begins at the firs quartile, 1, and ends at the third quartile,6. A vertical, dashed line marks the median at 3. The second whisker extends from 6 on.
- The medians for all three graphs are the same.
- We cannot determine if any of the means for the three graphs is different.
- The standard deviation for graph b is larger than the standard deviation for graph a.
- We cannot determine if any of the third quartiles for the three graphs is different.
Solution
- True
- True
- True
- False
In a recent issue of the IEEE Spectrum, 84 engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days. Let X = the length (in days) of an engineering conference.
- Organize the data in a chart.
- Find the median, the first quartile, and the third quartile.
- Find the 65th percentile.
- Find the 10th percentile.
- Construct a box plot of the data.
- The middle 50% of the conferences last from [latex]\underline{\hspace{2cm}}[/latex] days to [latex]\underline{\hspace{2cm}}[/latex] days.
- Calculate the sample mean of days of engineering conferences.
- Calculate the sample standard deviation of days of engineering conferences.
- Find the mode.
- If you were planning an engineering conference, which would you choose as the length of the conference: mean; median; or mode? Explain why you made that choice.
- Give two reasons why you think that three to five days seem to be popular lengths of engineering conferences.
[latex]X[/latex] = the number of days per week that 100 clients use a particular exercise facility.
| x | Frequency |
|---|---|
| 0 | 3 |
| 1 | 12 |
| 2 | 33 |
| 3 | 28 |
| 4 | 11 |
| 5 | 9 |
| 6 | 4 |
- The 80th percentile is [latex]\underline{\hspace{2cm}}[/latex].
- 5
- 80
- 3
- 4
- The number that is 1.5 standard deviations BELOW the mean is approximately [latex]\underline{\hspace{2cm}}[/latex].
- 0.7
- 4.8
- –2.8
- Cannot be determined
Suppose that a publisher conducted a survey asking adult consumers the number of fiction paperback books they had purchased in the previous month. The results are summarized in the table below.
| # of books | Freq. | Rel. Freq. |
|---|---|---|
| 0 | 18 | |
| 1 | 24 | |
| 2 | 24 | |
| 3 | 22 | |
| 4 | 15 | |
| 5 | 10 | |
| 7 | 5 | |
| 9 | 1 |
- Are there any outliers in the data? Use an appropriate numerical test involving the IQR to identify outliers, if any, and clearly state your conclusion.
- If a data value is identified as an outlier, what should be done about it?
- Are any data values further than two standard deviations away from the mean? In some situations, statisticians may use this criteria to identify data values that are unusual, compared to the other data values. (Note that this criteria is most appropriate to use for data that is mound-shaped and symmetric, rather than for skewed data.)
- Do parts a and c of this problem give the same answer?
- Examine the shape of the data. Which part, a or c, of this question gives a more appropriate result for this data?
- Based on the shape of the data which is the most appropriate measure of center for this data: mean, median, or mode?
The population parameters below describe the full-time equivalent number of students (FTES) each year at Lake Tahoe Community College from 1976–1977 through 2004–2005.
- [latex]\mu = 1000 \text{FTES}[/latex]
- median = 1,014 FTES
- [latex]\sigma = 474 \text{FTES}[/latex]
- first quartile = 528.5 FTES
- third quartile = 1,447.5 FTES
- [latex]n = 29 \text{years}[/latex]
- A sample of 11 years is taken. About how many are expected to have a FTES of 1014 or above? Explain how you determined your answer.
- 75% of all years have an FTES:
- at or below: [latex]\underline{\hspace{2cm}}[/latex]
- at or above: [latex]\underline{\hspace{2cm}}[/latex]
- The population standard deviation = [latex]\underline{\hspace{2cm}}[/latex]
- What percentage of the FTES was from 528.5 to 1447.5? How do you know?
- What is the IQR? What does the IQR represent?
- How many standard deviations away from the mean is the median?
Three students were applying to the same graduate school. They came from schools with different grading systems.
| Student | GPA | School Average GPA | School Standard Deviation |
|---|---|---|---|
| Thuy | 2.7 | 3.2 | 0.8 |
| Vichet | 87 | 75 | 20 |
| Kamala | 8.6 | 8 | 0.4 |
Which student had the best GPA when compared to other students at his school? Explain how you determined your answer.
A music school has budgeted to purchase three musical instruments. They plan to purchase a piano costing $3,000, a guitar costing $550, and a drum set costing $600. The mean cost for a piano is $4,000 with a standard deviation of $2,500. The mean cost for a guitar is $500 with a standard deviation of $200. The mean cost for drums is $700 with a standard deviation of $100. Which cost is the lowest, when compared to other instruments of the same type? Which cost is the highest when compared to other instruments of the same type. Justify your answer.
Solution
For pianos, the cost of the piano is 0.4 standard deviations BELOW the mean. For guitars, the cost of the guitar is 0.25 standard deviations ABOVE the mean. For drums, the cost of the drum set is 1.0 standard deviations BELOW the mean. Of the three, the drums cost the lowest in comparison to the cost of other instruments of the same type. The guitar costs the most in comparison to the cost of other instruments of the same type.
An elementary school class ran one mile with a mean of 11 minutes and a standard deviation of three minutes. Rachel, a student in the class, ran one mile in eight minutes. A junior high school class ran one mile with a mean of nine minutes and a standard deviation of two minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes. A high school class ran one mile with a mean of seven minutes and a standard deviation of four minutes. Nedda, a student in the class, ran one mile in eight minutes.
- Why is Kenji considered a better runner than Nedda, even though Nedda ran faster than him?
- Who is the fastest runner with respect to his or her class? Explain why.
The most obese countries in the world have obesity rates that range from 11.4% to 74.6%. This data is summarized in the table below.
| Percent of Population Obese | Number of Countries |
|---|---|
| 11.4–20.45 | 29 |
| 20.45–29.45 | 13 |
| 29.45–38.45 | 4 |
| 38.45–47.45 | 0 |
| 47.45–56.45 | 2 |
| 56.45–65.45 | 1 |
| 65.45–74.45 | 0 |
| 74.45–83.45 | 1 |
What is the best estimate of the average obesity percentage for these countries? What is the standard deviation for the listed obesity rates? The United States has an average obesity rate of 33.9%. Is this rate above average or below? How “unusual” is the United States’ obesity rate compared to the average rate? Explain.
Solution
- [latex]\overline{x}=23.32[/latex]
- Using the TI 83/84, we obtain a standard deviation of: [latex]{s}_{x}=12.95.[/latex]
- The obesity rate of the United States is 10.58% higher than the average obesity rate.
- Since the standard deviation is 12.95, we see that 23.32 + 12.95 = 36.27 is the obesity percentage that is one standard deviation from the mean. The United States obesity rate is slightly less than one standard deviation from the mean. Therefore, we can assume that the United States, while 34% obese, does not have an unusually high percentage of obese people.
References
Data from Microsoft Bookshelf.
King, Bill. “Graphically Speaking.” Institutional Research, Lake Tahoe Community College. Available online at http://www.ltcc.edu/web/about/institutional-research (accessed April 3, 2013).
mean of the squared deviations from the mean, or the square of the standard deviation; for a set of data, a deviation can be represented as x – [latex]\overline{x}[/latex] where x is a value of the data and [latex]\overline{x}[/latex] is the sample mean. The sample variance is equal to the sum of the squares of the deviations divided by the difference of the sample size and one.
a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation
