2.7 Measures of the Spread of the Data

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 all the data are 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 whereas 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 both Rosa and Binh 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. A z-score is a standardized score that lets us compare data sets. It tells us how many standard deviations a data value is from the mean and is calculated as the ratio of the difference in a particular score and the population mean to the population standard deviation.

We can use the given information to create the table below.

Since Rosa and Binh only shop at Supermarket A, we can ignore the row for Supermarket B.

We need the values from the first row to determine the number of standard deviations above or below the mean each individual wait time is; we can do so by calculating two different z-scores.

Rosa waited for seven minutes, so the z-score representing this deviation from the population mean may be calculated as

The z-score of one tells us that Rosa’s wait time is one standard deviation above the mean wait time of five minutes.

Binh waited for one minute, so the z-score representing this deviation from the population mean may be calculated as

The z-score of −2 tells us that Binh’s wait time is two standard deviations below the mean wait time 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 they are 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 farther away than two standard deviations. You will learn more about this in later chapters.

The 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 5 + (1)(2) = 7.

If one were also part of the data set, then one is two standard deviations to the left of five because 5 + (–2)(2) = 1.

Figure 2.26

• 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 1 = 5 + (–2)(2).

The equation value = mean + (#ofSTDEVs)(standard deviation) can be expressed for a sample and for a population as follows:

• Sample: $x\text{ = }\overline{x}\text{ + }(\#ofSTDEV)(s)$
• Population: $x=\mu +(\#ofSTDEV)(\sigma )\text{.}$

The lowercase letter s represents the sample standard deviation and the Greek letter σ (lower case) represents the population standard deviation.

The symbol $\overline{x}$ is the sample mean, and the Greek symbol $\mu$ is the population mean.

Calculating the Standard Deviation

If x 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 x – μ. For sample data, in symbols, a deviation is x – $\overline{x}$.

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 lowercase letter s represents the sample standard deviation and the Greek letter σ (lowercase sigma) represents the population standard deviation. If the sample has the same characteristics as the population, then s should be a good estimate of σ.

To calculate the standard deviation, we need to calculate the variance first. The variance is the average of the squares of the deviations (the x – $\overline{x}$ values for a sample or the x – μ values for a population). The symbol σ² represents the population variance; the population standard deviation σ is the square root of the population variance. The symbol s² represents the sample variance; the sample standard deviation s 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 N, 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 n – 1, one less than the number of items in the sample.

Formulas for the Sample Standard Deviation

• $s=\sqrt{\frac{\Sigma {(x-\overline{x})}^2}{n-1}}$ or $s=\sqrt{\frac{\Sigma f{(x-\overline{x})}^2}{n-1}}$
• For the sample standard deviation, the denominator is n−; that is, the sample size minus 1.

Formulas for the Population Standard Deviation

• $\sigma = \sqrt{\frac{\Sigma {(x-\mu )}^2}{N}}$ or $\sigma = \sqrt{\frac{\Sigma f{(x–\mu )}^2}{N}}$
• For the population standard deviation, the denominator is N, the number of items in the population.

In these formulas, f represents the frequency with which a value appears. For example, if a value appears once, f is one. If a value appears three times in the data set or population, f is three.

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 .97 and .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. We can sum the products of the frequencies and deviations to show that the sum of the deviations is always zero. $$1\left(-1.525\right)+2\left(-1.025\right)+4\left(-.525\right)+4\left(-.025\right)+6\left(.475\right)+3\left(.975\right)=0$$ For Example 2.33, there are n = 20 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 n = 20, the calculation divided by n – 1 = 20 – 1 = 19 because the data is a sample. For the sample variance, we divide by the sample size minus one (n – 1). Why not divide by n? 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 (n – 1) gives a better estimate of the population variance.

The standard deviation, s or σ, is either zero or larger than zero. Describing the data with reference to the spread is called variability. The variability in data depends on the method by which the outcomes are obtained, for example, by measuring or by random sampling. 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 all the data are concentrated close to the mean and 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 s or σ 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 2.34

Problem

Use the following data (first exam scores) from Susan Dean's spring precalculus 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

1. Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places.
2. Calculate the following to one decimal place using a TI-83+ or TI-84 calculator:
   1. The sample mean
   2. The sample standard deviation
   3. The median
   4. The first quartile
   5. The third quartile
   6. IQR
3. 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

1. See Table 2.33.
2. Entering the data values into a list in your graphing calculator and then selecting Stat, Calc, and 1-Var Stats will produce the one-variable statistics you need.
3. The x-axis goes from 32.5 to 100.5; the y-axis goes from –2.4 to 15 for the histogram. The number of intervals is 5, so the width of an interval is (100.5 – 32.5) divided by 5, 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.

Figure 2.27

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 percent is greater (73 – 33 = 40) than the spread in the upper 50 percent (100 – 73 = 27). 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 percent of the exam scores (IQR = 29) are Ds, Cs, and Bs. The box plot also shows us that the lower 25 percent of the exam scores are Ds and Fs.