2.8 Descriptive Statistics
SLO 2
Identify appropriate graphs and summary statistics for variables and relationships between them and correctly interpret information from graphs and summary statistics.
This lab is where you pick the graphs and statistics yourself, on data you collected: a histogram and a box plot of your class's shoe counts, the mean, standard deviation, and five-number summary beside them, and a written reading of what the two pictures agree on.
Learning Objectives
By the end of this section, you will be able to:
- collect a real univariate data set from your own class and record it in an organized table;
- construct a histogram and a box plot from that data set by hand, choosing and scaling your own intervals;
- calculate the univariate statistics that summarize it — the mean, the standard deviation, and the five-number summary;
- read both graphs to say what the data implies about center, spread, shape, and potential outliers.
2.8.1 Stats Lab: Descriptive Statistics
This section is a lab, not a reading. Every other section of this chapter handed you a data set and asked you to describe it. Here you go get the data yourself, and then run the whole chapter over it: a histogram from §2.2, quartiles and the outlier test from §2.3, a box plot from §2.4, the mean from §2.5, the shape language from §2.6, and the standard deviation from §2.7.
Class Time:
Names:
The question is one you can answer without any equipment: how many pairs of shoes do you own? Count yours before you read on, so your own value goes into the data set honestly rather than after you have seen everyone else's.
Every number in this lab arrives through a person answering a question about themselves, so the usual data problems show up at once — someone counts flip-flops, someone else does not, someone rounds thirty-something down to thirty. That does not make the data unusable. It makes the definition part of the result.
Fix the counting rule with your group before the first person answers. Does a pair of slippers count? Cleats you have not worn since high school? A single shoe whose partner is missing? There is no right answer, and the statistics do not care which convention you pick — but they care very much whether all thirty people used the same one. A data set collected under thirty slightly different definitions has a spread that is partly real and partly an artifact of the question, and nothing you compute afterward can separate the two. This is the difference between a variable and a well-defined variable, and it is settled before collection, never after. Write the rule at the top of your data sheet in one sentence, and read it to each person you survey rather than paraphrasing it from memory — a rule restated thirty times from memory is thirty slightly different rules.
Pao Xiong's group and a second group run this lab in the same classroom. Pao tells their classmates to count every pair of shoes they own; the second group tells its classmates to count only the pairs they have worn in the last year. Both report a mean near 12, but Pao's standard deviation is nearly twice the other group's. Give a reason for the difference that has nothing to do with sampling error.
Solution
Step 1 — compare what the two questions admit. The second group's rule caps what any one person can report: there are only so many pairs anybody actually wears in a year. Pao's rule has no such cap — a person with forty pairs in a closet reports forty.
Step 2 — connect that to spread. Standard deviation measures how far values sit from their mean. Pao's question lets the high end run much further from the center, so their deviations are larger even when the two groups survey the same people.
Answer: the two groups measured different variables under the same name. Pao's larger standard deviation is a property of their wider definition, not evidence of a more variable population or a worse sample. This is why "describe how you collected the data" is not busywork — a reader handed only \(\bar{x}\) and \(s\) has no way to notice that two studies of "pairs of shoes" were never measuring the same thing.
A demonstration data set
Every step of this lab is easier to do on your own data once you have watched it done on somebody else's. The 30 values below are a class that ran this lab before you, already sorted from smallest to largest. Each Try It Now that follows works one stage of the lab on these numbers, so keep the table where you can see it.
| Position | Values |
|---|---|
| 1-10 | 2, 3, 4, 4, 5, 5, 6, 6, 6, 7 |
| 11-20 | 7, 8, 8, 8, 9, 9, 10, 10, 11, 12 |
| 21-30 | 12, 13, 14, 15, 16, 18, 20, 22, 25, 32 |
Sorting is not one of the seven things the lab asks you to do, which is exactly why it gets skipped. Do it anyway, and do it first. The median, both quartiles, the minimum, and the maximum are all read straight off a sorted list by counting positions — on an unsorted list every one of them turns into a hunt, and a hunt is where the arithmetic mistakes come from. The mean and the standard deviation do not care about order, so sorting costs you nothing there. Sort once, then never reorder again: if you renumber halfway through, the position counts you already wrote down stop pointing at the values you thought they did.
Alondra Reyes checks the demonstration data before she computes anything with it. Using Table 2.8.1, confirm for her that the count is 30, then name the value in position 15 and the value in position 16. Why are those two positions the ones worth finding first?
Solution
Step 1 — count. Three rows of ten values each gives \(3 \times 10 = 30\) responses, which matches the 30 classmates the lab asks you to survey.
Step 2 — read positions 15 and 16. The second row holds positions 11 through 20: 7, 8, 8, 8, 9, 9, 10, 10, 11, 12. Position 15 is the fifth entry of that row, 9, and position 16 is the sixth, also 9.
Answer: \(n = 30\), and both position 15 and position 16 hold the value 9. With an even count there is no single middle value, so the median is the mean of the two middle positions — those two. Finding them first pins the center of the data before you compute anything, and it gives you a number to sanity-check the mean against later.
Try it in rāSHio
Open rāSHio, paste your class’s 30 shoe counts into the spreadsheet with File → Delimited List… (it accepts commas, semicolons, tabs, or one value per line), then choose Transform → Sort to put them in ascending order — every position count in this lab, the median at positions 15 and 16 included, is read straight off that sorted column.
Figure 2.8.1 — Putting the lab’s values in ascending order in rāSHio: Transform → Sort.
2.8.2 Collect the Data
A data value is a potential outlier when it falls outside the two fences built from the quartiles:
$$ \text{lower fence} = Q_1 - 1.5(IQR) \qquad \text{upper fence} = Q_3 + 1.5(IQR) $$where \(IQR = Q_3 - Q_1\). A value below the lower fence or above the upper fence is flagged as a potential outlier; a value inside both fences is not, however far from the mean it may feel.
The word potential is load-bearing. The rule flags a value for inspection; it does not convict it. A student who genuinely owns 32 pairs of shoes is a real member of the population, and deleting them because a formula raised its hand would make your sample less representative, not more. What the flag earns is a look: is this a recording error, a person who misread the question, or a real and interesting extreme? Only the first two are grounds for removing anything, and whichever you decide, the write-up says what you found and what you did about it. An outlier quietly dropped is the single easiest way to make an honest data set lie.
Record the number of pairs of shoes you own.
1. Randomly survey 30 classmates about the number of pairs of shoes they own. Record their values in the table below.
| Pairs of shoes owned — 30 responses | ||||
|---|---|---|---|---|
2. Construct a histogram. Make five to six intervals. Sketch the graph using a ruler and pencil and scale the axes. Title it Number of Pairs of Shoes, label the horizontal axis with the number of pairs, and label the vertical axis with the frequency.
Choosing the intervals is the only real decision in step 2, and it is worth a minute rather than a guess. Take your largest value minus your smallest, divide by the number of intervals you want, and round the result up to something a reader can hold in their head — a width of 5 beats a width of 4.83. Then set the first boundary just below your smallest value, at a half unit, so that no data point lands exactly on a boundary and has to be argued about. Five to six bars is the range the lab asks for because it is wide enough to show a shape and narrow enough that each bar still holds several people; with thirty values, ten bars would leave most of them nearly empty and the shape would dissolve into noise.
The demonstration data runs from 2 to 32 pairs. Find an interval width that fits the whole data set into six intervals, and give the six boundaries. Show why a width of 5 does not work here.
Solution
Step 1 — take the range and divide.
$$ \frac{32 - 2}{6} = \frac{30}{6} = 5 $$Step 2 — test that width against the actual data. Starting a half unit below the minimum, at 1.5, six intervals of width 5 run 1.5 to 31.5. The largest value, 32, falls outside — the last interval ends before the data does.
Step 3 — round the width up. A width of 6, starting at 1.5, gives boundaries at
$$ 1.5,\ 7.5,\ 13.5,\ 19.5,\ 25.5,\ 31.5,\ 37.5 $$which covers 1.5 through 37.5 and holds every value.
Answer: use a width of 6, with boundaries 1.5, 7.5, 13.5, 19.5, 25.5, 31.5, and 37.5. This is why the rule says round the width up rather than to the nearest whole number: dividing the range by the bar count gives the smallest width that could work, and any rounding down leaves data outside the last bar. The half-unit start is what keeps a value like 8 from sitting on a boundary, where two different students would put it in two different bars.
Try it in rāSHio
With your values in rāSHio, choose Graph → Histogram and set the bin width you just worked out — the tool bins and draws the bars for you, so you can try a width of 5 against a width of 6 and watch your largest value fall outside the last bar in one and inside it in the other.
Ethan Doyle is building the frequency table for the demonstration data using the six intervals from Try It Now 2.8.3. Work it out alongside him and check his total.
Solution
Step 1 — tally each interval. Walk the sorted list once, dropping each value into the interval that contains it.
| Interval | Values in it | Frequency |
|---|---|---|
| 1.5-7.5 | 2, 3, 4, 4, 5, 5, 6, 6, 6, 7, 7 | 11 |
| 7.5-13.5 | 8, 8, 8, 9, 9, 10, 10, 11, 12, 12, 13 | 11 |
| 13.5-19.5 | 14, 15, 16, 18 | 4 |
| 19.5-25.5 | 20, 22, 25 | 3 |
| 25.5-31.5 | none | 0 |
| 31.5-37.5 | 32 | 1 |
Step 2 — check the total.
$$ 11 + 11 + 4 + 3 + 0 + 1 = 30 $$Answer: the six frequencies are 11, 11, 4, 3, 0, 1, and they total 30 — the check that no value was dropped or double-counted. Do not be tempted to delete the empty fifth interval to make the picture tidier. That gap is data: it says nobody in the class owns between 26 and 31 pairs, which is precisely what separates the lone value of 32 from the rest and is the first visual hint that you have an outlier on your hands.
Try it in rāSHio
To check the tallies you just made by hand, open Graph → Frequency Table in rāSHio and read the counts back — they still have to total 30, and an interval showing 0 is data about your class, not an error to bin away.
Figure 2.8.2 — Reading the interval counts back off rāSHio: Graph → Frequency Table.
3. Calculate the following values.
a. \(\bar{x} =\) ________
b. \(s =\) ________
Nathan Reed wants the sample mean of the demonstration data so that he can compare it to the median located in Try It Now 2.8.2. Find both for him.
Solution
Step 1 — add the 30 values. Row by row from Table 2.8.1: the first ten sum to 48, the second ten to 92, and the third ten to 187.
$$ 48 + 92 + 187 = 327 $$Step 2 — divide by the count.
$$ \bar{x} = \frac{327}{30} = 10.9 $$Step 3 — compare with the median. Positions 15 and 16 both hold 9, so the median is 9. The mean sits 1.9 pairs above it.
Answer: \(\bar{x} = \mathbf{10.9}\) pairs, against a median of 9. The mean being pulled above the median is the arithmetic fingerprint of a right-skewed data set from §2.6: the single value of 32 adds to the sum and drags the mean up, while the median only cares about which position it sits in and barely moves. Neither number is wrong — they are answering different questions, and the gap between them is itself a description of the shape.
Find the sample standard deviation of the demonstration data. The sum of the squared values is \(\sum x^2 = 4{,}967\).
Solution
Step 1 — find the total squared deviation. With \(\sum x = 327\) and \(n = 30\),
$$ \sum (x - \bar{x})^2 = \sum x^2 - \frac{(\sum x)^2}{n} = 4{,}967 - \frac{327^2}{30} = 4{,}967 - 3{,}564.3 = 1{,}402.7 $$Step 2 — divide by \(n - 1\) to get the sample variance.
$$ s^2 = \frac{1{,}402.7}{29} \approx 48.37 $$Step 3 — take the square root.
$$ s = \sqrt{48.37} \approx 6.95 $$Answer: \(s \approx \mathbf{6.95}\) pairs. Notice how large that is next to a mean of 10.9 — the typical distance from the center is nearly two thirds of the center itself. That is a warning to read before you interpret anything else: in a data set this spread out relative to its mean, quoting "the average student owns about 11 pairs" without the standard deviation beside it tells a reader almost nothing about what any individual student actually owns.
Try it in rāSHio
Rather than square thirty deviations by hand every time, choose Stats → Summary Statistics in rāSHio and read the mean and the sample standard deviation off one panel — work the arithmetic yourself once, as you just did, then let the tool carry it for the rest of the lab.
Figure 2.8.3 — The mean and the sample standard deviation on one panel in rāSHio: Stats → Summary Statistics.
4. Are the data discrete or continuous? How do you know?
Cruz Aguilar cannot decide whether their shoe counts are discrete or continuous data. Settle it, name the test you used, and then say what would change if the survey had asked for each classmate's shoe size instead.
Solution
Step 1 — apply the test. Ask whether the values come from counting or from measuring. You count pairs of shoes; you cannot own 6.4 pairs, and there is no value between 6 and 7 that anyone could report.
Step 2 — apply the same test to shoe size. Shoe size comes from measuring a foot. The half sizes on the box are a rounding convention, not a limit on the foot — a foot can be any length in between.
Answer: the shoe counts are discrete, because they are whole-number counts with nothing in between. Shoe size would be continuous, because the underlying quantity is a length that can take any value in a range and is only reported in rounded steps. The reason the lab asks this in the middle of building a histogram is that the answer justifies the half-unit interval boundaries: with discrete counts, a boundary at 7.5 can never have a data point sitting on it, so the ambiguity is designed out rather than resolved by a convention.
5. In complete sentences, describe the shape of the histogram.
"Skewed right" earns its keep only when you can say where you saw it — which side the long tail runs off, which bars are tall, where the gap is. A reader who cannot rebuild your picture from your sentence has not been told the shape.
A good shape sentence has three parts, and you can assemble them straight off your own histogram. Start with where the bars are tallest, because that is where most of the class lives — say it as a range of values rather than as a bar number, since a reader who does not have your intervals in front of them cannot decode "the second bar." Then say which direction the bars fall away in and how far they run before they stop; that is the tail, and its direction is what the words skewed left and skewed right are naming. Finish with anything that breaks the pattern — a gap where nobody landed, a second pile far from the first, a single isolated bar out at one end. Write those three parts in order and you have described the shape without using a word the reader has to take on faith. Then check the sentence against your mean and median, which should be telling the same story in numbers.
Claire Whitfield has to describe the shape of the demonstration data's histogram in her write-up, using the frequencies 11, 11, 4, 3, 0, 1 from Try It Now 2.8.4. Write the description she needs: name the shape and point at the evidence for it.
Solution
Step 1 — find where the data piles up. The first two intervals hold 11 classmates each, so \(22\) of the 30 — over two thirds of the class — own 13 pairs or fewer.
Step 2 — follow the bars to the right. After that pile the heights fall away steadily: 4, then 3, then 0, then a single lone bar at the far right holding the value 32.
Step 3 — name the shape. The tail runs off to the right, so the greater values are the ones spread out.
Answer: the histogram is skewed to the right. Two tall bars of 11 sit at the low end, the bars fall off through 4 and 3, an entire interval from 25.5 to 31.5 is empty, and one isolated classmate sits in the final interval. The mean of 10.9 lying above the median of 9 says the same thing in numbers. Both descriptions have to agree — if your sentence says "roughly symmetric" while your mean sits well above your median, one of the two was computed or drawn wrong, and it is usually the interval boundaries.
6. Are there any potential outliers? List the value(s) that could be outliers. Use a formula to check the end values to determine if they are potential outliers.
The formula item 6 is asking for is the 1.5 x IQR rule of Definition 2.8.1. It is the same rule you met in §2.3, restated here because this is the step of the lab students most often skip — an unusually large value looks like an outlier, and looking is not a test.
The demonstration data has \(Q_1 = 6\) and \(Q_3 = 14\). Test both end values, 2 and 32, against the fences and say what you have shown.
Solution
Step 1 — find the IQR.
$$ IQR = Q_3 - Q_1 = 14 - 6 = 8 $$Step 2 — build both fences.
$$ \text{lower} = 6 - 1.5(8) = 6 - 12 = -6 \qquad \text{upper} = 14 + 1.5(8) = 14 + 12 = 26 $$Step 3 — test the two end values. The minimum, 2, sits above \(-6\), so it is inside the lower fence. The maximum, 32, sits above 26, so it is outside the upper fence.
Answer: the minimum of 2 is not a potential outlier; the maximum of 32 is. Notice that the lower fence came out negative, which no shoe count can ever be — with a distribution pushed up against zero on the left, the IQR rule can only ever flag values on the high side. That is a property of this data set, not a failure of the test, and it is worth saying out loud in your write-up rather than leaving a reader to wonder why every flagged value is large. Note too that 25 sits inside the upper fence and is therefore not flagged, even though it is far above the mean — the test is about the quartiles, not about how surprising a value feels.
2.8.3 Analyze the Data
1. Determine the following values.
a. \(\text{Min} =\) ________
b. \(M =\) ________ (the median)
c. \(\text{Max} =\) ________
d. \(Q_1 =\) ________
e. \(Q_3 =\) ________
f. \(IQR =\) ________
Produce the whole five-number summary for the demonstration data, plus the IQR, reading everything off Table 2.8.1 by position.
Solution
Step 1 — read the ends. The first value is 2 and the thirtieth is 32, so \(\text{Min} = 2\) and \(\text{Max} = 32\).
Step 2 — take the median. With \(n = 30\), average positions 15 and 16, which both hold 9.
$$ M = \frac{9 + 9}{2} = 9 $$Step 3 — quarter the halves. The lower half is positions 1 through 15, and its median is position 8, which holds 6. The upper half is positions 16 through 30, and its median is position 23, which holds 14.
$$ Q_1 = 6 \qquad Q_3 = 14 $$Step 4 — subtract for the IQR.
$$ IQR = 14 - 6 = 8 $$Answer: the five-number summary is Min 2, \(Q_1\) 6, \(M\) 9, \(Q_3\) 14, Max 32, with an \(IQR\) of 8. Every one of those came from counting positions in a sorted list rather than from a calculation, which is why sorting first was worth the minute it cost. Look at the spacing before you move on: the distance from the minimum up to \(Q_1\) is 4, while the distance from \(Q_3\) up to the maximum is 18. The picture is already lopsided, and the box plot is about to make that visible.
Try it in rāSHio
Choose Graph → Box Plot in rāSHio to draw the five-number summary you just read off the table — the tool marks the quartiles, runs each whisker out to the last value inside the fences, and plots anything past them as its own dot, which is exactly the picture step 2 of Analyze the Data asks you to construct.
Figure 2.8.4 — Drawing the five-number summary in rāSHio: Graph → Box Plot.
2. Construct a box plot of the data. Draw it above a scaled number line that covers your minimum through your maximum, mark the five-number summary from step 1, and draw the whiskers out to the smallest and largest values that are not potential outliers.
Every piece of the box covers exactly a quarter of your classmates, so a short piece means those people are packed tightly into a narrow range and a long piece means the same number of people are stretched thin across a wide one. Width is crowding, not headcount.
That reading is what makes the box plot answer question 3 directly, and it is worth walking once on your own picture before you write the sentence. Find the box: the middle half of your class lives inside it, so a narrow box says half of everyone owns nearly the same number of pairs. Then compare the two whiskers. A long right whisker beside a short left one means the top quarter of the class is spread across a much wider range of values than the bottom quarter — a few people with a lot of shoes, pulling the picture to the right. That is the same right-skew you named from the histogram, seen from a different angle, and the two descriptions should agree.
3. What does the shape of the box plot imply about the concentration of the data? Use complete sentences.
4. Using the box plot, how can you determine if there are potential outliers?
5. How does the standard deviation help you to determine concentration of the data and whether or not there are potential outliers?
6. What does the \(IQR\) represent in this problem?
Using the five-number summary from Try It Now 2.8.10, describe the concentration of the demonstration data in complete sentences, and say how the drawn box plot lets a reader spot the outlier without doing any arithmetic.
Solution
Step 1 — measure the four pieces. Left whisker \(6 - 2 = 4\); left half of the box \(9 - 6 = 3\); right half of the box \(14 - 9 = 5\); right whisker out to the last non-outlier value, 25, giving \(25 - 14 = 11\). The flagged value 32 is plotted as its own dot beyond the whisker.
Step 2 — say what equal quarters over unequal widths means. Each of those four pieces holds about 7 or 8 classmates. The middle half of the class is squeezed into the 8 pairs between 6 and 14, while the top quarter alone is spread across the 11 pairs from 14 to 25.
Step 3 — read the outlier off the picture. The whisker stops at 25 and a single dot sits well beyond it at 32, separated by a visible gap.
Answer: the data is concentrated at the low end — half the class owns between 6 and 14 pairs, and the box covering that half is narrower than the right whisker beside it, so the upper quarter is far more spread out than the middle. A reader spots the outlier because the whisker stops short and a lone dot sits past it: the whisker is drawn only to the most extreme value still inside the fences, so anything plotted beyond it has already failed the \(1.5 \times IQR\) test. The standard deviation of 6.95 tells the same story a different way — it is large next to a mean of 10.9, and a large standard deviation with a narrow box means the spread is coming from a few extreme values rather than from the bulk of the class.
7. Show your work to find the value that is 1.5 standard deviations:
a. above the mean.
b. below the mean.
Danielle Tallchief has finished her calculations for the demonstration data and has \(\bar{x} = 10.9\) and \(s \approx 6.95\). Find the values 1.5 standard deviations above and below the mean, and say what the lower one tells her about the shape.
Solution
Step 1 — go 1.5 standard deviations up.
$$ \bar{x} + 1.5s = 10.9 + 1.5(6.95) = 10.9 + 10.425 \approx 21.3 $$Step 2 — go 1.5 standard deviations down.
$$ \bar{x} - 1.5s = 10.9 - 1.5(6.95) = 10.9 - 10.425 \approx 0.5 $$Step 3 — read the lower value. Half a pair of shoes is impossible, and yet it is only 1.5 standard deviations below the mean. The data cannot reach that far down because it stops at zero, while nothing stops it from running upward — the largest value sits well past 21.3.
Answer: the two values are about 21.3 pairs and 0.5 pairs. The lower one lands essentially on the floor of the variable, which says the standard deviation is large relative to how much room the data has on its left: the signature of a distribution skewed to the right. Compare this with the fences in Try It Now 2.8.9, which flagged only high values for the same reason. Two different tools, built from two different measures of center and spread, are describing one lopsided data set — and when they disagree on your own data, that disagreement is the thing worth writing about.
Key Terms
box plot — a graph that gives a quick picture of the middle 50% of the data.
first quartile — the value that is the median of the lower half of the ordered data set.
frequency — the number of times a value of the data occurs.
frequency polygon — looks like a line graph but uses intervals to display ranges of large amounts of data.
frequency table — a data representation in which grouped data is displayed along with the corresponding frequencies.
histogram — a graphical representation, in \(x\)-\(y\) form, of the distribution of data in a data set; \(x\) represents the data and \(y\) represents the frequency or relative frequency, and the graph consists of contiguous rectangles.
interquartile range — written \(IQR\), the range of the middle 50 percent of the data values; found by subtracting the first quartile from the third quartile.
interval — also called a class interval; a range of data used when displaying large data sets.
mean — a number that measures the central tendency of the data, commonly called the average; the sample mean \(\bar{x}\) is the sum of all values in the sample divided by the number of values in the sample, and the population mean \(\mu\) is the sum of all values in the population divided by the number of values in the population.
median — a number that separates ordered data into halves; half the values are the same number or smaller than the median and half are the same number or larger. The median may or may not be part of the data.
midpoint — the mean of an interval in a frequency table.
mode — the value that appears most frequently in a set of data.
outlier — an observation that does not fit the rest of the data.
paired data set — two data sets of the same size that have a one-to-one relationship, so each data point in one set is matched with exactly one point from the other.
percentile — a number that divides ordered data into hundredths; the median is the second quartile and the \(50^{\text{th}}\) percentile, and the first and third quartiles are the \(25^{\text{th}}\) and \(75^{\text{th}}\) percentiles.
quartiles — the numbers that separate the data into quarters; the second quartile is the median of the data.
relative frequency — the ratio of the number of times a value of the data occurs to the number of all outcomes.
skewed — used to describe data that is not symmetrical; when the right side of a graph looks chopped off compared to the left side we say it is skewed to the left, and when the left side looks chopped off compared to the right side we say it is skewed to the right.
standard deviation — a number equal to the square root of the variance, measuring how far data values sit from their mean; written \(s\) for a sample and \(\sigma\) for a population.
variance — the mean of the squared deviations from the mean, or the square of the standard deviation; a deviation is \(x - \bar{x}\), and the sample variance is the sum of the squared deviations divided by one less than the sample size.