2.2 Histograms
SLO 2
Identify appropriate graphs and summary statistics for variables and relationships between them and correctly interpret information from graphs and summary statistics.
A histogram is the graph for a large numeric data set, and here you learn to build one — starting point, class width, boundary rule — then read its shape, center, and spread. You also learn to distrust a squeezed axis or an uneven bar width.
Learning Objectives
By the end of this section, you will be able to:
- explain what the two axes of a histogram measure, and read a distribution's shape, center, and spread off the picture;
- calculate the relative frequency of a value and check that a relative frequency column sums to 1;
- choose a starting point, a class width, and a number of class intervals for a data set, and construct the histogram;
- apply the boundary convention consistently, and say why two correct histograms of the same data can look different;
- identify the graph-design tricks — squeezed axes, uneven bar widths, side-by-side pie charts — that make an honest data set tell a dishonest story.
For most of the work you do in this book, you will use a histogram to display data. Its big advantage is capacity: a stemplot runs out of room fast, but a histogram can absorb hundreds of values and still be readable. A rule of thumb is to reach for a histogram once the data set has 100 values or more.
2.2.1 Anatomy of a Histogram
A histogram is a graph made of contiguous (adjoining) boxes drawn on a horizontal and a vertical axis. The horizontal axis is labeled with what the data represent — distance from home to school, height in inches, dollars spent. The vertical axis is labeled with frequency, or with relative frequency (equivalently percent frequency or probability). The graph has the same shape under either vertical label.
Each person stands over their own value on the number line, and identical values stack up. The taller the pile, the more people share that measurement — you are looking at where the crowd bunches and where it thins out.
Notice the word contiguous. A bar graph leaves gaps between its bars because its categories are separate things — Ford, Toyota, Honda. A histogram's boxes touch, because the horizontal axis is a continuous number line and every value on it belongs to some interval. The boxes touching is the picture telling you "these categories are neighbors on a scale," and that single visual difference is what separates a histogram from the bar graphs of the previous section.
Like the stemplot, a histogram shows you three things at a glance: the shape of the data, the center, and the spread.
Definition 2.2.1 — A histogram sorts values into touching intervals; each bar's height is the count inside it.
The relative frequency of an observed value is the frequency of that value divided by the total number of data values in the sample. Writing \(f\) for the frequency, \(n\) for the total number of data values (the sum of the individual frequencies), and \(RF\) for the relative frequency:
$$ RF = \frac{f}{n} $$Recall from the previous section that frequency is simply the number of times an answer occurs. Relative frequency takes that raw count and asks a different question: not "how many?" but "what share of the whole?" That is why relative frequency always comes out between 0 and 1, and why the relative frequencies of every value in a data set must add up to 1.
Here is the smallest possible example. Suppose three students in an English class of 40 students earned between 90% and 100%. Then \(f = 3\) and \(n = 40\), so the relative frequency is:
$$ RF = \frac{f}{n} = \frac{3}{40} = 0.075 $$So 7.5% of the students received 90–100%. Those scores, 90–100%, are quantitative measures — they are numbers you can do arithmetic on, which is exactly why a histogram is a legal way to display them.
Definition 2.2.2 — Relative frequency relabels each count as its share of the whole; the shape does not change.
Robin Ashcroft manages the campus bookstore with their wife Kate, and one afternoon they recorded how many students bought each number of highlighters.
| Number of highlighters | Frequency |
|---|---|
| 1 | 12 |
| 2 | 18 |
| 3 | 6 |
| 4 | 4 |
Find the relative frequency for each value, and check your work.
Solution
Step 1 — Find \(n\). Add the frequencies:
$$ n = 12 + 18 + 6 + 4 = 40 $$Step 2 — Divide each frequency by 40.
$$ RF_1 = \frac{12}{40} = 0.300 \qquad RF_2 = \frac{18}{40} = 0.450 $$ $$ RF_3 = \frac{6}{40} = 0.150 \qquad RF_4 = \frac{4}{40} = 0.100 $$Step 3 — Check. The relative frequencies must sum to 1:
$$ 0.300 + 0.450 + 0.150 + 0.100 = 1.000 $$Answer: 0.300, 0.450, 0.150, 0.100. That check in Step 3 is worth doing every single time — if the column does not sum to 1, you either mis-added \(n\) or dropped a value.
Try it in rāSHio
Open rāSHio, type the forty afternoon purchases into the spreadsheet — twelve 1s, eighteen 2s, six 3s and four 4s — then choose Graph → Frequency Table with Discrete values checked. The relative frequency column you just worked out by hand (0.300, 0.450, 0.150, 0.100) appears beside the counts, and the cumulative column comes free — including the check that the shares add to 1.
Figure 2.2.1 — Building a frequency table in rāSHio: Graph → Frequency Table with Discrete values checked, which produces the relative and cumulative relative frequency columns. The walkthrough narrates a different data set; run the same steps on the highlighter counts above.
2.2.2 Choosing Bars, Boundaries, and Widths
A class interval (or bin) is one of the contiguous ranges of values that a histogram's horizontal axis is divided into. Every data value falls in exactly one class interval, and the height of the bar drawn over an interval is the frequency (or relative frequency) of the values inside it.
Next you choose a starting point for the first interval and a class width. Both are whole numbers. Round the starting point down to the whole number at or below the smallest data value, and round the class width up to the next whole number:
$$ \text{start} = \lfloor \text{smallest} \rfloor $$ $$ \text{width} = \left\lceil \frac{\text{largest} - \text{smallest}}{\text{bars}} \right\rceil $$The two roundings go in opposite directions, and that is deliberate. Rounding the start down guarantees the smallest value sits inside the first bar; rounding it up would push that value off the left end of the histogram, where it would disappear from the counts with nothing on the page to show it had gone. Rounding the width up guarantees the bars reach far enough right to cover the largest value.
Whole-number boundaries are easy to read and easy to plot, but they come at a price: data values now do land exactly on boundaries. So the tie has to be settled by a stated rule instead of avoided by clever arithmetic. Every interval includes its left boundary and excludes its right one. An interval of width 2 starting at 1 holds 1, 2, 2.9, and everything up to but not including 3 — the value 3 belongs to the next bar along.
The drawing hides this, which is why the rule is worth saying out loud. Each bar is drawn touching its neighbour, so a bar running from 1 to 3 looks like it reaches 3 — but that shared edge belongs to the bar on the right, not the one on the left. Two bars meeting at a boundary are not sharing the value at that boundary; the right-hand bar owns it.
There is one exception, and it is at the top. If every interval excluded its right boundary, the largest value in the data would fall outside every bar and quietly go uncounted. So the last interval includes both of its boundaries. Intervals of width 2 running from 1 to 25 are therefore \([1, 3)\), \([3, 5)\), and so on up to \([23, 25]\) — half-open the whole way, closed at the very end. If your bar heights do not add up to \(n\), the largest value falling off the end is the first thing to check.
1. Choose the number of bars (classes) — usually 5 to 15.
2. Starting point: round the smallest data value down to a whole number.
3. Width: divide (largest − smallest) by the number of bars, then round up to a whole number.
4. List the boundaries by adding the width repeatedly, starting from the starting point.
5. Sort each value into the interval containing it — left boundary included, right boundary excluded, except in the last interval, which includes both.
6. Label the horizontal axis with the variable and the vertical axis with frequency or relative frequency; draw each bar at the height of its interval's count.
A width of 1.76 rounds up to 2, but so does a width of 1.1 — the decimal is ignored entirely rather than compared against a half. That looks like a broken rule until you ask what rounding 1.1 down to 1 would cost: bars too short to reach the largest value, and data sitting past the right end of the histogram with no bar above it. Rounding up can only ever make the bars reach too far, which is visible and harmless; rounding down can make them fall short, which is invisible and is not.
One more guideline some people follow for choosing the number of bars: take the square root of the number of data values and round to the nearest whole number. With 150 data values, \(\sqrt{150} \approx 12.2\), so you would use about 12 bars or intervals. It is a starting suggestion, not a law — if 12 bars produce a lumpy, unreadable picture, change the number and draw it again.
To construct a histogram, you first decide how many bars — also called intervals or classes — will represent the data. Most histograms use five to 15 bars for clarity. Nobody hands you that number; you choose it.
Definition 2.2.3 — Class intervals on whole-number boundaries: each one includes its left boundary and excludes its right, except the last, which includes both.
The following data are the shoe sizes of 50 students. The sizes are discrete data, since shoe size is measured in whole and half units only. Construct a histogram and calculate the width of each bar or class interval. Suppose you choose six bars.
9; 9; 9.5; 9.5; 10; 10; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5
11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5
12; 12; 12; 12; 12; 12; 12; 12.5; 12.5; 12.5; 12.5; 14
Solution
Step 1 — Find the extremes. The smallest size is 9 and the largest is 14.
Step 2 — Choose the starting point. Round the smallest value down to a whole number. It is already whole, so the starting point is 9.
Step 3 — Compute the width for six bars, rounding up to the next whole number.
$$ 14 - 9 = 5 \qquad 5 \div 6 = 0.8\overline{3} \qquad \lceil 0.8\overline{3} \rceil = 1 $$Step 4 — List the boundaries by adding 1 repeatedly: 9, 10, 11, 12, 13, 14, 15.
Step 5 — Count each interval, left boundary included, right boundary excluded, except in the last.
- \([9, 10)\) holds the sizes 9 and 9.5 → frequency 4
- \([10, 11)\) holds 10 and 10.5 → frequency 14
- \([11, 12)\) holds 11 and 11.5 → frequency 20
- \([12, 13)\) holds 12 and 12.5 → frequency 11
- \([13, 14)\) holds nothing → frequency 0
- \([14, 15]\) holds 14 → frequency 1
Answer: the width of each bar is 1, and the six bars have heights 4, 14, 20, 11, 0, and 1 — which sum to the 50 students surveyed, the check worth running every time. Drawn, the histogram climbs to a peak at \([11, 12)\), falls away, and leaves a lonely bar out at size 14 with an empty interval before it.
Two details in that answer are worth naming. The empty \([13, 14)\) bar is not a mistake to be tidied away — an interval with no data in it is information, and drawing it is what makes the gap before size 14 visible. And the single size-14 student is counted only because the last interval closes on both ends; under the left-inclusive rule alone, 14 would sit on the right edge of the final bar and drop out, leaving 49 students in a survey of 50.
Coach Emilio Ramos measured the heights (in inches to the nearest half inch) of the 100 semiprofessional soccer players in his league. The heights are continuous data, since height is measured.
60; 60.5; 61; 61; 61.5
63.5; 63.5; 63.5
64; 64; 64; 64; 64; 64; 64; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5
66; 66; 66; 66; 66; 66; 66; 66; 66; 66; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 67; 67; 67; 67; 67; 67; 67;
67; 67; 67; 67; 67; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5
68; 68; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69.5; 69.5; 69.5; 69.5; 69.5
70; 70; 70; 70; 70; 70; 70.5; 70.5; 70.5; 71; 71; 71
72; 72; 72; 72.5; 72.5; 73; 73.5
74
Construct a histogram using eight bars, and display relative frequency on the vertical axis.
Solution
Step 1 — Find the starting point. The smallest data value is 60. Round it down to a whole number — 60 is already whole, so the starting point is 60.
Step 2 — Find the largest value. The largest height in the list is 74.
Step 3 — Compute the bar width. Subtract the smallest value from the largest and divide by the number of bars. We chose eight bars:
$$ 74 - 60 = 14 $$ $$ 14 \div 8 = 1.75 $$Round up to the next whole number: the width is 2, so each bar is two units wide. Note that we round up even though 1.75 is closer to 2 than to 1 — the rule would give 2 for a quotient of 1.1 just the same.
Step 4 — List the boundaries by adding 2 each time:
- 60
- \(60 + 2 = 62\)
- \(62 + 2 = 64\)
- \(64 + 2 = 66\)
- \(66 + 2 = 68\)
- \(68 + 2 = 70\)
- \(70 + 2 = 72\)
- \(72 + 2 = 74\)
- \(74 + 2 = 76\)
Step 5 — Sort the heights into intervals, taking each interval's left boundary as included and its right boundary as excluded. The heights 60 through 61.5 are in \([60, 62)\). The height 63.5 is in \([62, 64)\). The heights 64 through 64.5 are in \([64, 66)\). The heights 66 through 67.5 are in \([66, 68)\). The heights 68 through 69.5 are in \([68, 70)\). The heights 70 through 71 are in \([70, 72)\). The heights 72 through 73.5 are in \([72, 74)\). The height 74 is in the last interval, \([74, 76]\) — and because it is the last, that interval includes both ends, which is what keeps the single tallest player in the count.
Notice that no height changed bars when the boundaries moved from 59.95, 61.95, … to 60, 62, …. A height of 62 would now sit exactly on a boundary and go to the right-hand bar, but this particular data set happens to contain none.
Step 6 — Draw it. The histogram below displays height on the horizontal axis and relative frequency on the vertical axis. Because \(n = 100\), each relative frequency is simply the count divided by 100.
Figure 2.2.2 — Relative frequency histogram of the heights of 100 semiprofessional soccer players, in 2-inch class intervals starting at 60.
Answer: eight bars of width 2, running from 60 to 76, with the tallest bars in the \([66, 68)\) and \([68, 70)\) intervals.
2.2.3 Histograms for Discrete Data
Discrete data — counts of things — get one useful simplification, and it falls straight out of the rule you already have. Take a width of 1. Because the data are integers and every interval includes its left boundary but not its right, the interval \([3, 4)\) contains the value 3 and nothing else: 4 has already gone to the next bar along. So each bar holds exactly one count, and the bar reads as "this many people answered exactly 3."
Older treatments shift the boundaries by half a unit, starting at 0.5 so that 3 sits dead center in the interval 2.5 to 3.5. That works, but it is a second convention to remember and it puts fractions on an axis where the data have none. The left-inclusive rule already gives you one value per bar, so discrete data needs no special case at all.
The starting-point rule needs no special case either: round the smallest value down, which for integer data is the value itself. If the data run from 1 to 6, start at 1 and the boundaries are 1, 2, 3, 4, 5, 6, 7 — six bars, the last one \([6, 7]\) closed on both ends so the largest value is counted.
One thing does change when you draw it. The tick marks now sit at the edges of the bars rather than under their centers, so label each bar with the value it holds instead of reading the value off the nearest tick. The ticks tell you where the bars break; the labels tell you what each bar counts.
The following data are the number of sports played by 50 student athletes. The number of sports is discrete data since sports are counted.
1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1
2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2; 2
3; 3; 3; 3; 3; 3; 3; 3
Twenty student athletes play one sport. Twenty-two student athletes play two sports. Eight student athletes play three sports. Since the data consist of the numbers 1, 2, 3, the starting point is 1 and a width of one puts the 1 alone in the interval \([1, \_\_\_)\), the 2 alone in \([\_\_\_, \_\_\_)\), and the 3 alone in the closed last interval \([\_\_\_, \_\_\_]\). Fill in the blanks and construct the histogram.
Solution
Step 1 — Fill in the boundaries. With a starting point of 1 and a width of 1, the boundaries step up by one each time: 1, 2, 3, 4.
- The 1 is alone in the interval \([1, \mathbf{2})\).
- The 2 is alone in the interval \([\mathbf{2}, \mathbf{3})\).
- The 3 is alone in the closed last interval \([\mathbf{3}, \mathbf{4}]\).
Each interval holds one value because the width is 1 and every interval excludes its right boundary — the 2 does not sneak into the first bar, it starts the second one.
Step 2 — Read off the heights. The frequencies are 20, 22, and 8.
Step 3 — Check. \(20 + 22 + 8 = 50\), which matches the 50 athletes surveyed.
Answer: three bars of width 1, spanning \([1, 2)\), \([2, 3)\) and \([3, 4]\), with heights 20, 22, and 8. The tallest bar is the one holding 2, so playing two sports is the most common answer.
Dr. Claire Nesbitt, who teaches at ABC College alongside her wife, asked 50 part-time students there how many books they bought this term. Create a histogram for her data. The number of books is discrete data, since books are counted.
1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1
2; 2; 2; 2; 2; 2; 2; 2; 2; 2
3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3
4; 4; 4; 4; 4; 4
5; 5; 5; 5; 5
6; 6
Solution
Step 1 — Count each value. Eleven students buy one book. Ten students buy two books. Sixteen students buy three books. Six students buy four books. Five students buy five books. Two students buy six books.
Step 2 — Set the starting point. Round the smallest data value down to a whole number. It is 1, already whole, so the starting point is 1.
Step 3 — Choose the width. The data are the integers 1 through 6, so take a width of 1. The boundaries are then 1, 2, 3, 4, 5, 6, 7, and because every interval excludes its right boundary, each one holds exactly one book count: \([1, 2)\) holds the 1s, \([2, 3)\) holds the 2s, and so on to \([6, 7]\), which is closed at both ends and holds the 6s.
Step 4 — Calculate the number of bars.
$$ 7 - 1 = 6 \qquad 6 \div 1 = 6 $$Here 1 is the width of a bar, so the number of bars is 6.
Step 5 — Draw it. The histogram below displays the number of books on the horizontal axis and the frequency on the vertical axis.
Figure 2.2.3 — Frequency histogram of the number of books bought by 50 part-time students, one bar per book count.
Answer: six bars of width 1 spanning \([1, 2)\) through \([6, 7]\), with heights 11, 10, 16, 6, 5, and 2.
Try it in rāSHio
Open rāSHio, enter the fifty students' book counts into the spreadsheet — eleven 1s, ten 2s, sixteen 3s, six 4s, five 5s and two 6s — then choose Graph → Histogram and set the bin start to 1 with a width of 1. That is exactly the scheme Step 3 derived, so each book count lands alone in its own bar and the six heights 11, 10, 16, 6, 5 and 2 read straight off the chart. Try nudging the bin start to 0.5 afterwards: the bars slide half a unit and the counts stay put, which is the clearest way to see that the boundary convention moves the picture without changing the data.
2.2.4 When a Value Lands on a Boundary
Sometimes you inherit a data set and an interval scheme that were not designed together, and values do land on boundaries. There is a convention for that.
A value is counted in a class interval if it falls on the left boundary of that interval, but not if it falls on the right boundary. Different researchers may set up histograms for the same data in different ways — there is more than one correct way to set up a histogram, so long as you state which convention you used and apply it consistently.
The phrase "more than one correct way" is doing real work in that box, and it is worth pausing on before you meet the next example. Two analysts can take the same 25 numbers, pick different starting points, pick different widths, and produce two histograms that look genuinely different — one smooth and mound-shaped, one lumpy with a gap in the middle — and neither of them has made a mistake. What makes a histogram wrong is not a choice you disagree with; it is an inconsistency, like counting one boundary value to the left and the next boundary value to the right. So when you construct one, write down your starting point, your width, and your boundary convention before you count a single value. That record is what lets a reader reproduce your picture, and reproducibility is the only standard a histogram can be held to.
The following data represent the number of employees at various restaurants in New York City. Using this data, create a histogram. Use 10–19 as the first interval.
22; 35; 15; 26; 40; 28; 18; 20; 25; 34; 39; 42; 24; 22; 19; 27; 22; 34; 40; 20; 38; 28
Solution
Step 1 — Set up the intervals. We are told to start with 10–19, so the width is 10 and the intervals are 10–19, 20–29, 30–39, 40–49.
Step 2 — Tally each interval.
- 10–19: 15, 18, 19 → frequency 3
- 20–29: 22, 26, 28, 20, 25, 24, 22, 27, 22, 20, 28 → frequency 11
- 30–39: 35, 34, 39, 34, 38 → frequency 5
- 40–49: 40, 42, 40 → frequency 3
Step 3 — Check. \(3 + 11 + 5 + 3 = 22\), which matches the 22 restaurants.
Answer: four bars of width 10, with heights 3, 11, 5, and 3. The distribution has a single strong peak in the 20s and thin tails on both sides — most of these restaurants employ twenty-something people.
Tou Xiong asked his classmates how many hours they spent playing video games on weekends, and recorded their answers below. Using his data set, construct a histogram.
| 9.95 | 10 | 2.25 | 16.75 | 0 |
| 19.5 | 22.5 | 7.5 | 15 | 12.75 |
| 5.5 | 11 | 10 | 20.75 | 17.5 |
| 23 | 21.9 | 24 | 23.75 | 18 |
| 20 | 15 | 22.9 | 18.8 | 20.5 |
Solution
Step 1 — Look at the range. The values run from 0 to 24, so intervals of width 5 running 0–5, 5–10, 10–15, 15–20, 20–25 give five bars, which is at the low end of the usual 5-to-15 range but perfectly readable for 25 values.
Step 2 — Apply the boundary rule. Some values in this data set fall on boundaries for the class intervals — 10, 15, 20 all sit exactly on a fence. A value is counted in a class interval if it falls on the left boundary, but not if it falls on the right boundary. So 10 counts in 10–15, not in 5–10.
Step 3 — Draw it.
Figure 2.2.4 — Histogram of hours spent playing video games on weekends, with values on a boundary counted in the interval to the right.
Answer: the histogram above. Different researchers may set up histograms for the same data in different ways — there is more than one correct way to set up a histogram, provided the boundary convention is stated and used consistently.
Try it in rāSHio
Open rāSHio, paste the twenty-five weekend hours (9.95; 10; 2.25; 16.75; 0; …) into File → Delimited List… — the semicolons parse as-is — then choose Graph → Histogram with a bin start of 0 and a width of 5. Now watch the three boundary values 10, 15 and 20: nudge the bin start to 0.5 and each of them moves a bar, and the picture changes. That is the whole reason this section asks you to write your convention down.
Count the money (bills and change) in your pocket or purse. Your instructor will record the amounts. As a class, construct a histogram displaying the data. Discuss how many intervals you think is appropriate. You may want to experiment with the number of intervals.
2.2.5 How NOT to Lie with Statistics
The reason we build so many ways to display data is to get a sense of what the observations are telling us. Are the values all much alike, or spread across a wide range? Bunched at one end, or spread evenly? A picture answers those questions faster than a column of numbers ever will. But the same picture that can clarify can also mislead — sometimes on purpose, and often by accident. Formal measures come later in this chapter; for now, the goal is to make the picture honest.
Pie charts run into trouble first. Too many slices and the reader gives up: past five or six categories nobody can tell which piece matters most, which is the only question a pie chart is good at answering. Fold the small ones into an "other" slice, or pick a different chart. A pie chart also cannot show change over time, though it gets asked to constantly. Two pies side by side — this year's revenue and last year's — look comparable and are not, because both pies are drawn the same size while the totals behind them differ. A slice that grew in dollars can shrink on the page, and nothing in the picture tells you which happened.
Histograms have their own version of the trick, and it is the bar width. Because a histogram invites you to compare areas, widening one category's bar hands that category extra area it did not earn, and the eye reads the extra area as extra probability. Equal widths are what make the comparison mean anything; unequal widths make it meaningless while still looking like a chart.
Graphs that plot a quantity over time invite a subtler kind of manipulation, and it is worth recognizing even if you never draw one. The numbers on such a graph are usually correct; the deception lives in the axes. Stretching or compressing the time axis makes the same trend look violent or calm at will. Choosing units that are small enough, or starting the vertical axis well above zero, turns an unremarkable change into a cliff. None of these tricks alters a single value, which is exactly why checking the arithmetic will never catch them and reading the axes will.
Every graph described above plots correct data. The lie lives in the axis, the bar width, or the fixed-size pie — which is why checking someone's arithmetic will not catch it, and looking at their axes will.
Deliberate manipulation of this kind is fraud. But the same errors get made honestly all the time, and the result on the reader's understanding is identical. The goal of descriptive statistics is a visual that tells the story the data actually tell; anything else, intended or not, is a wrong answer.
A city council presents a chart of its annual budget. The vertical axis starts at $9,800,000 and ends at $10,200,000. Over five years the budget rose from $9.9 million to $10.1 million — about a 2% increase — but on the chart the last bar is roughly three times the height of the first. What has been done, and how would you redraw the chart honestly?
Solution
Step 1 — Name the trick. The vertical axis does not start at zero. It starts at $9,800,000, so only the top $400,000 of a $10 million budget is on the page at all.
Step 2 — See why the bars mislead. A bar's height is supposed to stand for the whole quantity. With a truncated axis, the height stands only for "how far above $9,800,000 this value is." The first bar represents \(9{,}900{,}000 - 9{,}800{,}000 = \$100{,}000\) of visible height and the last represents \(10{,}100{,}000 - 9{,}800{,}000 = \$300{,}000\), so the last bar is three times as tall:
$$ \frac{10{,}100{,}000 - 9{,}800{,}000}{9{,}900{,}000 - 9{,}800{,}000} = \frac{300{,}000}{100{,}000} = 3 $$Step 3 — Compare to the real change. The actual increase is this, a little over two percent:
$$ \frac{10{,}100{,}000 - 9{,}900{,}000}{9{,}900{,}000} \approx 0.020 = 2.0\% $$A 2% rise is being drawn as a bar three times taller — a 200% jump in height.
Answer: the vertical axis has been truncated. Redraw it starting at zero, and the five bars become nearly indistinguishable — which is the honest picture of a 2% change. If the small change is genuinely the story, say so in words or plot the change itself on its own axis; do not let a cropped axis make the argument silently. Notice that every number on the original chart was correct.
Problem Set 2.2
Problem 1. Beatriz Salcedo and her wife run a dealership group. She asked sixty-five randomly selected car salespersons the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. Complete the table.
| Data Value (# cars) | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 3 | |||
| 4 | |||
| 5 | |||
| 6 | |||
| 7 |
Solution
Step 1 — Fill in the frequency column. The counts are given in words: fourteen sell three cars, nineteen sell four, twelve sell five, nine sell six, eleven sell seven. So the frequencies are 14, 19, 12, 9, 11.
Step 2 — Find \(n\). Add the frequencies:
$$ n = 14 + 19 + 12 + 9 + 11 = 65 $$That matches the 65 salespersons Beatriz surveyed, which is the first check worth making.
Step 3 — Divide each frequency by 65 to get the relative frequency, rounded to four decimal places.
$$ \frac{14}{65} = 0.2154 \qquad \frac{19}{65} = 0.2923 $$ $$ \frac{12}{65} = 0.1846 \qquad \frac{9}{65} = 0.1385 $$ $$ \frac{11}{65} = 0.1692 $$Step 4 — Build the cumulative column by running a total down the relative frequency column:
$$ 0.2154 + 0.2923 = 0.5077 $$ $$ 0.5077 + 0.1846 = 0.6923 $$ $$ 0.6923 + 0.1385 = 0.8308 $$ $$ 0.8308 + 0.1692 = 1.0000 $$Answer: the completed table.
| # cars | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 3 | 14 | 0.2154 | 0.2154 |
| 4 | 19 | 0.2923 | 0.5077 |
| 5 | 12 | 0.1846 | 0.6923 |
| 6 | 9 | 0.1385 | 0.8308 |
| 7 | 11 | 0.1692 | 1.0000 |
The last cumulative entry landing exactly on 1.0000 is the check that the whole table is consistent.
Problem 2. What does the frequency column in Table 2.2.3 sum to? Why?
Solution
Step 1 — Add the column.
$$ 14 + 19 + 12 + 9 + 11 = 65 $$Step 2 — Say why that number and no other. A frequency column counts people, and every one of the 65 salespersons gave exactly one answer. Each person is therefore tallied in exactly one row — never skipped, never double-counted — so the column has to add back up to the number of people surveyed.
Answer: the frequency column sums to 65, which is \(n\), the total number of salespersons surveyed. If it had summed to anything else, a response was either lost or counted twice.
Problem 3. What does the relative frequency column in Table 2.2.3 sum to? Why?
Solution
Step 1 — Add the column.
$$ 0.2154 + 0.2923 + 0.1846 $$ $$ +\; 0.1385 + 0.1692 = 1.0000 $$Step 2 — Say why it must. Each entry is \(f/n\), so adding them is the same as adding the frequencies first and then dividing once:
$$ \sum \frac{f}{n} = \frac{\sum f}{n} = \frac{65}{65} = 1 $$Answer: the relative frequency column sums to 1, because the frequencies sum to \(n\) and every entry is that frequency divided by the same \(n\). A relative frequency column is the whole sample split into shares, and the shares of a whole always add to one whole. If yours does not, expect a rounding artifact in the last decimal place or an arithmetic slip — anything larger is a real error.
Problem 4. What is the difference between relative frequency and frequency for each data value in Table 2.2.3?
Solution
Step 1 — Name what each one measures. The frequency of a data value is a raw count: how many salespersons gave that answer. The relative frequency is that same count expressed as a share of the whole sample, \(RF = f/n\).
Step 2 — Compare them row by row. For 3 cars the frequency is 14 and the relative frequency is \(14/65 = 0.2154\); for 4 cars, 19 and \(19/65 = 0.2923\); and so on. Every relative frequency here is the frequency in the same row divided by 65.
Step 3 — Note the practical difference. Frequency answers "how many?" and depends on how big the survey was. Relative frequency answers "what share?" and does not — 14 out of 65 is 21.5% whether the survey had 65 people or 6,500 with the same proportions, which is what makes relative frequency the right column when you want to compare two samples of different sizes.
Answer: frequency is the count \(f\); relative frequency is that count divided by the sample size, \(f/65\). The frequency column sums to 65, the relative frequency column sums to 1.
Problem 5. What is the difference between cumulative relative frequency and relative frequency for each data value?
Solution
Step 1 — Define each one. The relative frequency of a value is that value's own share of the sample, \(f/n\). The cumulative relative frequency of a value is the sum of the relative frequencies of that value and every value below it.
Step 2 — See it in the table. For 5 cars, the relative frequency is 0.1846 — the share who sell exactly five. The cumulative relative frequency is
\(0.2154 + 0.2923 + 0.1846 = 0.6923\) — the share who sell five cars or fewer.
Step 3 — Note the two consequences. A relative frequency column can go up and down as you read down it; a cumulative column can only ever climb, because each entry adds a non-negative amount to the one above. And the last cumulative entry is always 1, since by then every value has been included.
Answer: relative frequency is one value's share on its own ("exactly this many"); cumulative relative frequency is the running total of all shares up to and including that value ("this many or fewer"). The cumulative column never decreases and ends at 1.0000.
Problem 6. To construct the histogram for the data in Table 2.2.3, determine appropriate minimum and maximum horizontal and vertical values and the scaling. Sketch the histogram. Label the horizontal and vertical axes with words. Include numerical scaling.
Solution
Step 1 — Choose the horizontal scale. The data values are the whole numbers 3 through 7. Round the smallest value down — it is already whole, so the starting point is 3 — and take a width of 1, which puts one car count in each bar. The boundaries are then 3, 4, 5, 6, 7, 8, so the horizontal axis needs to run from 3 to 8.
Step 2 — Choose the vertical scale. The tallest bar is the 19 salespersons who sell four cars, so run the vertical axis from 0 to 20 and mark it every 2 units. Starting at 0 is not optional here: a bar's height stands for its whole count, so cutting the axis off above zero would misreport every bar's size.
Step 3 — Label both axes in words. Horizontal: Number of cars sold per week. Vertical: Frequency (number of salespersons). Numerical scaling: horizontal ticks at 3, 4, 5, 6, 7, 8; vertical ticks at 0, 2, 4, …, 20.
Step 4 — Draw the bars, touching, at the heights from the frequency column.
- \([3, 4)\) → height 14
- \([4, 5)\) → height 19
- \([5, 6)\) → height 12
- \([6, 7)\) → height 9
- \([7, 8]\) → height 11 (the last interval closes on both ends)
Answer: five contiguous bars of width 1 spanning 3 to 8, with heights 14, 19, 12, 9, and 11 on a vertical axis running 0 to 20. The heights sum to 65, matching the sample. The distribution peaks at four cars and then falls away, with a small rise again at seven.
Problem 7. Suppose that three book publishers were interested in the number of fiction paperbacks adult consumers purchase per month. Each publisher conducted a survey. In the survey, adult consumers were asked the number of fiction paperbacks they had purchased the previous month. The results are below.
| # of books | Freq. | Rel. Freq. |
|---|---|---|
| 0 | 10 | |
| 1 | 12 | |
| 2 | 16 | |
| 3 | 12 | |
| 4 | 8 | |
| 5 | 6 | |
| 6 | 2 | |
| 8 | 2 |
| # of books | Freq. | Rel. Freq. |
|---|---|---|
| 0 | 18 | |
| 1 | 24 | |
| 2 | 24 | |
| 3 | 22 | |
| 4 | 15 | |
| 5 | 10 | |
| 7 | 5 | |
| 9 | 1 |
| # of books | Freq. | Rel. Freq. |
|---|---|---|
| 0–1 | 20 | |
| 2–3 | 35 | |
| 4–5 | 12 | |
| 6–7 | 2 | |
| 8–9 | 1 |
a) Find the relative frequencies for each survey. Write them in the charts.
b) Using either a graphing calculator, computer, or by hand, use the frequency column to construct a histogram for each publisher's survey. For Publishers A and B, make bar widths of one. For Publisher C, make bar widths of two.
c) In complete sentences, give two reasons why the graphs for Publishers A and B are not identical.
d) Would you have expected the graph for Publisher C to look like the other two graphs? Why or why not?
e) Make new histograms for Publisher A and Publisher B. This time, make bar widths of two.
f) Now, compare the graph for Publisher C to the new graphs for Publishers A and B. Are the graphs more similar or more different? Explain your answer.
Solution
Step 1 (part a) — Find each publisher's sample size.
$$ n_A = 10 + 12 + 16 + 12 + 8 + 6 + 2 + 2 $$ $$ n_A = 68 $$ $$ n_B = 18 + 24 + 24 + 22 + 15 + 10 + 5 + 1 $$ $$ n_B = 119 $$ $$ n_C = 20 + 35 + 12 + 2 + 1 = 70 $$Step 2 (part a) — Divide each frequency by its own \(n\).
Publisher A: 0.1471, 0.1765, 0.2353, 0.1765, 0.1176, 0.0882, 0.0294, 0.0294.
Publisher B: 0.1513, 0.2017, 0.2017, 0.1849, 0.1261, 0.0840, 0.0420, 0.0084.
Publisher C: 0.2857, 0.5000, 0.1714, 0.0286, 0.0143.
Each column sums to 1.0000 up to rounding in the last decimal place.
Step 3 (part b) — Construct the three histograms. For A and B use a width of 1, so each bar holds a single book count; the bars stand over 0, 1, 2, … at the frequencies given. For C use a width of 2, giving five bars over 0–1, 2–3, 4–5, 6–7, 8–9 at heights 20, 35, 12, 2, 1.
Step 4 (part c) — Two reasons A and B are not identical. First, the surveys are different sizes: 68 people versus 119, so the same share of respondents produces a taller bar for B than for A. Second, the two samples do not even cover the same values — A has a bar at 8 books and none at 7 or 9, while B has bars at 7 and 9 and none at 8. The gaps sit in different places, so the two pictures cannot line up.
Step 5 (part d) — Would you expect C to match? No. C's data are already grouped into intervals two books wide, so it has five bars where A and B have eight. Grouping merges neighbouring counts and hides the detail between them, which changes the shape of the picture even when the underlying buying habits are similar. Comparing a five-bar histogram to an eight-bar one compares bin choices as much as data.
Step 6 (part e) — Redraw A and B with bars of width two. Combine adjacent counts.
Publisher A: 0–1 → \(10 + 12 = 22\); 2–3 → \(16 + 12 = 28\); 4–5 → \(8 + 6 = 14\); 6–7 → \(2 + 0 = 2\); 8–9 → \(2 + 0 = 2\). Total 68.
Publisher B: 0–1 → \(18 + 24 = 42\); 2–3 → \(24 + 22 = 46\); 4–5 → \(15 + 10 = 25\); 6–7 → \(5 + 0 = 5\); 8–9 → \(0 + 1 = 1\). Total 119.
Step 7 (part f) — Compare all three on the common scale. In relative frequency the five bars now read:
A: 0.324, 0.412, 0.206, 0.029, 0.029 B: 0.353, 0.387, 0.210, 0.042, 0.008 C: 0.286, 0.500, 0.171, 0.029, 0.014
Answer: the graphs are far more similar once all three use the same bar width. All three peak in the 2–3 interval, drop sharply after 4–5, and trail to almost nothing past 6. Most of the difference in part (d) came from the bin width, not from the readers — which is exactly the warning this section gives about bar width: change the width and you change the picture without changing a single data value.
Problem 8. Often, cruise ships conduct all on-board transactions, with the exception of gambling, on a cashless basis. At the end of the cruise, guests pay one bill that covers all onboard transactions. Suppose that 60 single travelers and 70 couples were surveyed as to their on-board bills for a seven-day cruise from Los Angeles to the Mexican Riviera. Following is a summary of the bills for each group.
| Amount (\$) | Frequency | Rel. Frequency |
|---|---|---|
| 51–100 | 5 | |
| 101–150 | 10 | |
| 151–200 | 15 | |
| 201–250 | 15 | |
| 251–300 | 10 | |
| 301–350 | 5 |
| Amount (\$) | Frequency | Rel. Frequency |
|---|---|---|
| 100–149 | 5 | |
| 150–199 | 5 | |
| 200–249 | 15 | |
| 250–299 | 15 | |
| 300–349 | 10 | |
| 350–399 | 10 | |
| 400–449 | 10 |
a) Fill in the relative frequency for each group.
b) Construct a histogram for the singles group. Scale the horizontal axis by $50 widths. Use relative frequency on the vertical axis.
c) Construct a histogram for the couples group. Scale the horizontal axis by $50 widths. Use relative frequency on the vertical axis.
d) Compare the two graphs. How do the graphs differ in shape, center, and spread?
e) Construct a new graph for the couples by hand. Since each couple is paying for two individuals, instead of scaling the horizontal axis by $50, scale it by $100. Use relative frequency on the vertical axis.
f) Compare the graph for the singles with the new graph for the couples. How do the graphs differ in shape, center, and spread?
g) How did scaling the couples graph differently change the way you compared it to the singles graph?
h) Based on the graphs, do you think that individuals spend the same amount, more or less, as singles as they do person-by-person as a couple? Explain why in one or two complete sentences.
Solution
Step 1 (part a) — Divide each frequency by its group's total. The singles group has \(n = 5 + 10 + 15 + 15 + 10 + 5 = 60\); the couples group has \(n = 5 + 5 + 15 + 15 + 10 + 10 + 10 = 70\).
Singles: 0.0833, 0.1667, 0.2500, 0.2500, 0.1667, 0.0833.
Couples: 0.0714, 0.0714, 0.2143, 0.2143, 0.1429, 0.1429, 0.1429.
Each column sums to 1.0000 up to rounding.
Step 2 (part b) — Singles histogram. Bars $50 wide spanning $51 to $350, relative frequency on the vertical axis, heights 0.0833, 0.1667, 0.2500, 0.2500, 0.1667, 0.0833. Label the horizontal axis On-board bill ($).
Step 3 (part c) — Couples histogram. Bars $50 wide spanning $100 to $449, relative frequency on the vertical axis, heights 0.0714, 0.0714, 0.2143, 0.2143, 0.1429, 0.1429, 0.1429.
Step 4 (part d) — Compare shape, center, and spread. The singles graph is symmetric: it rises to a flat pair of tallest bars and falls away evenly on both sides, centered around $200. The couples graph is flatter and does not fall away on the right — the last three bars are all the same height — so its center sits higher, around $275, and its bills stretch from $100 to $449 instead of $51 to $350. Couples spend more in total and their bills vary more.
Step 5 (part e) — Redraw the couples on $100 bars. Merge adjacent $50 bins: $100–199 → \(5 + 5 = 10\); $200–299 → \(15 + 15 = 30\); $300–399 → \(10 + 10 = 20\); $400–499 → 10. Total 70. In relative frequency: 0.143, 0.429, 0.286, 0.143.
Step 6 (part f) — Compare the singles graph to the new couples graph. Because a couple's bill covers two people, a $100 interval for a couple is about a $50 interval per person — so this graph and the singles graph are now on comparable per-person footing. Both are single-peaked with short tails. The couples' modal bin, $200–299 per couple, is roughly $100–150 per person, which sits below the singles' peak near $200.
Step 7 (part g) — What the rescaling changed. On the $50 scale the couples graph had more bars and looked more spread out, which was partly a fact about the couples and partly a fact about the bin width. Doubling the width put both graphs on the same per-person scale, so what was left to compare was the spending, not the drawing.
Answer (part h): a person appears to spend more as a single traveler than as one half of a couple. On the per-person scale the singles cluster near $200 while the couples cluster near $100–150 each. That is not surprising: couples share cabins, split some purchases, and buy fewer duplicate items than two people travelling alone would.
Problem 9. Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows.
| # of movies | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 0 | 5 | ||
| 1 | 9 | ||
| 2 | 6 | ||
| 3 | 4 | ||
| 4 | 1 |
a) Construct a histogram of the data.
b) Complete the columns of the chart.
Solution
Step 1 (part b) — Find \(n\). Adding the frequencies gives \(n = 5 + 9 + 6 + 4 + 1 = 25\), which matches the 25 students surveyed.
Step 2 (part b) — Divide each frequency by 25.
$$ \frac{5}{25} = 0.20 \qquad \frac{9}{25} = 0.36 $$ $$ \frac{6}{25} = 0.24 \qquad \frac{4}{25} = 0.16 $$ $$ \frac{1}{25} = 0.04 $$Step 3 (part b) — Run the cumulative total down the relative frequency column: 0.20, 0.56, 0.80, 0.96, 1.00.
| # of movies | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 0 | 5 | 0.20 | 0.20 |
| 1 | 9 | 0.36 | 0.56 |
| 2 | 6 | 0.24 | 0.80 |
| 3 | 4 | 0.16 | 0.96 |
| 4 | 1 | 0.04 | 1.00 |
Step 4 (part a) — Construct the histogram. The data are the whole numbers 0 through 4, so the starting point is 0 and a width of 1 puts one movie count in each bar. The boundaries are 0, 1, 2, 3, 4, 5, and the bars are \([0, 1)\), \([1, 2)\), \([2, 3)\), \([3, 4)\), and the closed last interval \([4, 5]\), at heights 5, 9, 6, 4, and 1. Label the horizontal axis Number of movies watched last week and the vertical axis Frequency, running 0 to 10.
Answer: five bars of width 1 with heights 5, 9, 6, 4, 1, summing to 25. The peak is at one movie, and the distribution has a longer tail to the right than to the left — most students watched none, one, or two, and only one student watched four.
For the next two exercises: suppose 111 people who shopped in a special T-shirt store were asked the number of T-shirts they own costing more than $19 each. Their answers were displayed in a relative frequency histogram with bars for 1, 2, 3, 4, and 5 shirts.
Problem 10. The percentage of people who own at most three T-shirts costing more than $19 each is approximately:
a) 21
b) 59
c) 41
d) Cannot be determined
Solution
Step 1 — Read the question. "At most three" means 1, 2, or 3 shirts — three of the histogram's five bars.
Step 2 — Know what to add. On a relative frequency histogram each bar's height is already that value's share of the 111 shoppers, so the percentage who own at most three is just the sum of the first three bar heights. No division by 111 is needed; that has already been done.
$$ \text{percent} = (RF_1 + RF_2 + RF_3) \times 100\% $$Step 3 — Add them. Reading the three bars off the store's histogram and summing gives approximately 0.41.
Answer: c. 41. Two sanity checks on the other options. Choice (d) is wrong because a relative frequency histogram carries everything the question needs — the shares are already on the page. And 21 and 59 are the traps: 21 is roughly the share owning at most two, and 59 is the complement, the share owning four or more, which is what you get by adding the wrong end of the graph. Note that this exercise refers to a histogram published with the original survey which is not reproduced here; the method above is what the exercise is testing.
Problem 11. If the data were collected by asking the first 111 people who entered the store, then the type of sampling is:
a) cluster
b) simple random
c) stratified
d) convenience
Solution
Step 1 — Read what the researcher actually did. They stood at the door and asked the first 111 people who walked in. Nothing about the selection was random, and nothing divided the shoppers into groups first.
Step 2 — Rule the other three out. Simple random would require every shopper to have an equal chance of being picked, which a first-come rule does not give — anyone who arrives after the 111th has no chance at all. Stratified would require splitting the population into subgroups and sampling within each. Cluster would require picking whole groups at random and surveying everyone inside them. None of those happened.
Step 3 — Name what is left. The sample was chosen because it was easy to reach, which is the definition of a convenience sample.
Answer: d. convenience. It is worth naming the risk this carries: early-in-the-day shoppers may differ systematically from evening shoppers, so the sample may not represent the store's customers even though 111 is a respectable number of responses.
Problem 12. Following are the rates of unmarried adults for the 50 U.S. states and Washington, DC. Construct a bar graph of the unmarried adult rates of your state and the four states closest to your state. Hint: label the horizontal axis with the states.
| State | Percent (%) | State | Percent (%) | State | Percent (%) |
|---|---|---|---|---|---|
| Alabama | 32.2 | Kentucky | 31.3 | North Dakota | 27.2 |
| Alaska | 24.5 | Louisiana | 31.0 | Ohio | 29.2 |
| Arizona | 24.3 | Maine | 26.8 | Oklahoma | 30.4 |
| Arkansas | 30.1 | Maryland | 27.1 | Oregon | 26.8 |
| California | 24.0 | Massachusetts | 23.0 | Pennsylvania | 28.6 |
| Colorado | 21.0 | Michigan | 30.9 | Rhode Island | 25.5 |
| Connecticut | 22.5 | Minnesota | 24.8 | South Carolina | 31.5 |
| Delaware | 28.0 | Mississippi | 34.0 | South Dakota | 27.3 |
| Washington, DC | 22.2 | Missouri | 30.5 | Tennessee | 30.8 |
| Florida | 26.6 | Montana | 23.0 | Texas | 31.0 |
| Georgia | 29.6 | Nebraska | 26.9 | Utah | 22.5 |
| Hawaii | 22.7 | Nevada | 22.4 | Vermont | 23.2 |
| Idaho | 26.5 | New Hampshire | 25.0 | Virginia | 26.0 |
| Illinois | 28.2 | New Jersey | 23.8 | Washington | 25.5 |
| Indiana | 29.6 | New Mexico | 25.1 | West Virginia | 32.5 |
| Iowa | 28.4 | New York | 23.9 | Wisconsin | 26.3 |
| Kansas | 29.4 | North Carolina | 27.8 | Wyoming | 25.1 |
Solution
Step 1 — Pick your state and its four nearest neighbours, then pull each rate from the table. Worked here with California; use your own state the same way.
- California — 24.0%
- Oregon — 26.8%
- Nevada — 22.4%
- Arizona — 24.3%
- Idaho — 26.5%
Step 2 — Set up the axes. The states are categories, not numbers, so they go on the horizontal axis in whatever order you choose — alphabetical, or sorted by rate to make the comparison easier to read. The vertical axis carries the percentage. Run it from 0 to 35 and mark it every 5 percentage points.
Step 3 — Start the vertical axis at zero. All five rates sit between 22% and 27%, so it is tempting to start the axis at 20 and let the differences fill the page. Do not: a bar's height stands for the whole rate, and cropping the axis would make a 4-point difference look like a threefold one. This is exactly the trick the section's last subsection warns about.
Step 4 — Draw the bars with gaps between them. This is a bar graph, not a histogram. The categories are separate states rather than neighbouring intervals on a number line, so the bars must not touch — the gaps are what tell the reader the horizontal axis is categorical.
Answer: a five-bar graph, states on the horizontal axis, percent unmarried on a vertical axis running 0 to 35, bars separated by gaps, at heights 24.0, 26.8, 22.4, 24.3, and 26.5. Read off it: Oregon has the highest rate of the five and Nevada the lowest, and the whole group spans only about 4.4 percentage points — a genuinely narrow range that an honest zero-based axis shows as a nearly level row of bars.
Key Terms
histogram — a graph of contiguous boxes over a numeric horizontal axis, with bar height showing the frequency or relative frequency of each class interval.
relative frequency — the frequency of a value divided by the total number of data values, \(RF = f/n\); the value's share of the whole sample.
class interval — one of the contiguous ranges the horizontal axis is divided into; also called a bin or a class.
class width — the common size of each class interval: the range divided by the number of bars, then rounded up to a whole number.
starting point — the whole number at or below the smallest data value, found by rounding that value down; it is the left boundary of the first class interval.