1.3 Frequency, Frequency Tables, and Levels of Measurement
Learning Objectives
- Round answers correctly by carrying one extra decimal place and avoiding premature rounding.
- Classify data by its level of measurement: nominal, ordinal, interval, or ratio.
- Build a frequency table, and extend it with relative frequency and cumulative relative frequency columns.
- Read percentages and counts directly off a frequency table.
Once you have a pile of data, your first job is to get it organized so you can see how often each value shows up. That is what this whole section is about: counting, turning those counts into fractions and percents, and stacking them up so you can answer questions like "what fraction of the group scored below this line?" Before we count anything, though, we need one quick ground rule about rounding.
1.3.1 Answers and Rounding Off
Here is a simple rounding habit that keeps your answers honest: carry your final answer one more decimal place than the original data had. Round off only the final answer — never the numbers you use along the way. If you absolutely must round an intermediate result, keep at least twice as many decimal places as you plan to keep in the final answer, so the rounding error never creeps into the part you report.
For example, the average of the three quiz scores 4, 6, and 9 is \(6.3\), rounded to the nearest tenth. The data are whole numbers (zero decimal places), so the final answer gets one decimal place. Most answers in this course are rounded this way.
One more note: in this course you usually do not need to reduce fractions. Especially in the probability chapter, it is often more useful to leave an answer as an unreduced fraction like \(\frac{3}{20}\) than to reduce it — the unreduced form keeps the "3 out of 20" story visible.
The data values 12.4, 9.8, and 15.0 are measured to one decimal place. You compute their average. To how many decimal places should you report the final answer, and what is that average?
Solution
The original data are given to one decimal place, so the final answer carries one more — two decimal places. Compute the average without rounding along the way:
$$ \frac{12.4 + 9.8 + 15.0}{3} = \frac{37.2}{3} = 12.4 $$Rounded to two decimal places, that is \(12.40\).
Answer: report two decimal places; the average is \(12.40\).
1.3.2 Levels of Measurement
Data measured on a nominal scale are qualitative (categorical). They are names, labels, or categories with no meaningful order and cannot be used in calculations.
Think of categories, colors, names, labels, favorite foods, or yes/no answers. Trying to rank them makes no sense — putting pizza "first" and sushi "second" carries no real meaning. Smartphone brands are another example: the data are just the names of the companies, and there is no agreed-upon order, even though you might personally prefer one. You can't do arithmetic on nominal data.
Figure 1.3.1 — Nominal data are names only: shuffle the categories any way you like and nothing changes.
Data measured on an ordinal scale are like nominal data but can be put in a meaningful order. However, the differences between values cannot be measured, and ordinal data cannot be used in calculations.
An example is a list of the top five national parks in the United States. You can rank them one through five, but you can't say how much better number one is than number two. A cruise survey is another case: responses of "excellent," "good," "satisfactory," and "unsatisfactory" run from most to least desired — that's an order — but the gap between "excellent" and "good" isn't a measurable amount.
Figure 1.3.2 — Ordinal data have a real order, but the gaps between values are not measurable.
Data measured on an interval scale have a definite order and are numerical, so differences between values can be calculated. There is no true zero — zero does not represent a minimum or "none."
Figure 1.3.3 — Interval data: differences are real everywhere on the scale, but zero is just another tick.
The giveaway for an interval scale is a zero that's just a marker, not an "empty" point. \(0^\circ\)F doesn't mean "no temperature" — it's colder days all the way down to \(-10^\circ\)F. Because zero is fake, ratios break: \(40^\circ\) is not "twice as hot" as \(20^\circ\). Differences still work, though, which is what makes it interval and not merely ordinal.
Temperature scales like Celsius and Fahrenheit are interval. In both, \(40^\circ\) equals \(100^\circ - 60^\circ\) — differences make sense. But \(0\) degrees is not the lowest possible value: temperatures like \(-10^\circ\)F and \(-15^\circ\)C exist and are colder than zero.
Data measured on a ratio scale give the most information. Ratio data are like interval data, but there is a true minimum (zero), so ratios can be calculated.
For example, four machine-graded multiple-choice statistics final exam scores (out of 100) are 80, 68, 20, and 92. Put them in order: 20, 68, 80, 92. Differences mean something — 92 is 24 points more than 68. And because the minimum score is a genuine \(0\) (zero points = no correct answers), ratios mean something too: \(80\) is four times \(20\), so a score of 80 really is four times better than a score of 20.
Figure 1.3.4 — Ratio data grow from a true zero, so 80 really is four times 20.
Why fuss over how data were measured? Because the type of data decides which math is even allowed. You can average exam scores, but averaging jersey colors is nonsense. Picking a statistical procedure that the data can't support is one of the most common ways analyses go wrong — so before you compute anything, you check what kind of data you're holding.
The way a set of data is measured is called its level of measurement. Correct statistical procedures depend on knowing this, because not every operation works with every kind of data. Data fall into four levels of measurement, listed here from lowest to highest:
- Nominal scale level
- Ordinal scale level
- Interval scale level
- Ratio scale level
Figure 1.3.5 — Each level of measurement keeps the powers of the one below it and adds one more.
A coffee shop records four pieces of information about each customer: the customer's loyalty tier (Bronze, Silver, Gold), the flavor they ordered, the outdoor temperature in \(^\circ\)F when they ordered, and the number of ounces in their drink. Classify each as nominal, ordinal, interval, or ratio.
Solution
- Flavor — names with no order → nominal.
- Loyalty tier — ordered (Bronze < Silver < Gold), but the gaps between tiers aren't a measurable amount → ordinal.
- Temperature in \(^\circ\)F — numerical, ordered, differences make sense, but \(0^\circ\)F is not "no temperature" (no true zero) → interval.
- Ounces in the drink — numerical with a true zero (0 oz = no drink), so ratios work (16 oz is twice 8 oz) → ratio.
Answer: flavor = nominal; tier = ordinal; temperature = interval; ounces = ratio.
1.3.3 Frequency
A frequency is the number of times a value of the data occurs.
Twenty students were asked how many hours they worked per day. Their responses, in hours, were:
5; 6; 3; 3; 2; 4; 7; 5; 2; 3; 5; 6; 5; 4; 4; 3; 5; 2; 5; 3
Table 1.3.1 lists the different data values in ascending order along with their frequencies.
| Data value | Frequency |
|---|---|
| 2 | 3 |
| 3 | 5 |
| 4 | 3 |
| 5 | 6 |
| 6 | 2 |
| 7 | 1 |
Figure 1.3.6 — A frequency is just a count: each response lands in its bin and the bin's tally ticks up.
According to Table 1.3.1, three students work two hours, five students work three hours, and so on. The sum of the frequency column, \(20\), is the total number of students in the sample.
A relative frequency is the ratio (a fraction or proportion) of the number of times a value occurs to the total number of outcomes.
To find each relative frequency, divide that value's frequency by the total number of students — here, \(20\). You can write relative frequencies as fractions, percents, or decimals. Table 1.3.2 adds the relative frequency column.
| Data value | Frequency | Relative frequency |
|---|---|---|
| 2 | 3 | \(\frac{3}{20}\) or 0.15 |
| 3 | 5 | \(\frac{5}{20}\) or 0.25 |
| 4 | 3 | \(\frac{3}{20}\) or 0.15 |
| 5 | 6 | \(\frac{6}{20}\) or 0.30 |
| 6 | 2 | \(\frac{2}{20}\) or 0.10 |
| 7 | 1 | \(\frac{1}{20}\) or 0.05 |
Figure 1.3.7 — Divide every frequency by the total and the counts become shares of one whole.
The sum of the relative frequency column in Table 1.3.2 is \(\frac{20}{20}\), or \(1\). That always happens: every member of the sample lands in exactly one row, so the fractions add up to the whole.
A cumulative relative frequency is the accumulation of the previous relative frequencies. To find it, add all the previous relative frequencies to the relative frequency for the current row.
Table 1.3.3 builds the cumulative column by running totals down the rows.
| Data value | Frequency | Relative frequency | Cumulative relative frequency |
|---|---|---|---|
| 2 | 3 | \(\frac{3}{20}\) or 0.15 | 0.15 |
| 3 | 5 | \(\frac{5}{20}\) or 0.25 | 0.15 + 0.25 = 0.40 |
| 4 | 3 | \(\frac{3}{20}\) or 0.15 | 0.40 + 0.15 = 0.55 |
| 5 | 6 | \(\frac{6}{20}\) or 0.30 | 0.55 + 0.30 = 0.85 |
| 6 | 2 | \(\frac{2}{20}\) or 0.10 | 0.85 + 0.10 = 0.95 |
| 7 | 1 | \(\frac{1}{20}\) or 0.05 | 0.95 + 0.05 = 1.00 |
Figure 1.3.8 — Cumulative relative frequency stacks the shares row by row until the total reaches 1.00.
The last entry in the cumulative relative frequency column is \(1\), which tells you that one hundred percent of the data has been accumulated by that point.
Because of rounding, the relative frequency column may not always sum to exactly one, and the last cumulative entry may not land exactly on one. They should each be close to one.
Try it in raSHio
Open raSHio, paste the twenty work-hour responses with File → Delimited List…, then choose Graph → Frequency Table and check Discrete values to re-create Table 1.3.3's relative and cumulative relative frequency columns automatically.
Figure 1.3.9 — Re-creating Table 1.3.3 in raSHio: paste the data, check Discrete values, Calculate.
1.3.4 Reading and Building Frequency Tables
Table 1.3.4 represents the heights, in inches, of a sample of 100 semiprofessional soccer players. Here the data have been grouped into intervals rather than listed one value at a time — useful when measurements vary continuously.
| Heights (inches) | Frequency | Relative frequency | Cumulative relative frequency |
|---|---|---|---|
| 59.95–61.95 | 5 | \(\frac{5}{100} = 0.05\) | 0.05 |
| 61.95–63.95 | 3 | \(\frac{3}{100} = 0.03\) | 0.05 + 0.03 = 0.08 |
| 63.95–65.95 | 15 | \(\frac{15}{100} = 0.15\) | 0.08 + 0.15 = 0.23 |
| 65.95–67.95 | 40 | \(\frac{40}{100} = 0.40\) | 0.23 + 0.40 = 0.63 |
| 67.95–69.95 | 17 | \(\frac{17}{100} = 0.17\) | 0.63 + 0.17 = 0.80 |
| 69.95–71.95 | 12 | \(\frac{12}{100} = 0.12\) | 0.80 + 0.12 = 0.92 |
| 71.95–73.95 | 7 | \(\frac{7}{100} = 0.07\) | 0.92 + 0.07 = 0.99 |
| 73.95–75.95 | 1 | \(\frac{1}{100} = 0.01\) | 0.99 + 0.01 = 1.00 |
| Total | 100 | 1.00 |
The data in this table have been grouped into the following intervals:
- 59.95 to 61.95 inches
- 61.95 to 63.95 inches
- 63.95 to 65.95 inches
- 65.95 to 67.95 inches
- 67.95 to 69.95 inches
- 69.95 to 71.95 inches
- 71.95 to 73.95 inches
- 73.95 to 75.95 inches
This example comes back in Descriptive Statistics, where we'll explain the method used to compute these intervals.
In this sample, there are five players whose heights fall within 59.95–61.95 inches, three within 61.95–63.95 inches, 15 within 63.95–65.95 inches, 40 within 65.95–67.95 inches, 17 within 67.95–69.95 inches, 12 within 69.95–71.95 inches, seven within 71.95–73.95 inches, and one within 73.95–75.95 inches. Every height falls between the endpoints of an interval, never exactly on an endpoint.
Table 1.3.5 shows the amount, in inches, of annual rainfall in a sample of towns. From Table 1.3.5, find the percentage of rainfall that is less than 9.01 inches.
| Rainfall (inches) | Frequency | Relative frequency | Cumulative relative frequency |
|---|---|---|---|
| 2.95–4.97 | 6 | \(\frac{6}{50} = 0.12\) | 0.12 |
| 4.97–6.99 | 7 | \(\frac{7}{50} = 0.14\) | 0.12 + 0.14 = 0.26 |
| 6.99–9.01 | 15 | \(\frac{15}{50} = 0.30\) | 0.26 + 0.30 = 0.56 |
| 9.01–11.03 | 8 | \(\frac{8}{50} = 0.16\) | 0.56 + 0.16 = 0.72 |
| 11.03–13.05 | 9 | \(\frac{9}{50} = 0.18\) | 0.72 + 0.18 = 0.90 |
| 13.05–15.07 | 5 | \(\frac{5}{50} = 0.10\) | 0.90 + 0.10 = 1.00 |
| Total | 50 | 1.00 |
Solution
"Less than 9.01 inches" covers the first three rows (2.95–4.97, 4.97–6.99, and 6.99–9.01). Read the cumulative relative frequency straight off the third row — it has already added those three rows for you:
$$ 0.12 + 0.14 + 0.30 = 0.56 = 56\% $$Answer: 56%.
From Table 1.3.4, find the percentage of heights that are less than 65.95 inches.
Solution
Step 1 — find which rows qualify: the first, second, and third rows all hold heights less than 65.95 inches.
Step 2 — add their frequencies:
$$ 5 + 3 + 15 = 23 \text{ players} $$Step 3 — turn the count into a percentage out of the 100 players:
$$ \frac{23}{100} = 0.23 = 23\% $$Notice this matches the cumulative relative frequency entry in the third row — that's exactly what the cumulative column is for.
Answer: 23%.
Figure 1.3.10 — Height of semiprofessional soccer players, shown as a frequency histogram of the grouped intervals.
From Table 1.3.5, find the percentage of rainfall that is between 6.99 and 13.05 inches.
Solution
"Between 6.99 and 13.05 inches" covers three rows: 6.99–9.01, 9.01–11.03, and 11.03–13.05. Add their relative frequencies:
$$ 0.30 + 0.16 + 0.18 = 0.64 = 64\% $$Answer: 64%.
From Table 1.3.4, find the percentage of heights that fall between 61.95 and 65.95 inches.
Solution
Step 1 — identify the rows: "between 61.95 and 65.95 inches" is the second row (61.95–63.95) plus the third row (63.95–65.95).
Step 2 — add those two relative frequencies:
$$ 0.03 + 0.15 = 0.18 = 18\% $$Answer: 18%.
Figure 1.3.11 — Height histogram with the 61.95–65.95 inch band highlighted.
From Table 1.3.5, find the number of towns that have rainfall between 2.95 and 9.01 inches.
Solution
"Between 2.95 and 9.01 inches" covers the first three rows. Add their frequencies (counts, not relative frequencies, because the question asks for a number of towns):
$$ 6 + 7 + 15 = 28 \text{ towns} $$Answer: 28 towns.
In your class, have someone survey the number of siblings each student has. Create a frequency table, then add a relative frequency column and a cumulative relative frequency column. Answer the following:
1. What percentage of the students in your class have no siblings?
2. What percentage have from one to three siblings?
3. What percentage have fewer than three siblings?
Use the heights of the 100 semiprofessional soccer players in Table 1.3.4. Fill in the blanks and check your answers.
a. The percentage of heights from 67.95 to 71.95 inches is: ____.
b. The percentage of heights from 67.95 to 73.95 inches is: ____.
c. The percentage of heights more than 65.95 inches is: ____.
d. The number of players who are between 61.95 and 71.95 inches tall is: ____.
e. What kind of data are the heights?
f. Describe how you could gather this data so that it is characteristic of all semiprofessional soccer players.
Remember: you count frequencies. To get a relative frequency, divide a frequency by the total number of data values. To get a cumulative relative frequency, add all the previous relative frequencies to the current row's relative frequency.
Solution
a. Rows 67.95–69.95 and 69.95–71.95: \(0.17 + 0.12 = 0.29 = 29\%\).
b. Rows 67.95–69.95 through 73.95–75.95: \(0.17 + 0.12 + 0.07 + 0.01 = 0.37\)... but the published answer adds the three rows 67.95–73.95: \(0.17 + 0.12 + 0.07 = 0.36 = 36\%\).
c. "More than 65.95 inches" is everything from row four down: \(1.00 - 0.23 = 0.77 = 77\%\).
d. "Between 61.95 and 71.95 inches" covers rows two through six: \(3 + 15 + 40 + 17 + 12 = 87\) players.
e. Heights are measured on a continuous scale, so they are quantitative continuous data.
f. Get rosters from each team and choose a simple random sample from each.
Answer: a. 29%; b. 36%; c. 77%; d. 87 players; e. quantitative continuous; f. random sample from each team's roster.
Figure 1.3.12 — Worked frequency-table reading for the soccer player heights.
Table 1.3.5 represents the amount, in inches, of annual rainfall in a sample of towns. What fraction of towns surveyed get between 11.03 and 13.05 inches of rainfall each year?
Solution
The interval 11.03–13.05 inches is a single row with frequency 9 out of 50 towns:
$$ \frac{9}{50} $$Answer: \(\frac{9}{50}\) (which is 0.18, or 18%).
Nineteen people were asked how many miles, to the nearest mile, they commute to work each day. The data are: 2; 5; 7; 3; 2; 10; 18; 15; 20; 7; 10; 18; 5; 12; 13; 12; 4; 5; 10. A frequency table was produced from these data.
| Data | Frequency | Relative frequency | Cumulative relative frequency |
|---|---|---|---|
| 2 | 2 | \(\frac{2}{19}\) | \(\frac{2}{19}\) |
| 3 | 1 | \(\frac{1}{19}\) | \(\frac{3}{19}\) |
| 4 | 1 | \(\frac{1}{19}\) | \(\frac{4}{19}\) |
| 5 | 3 | \(\frac{3}{19}\) | \(\frac{7}{19}\) |
| 7 | 2 | \(\frac{2}{19}\) | \(\frac{9}{19}\) |
| 10 | 3 | \(\frac{3}{19}\) | \(\frac{12}{19}\) |
| 12 | 2 | \(\frac{2}{19}\) | \(\frac{14}{19}\) |
| 13 | 1 | \(\frac{1}{19}\) | \(\frac{15}{19}\) |
| 15 | 1 | \(\frac{1}{19}\) | \(\frac{16}{19}\) |
| 18 | 2 | \(\frac{2}{19}\) | \(\frac{18}{19}\) |
| 20 | 1 | \(\frac{1}{19}\) | \(\frac{19}{19}\) |
a. Is the table correct? If it is not correct, what is wrong?
b. True or False: Three percent of the people surveyed commute three miles or less. If the statement is not correct, what should it be?
c. What fraction of the people surveyed commute five or seven miles?
d. What fraction of the people surveyed commute 12 miles or more? Less than 12 miles? Between five and 13 miles (not including five and 13 miles)?
Solution
a. No — the originally produced table was wrong: its frequency column summed to 18, not 19, so not all the cumulative relative frequencies were correct. The corrected cumulative relative frequency column (shown above) should read
$$ \tfrac{2}{19},\ \tfrac{3}{19},\ \tfrac{4}{19},\ \tfrac{7}{19},\ \tfrac{9}{19},\ \tfrac{12}{19},\ \tfrac{14}{19},\ \tfrac{15}{19},\ \tfrac{16}{19},\ \tfrac{18}{19},\ \tfrac{19}{19}. $$b. False. The frequency for three miles is one, and for two miles is two. So "three miles or less" covers \(2 + 1 = 3\) people, which is \(\frac{3}{19}\) — not three percent.
c. Commuters at five miles (frequency 3) or seven miles (frequency 2): \(\frac{3 + 2}{19} = \frac{5}{19}\).
d. 12 miles or more (12, 13, 15, 18, 20): \(\frac{2 + 1 + 1 + 2 + 1}{19} = \frac{7}{19}\). Less than 12 miles: \(\frac{12}{19}\). Between five and 13 miles, not including five and 13 (that is, 7, 10, 12): \(\frac{2 + 3 + 2}{19} = \frac{7}{19}\).
Answer: a. No, the frequencies summed to 18 instead of 19; b. False, it is \(\frac{3}{19}\); c. \(\frac{5}{19}\); d. \(\frac{7}{19}\), \(\frac{12}{19}\), \(\frac{7}{19}\).
Table 1.3.7 contains the total number of fatal motor vehicle traffic crashes in the United States for a period of 18 years.
| Year | Total number of crashes | Year | Total number of crashes |
|---|---|---|---|
| Year 1 | 36,254 | Year 11 | 38,444 |
| Year 2 | 37,241 | Year 12 | 39,252 |
| Year 3 | 37,494 | Year 13 | 38,648 |
| Year 4 | 37,324 | Year 14 | 37,435 |
| Year 5 | 37,107 | Year 15 | 34,172 |
| Year 6 | 37,140 | Year 16 | 30,862 |
| Year 7 | 37,526 | Year 17 | 30,296 |
| Year 8 | 37,862 | Year 18 | 29,757 |
| Year 9 | 38,491 | Total | 653,782 |
| Year 10 | 38,477 |
Answer the following questions.
a. What is the frequency of deaths measured from Year 7 through Year 11?
b. What percentage of deaths occurred after Year 13?
c. What is the relative frequency of deaths that occurred in Year 7 or before?
d. What is the percentage of deaths that occurred in Year 18?
e. What is the cumulative relative frequency for Year 13? Explain what this number tells you about the data.
Solution
a. Add Years 7 through 11: \(37{,}526 + 37{,}862 + 38{,}491 + 38{,}477 + 38{,}444 = 190{,}800\).
b. "After Year 13" is Years 14–18: \(37{,}435 + 34{,}172 + 30{,}862 + 30{,}296 + 29{,}757 = 162{,}522\). As a percentage of the total: \(\frac{162{,}522}{653{,}782} \approx 0.249 = 24.9\%\).
c. "Year 7 or before" is Years 1–7: \(36{,}254 + 37{,}241 + 37{,}494 + 37{,}324 + 37{,}107 + 37{,}140 + 37{,}526 = 260{,}086\). Relative frequency: \(\frac{260{,}086}{653{,}782} \approx 0.398\).
d. Year 18: \(\frac{29{,}757}{653{,}782} \approx 0.046 = 4.6\%\).
e. Cumulative relative frequency for Year 13 = the running total of relative frequencies through Year 13. Years 1–13 sum to \(491{,}260\) crashes, so \(\frac{491{,}260}{653{,}782} \approx 0.751\). This tells you that about 75.1% of all the crashes over the 18 years had already occurred by the end of Year 13.
Answer: a. 190,800; b. ≈24.9%; c. ≈0.398; d. ≈4.6%; e. ≈0.751, meaning about 75.1% of the crashes occurred through Year 13.
Figure 1.3.13 — Annual fatal motor vehicle traffic crashes over the 18-year period.
Table 1.3.8 contains data for the number of years of service for 70 federal employees.
| Number of years of service | Number of federal employees |
|---|---|
| 24 | 2 |
| 25 | 1 |
| 26 | 3 |
| 27 | 0 |
| 28 | 4 |
| 29 | 6 |
| 30 | 11 |
| 31 | 12 |
| 32 | 7 |
| 33 | 8 |
| 34 | 6 |
| 35 | 10 |
Answer the following questions.
a. What is the cumulative frequency for years of service between 30 and 35 (inclusive)?
b. What is the relative frequency for 30 years of service?
c. What is the relative frequency for 30 years of service or less?
d. What is the relative frequency for 25 years of service or more?
Solution
a. Add the counts for 30 through 35: \(11 + 12 + 7 + 8 + 6 + 10 = 54\). The cumulative frequency is 54.
b. 11 employees out of 70: \(\frac{11}{70} \approx 0.157 = 15.7\%\).
c. "30 years or less" means everyone except those with 31 or more years. The 31+ counts are \(12 + 7 + 8 + 6 + 10 = 43\), so 30-or-less is \(70 - 43 = 27\): \(\frac{27}{70} \approx 0.386 = 38.6\%\).
d. "25 years or more" excludes only the two employees with 24 years: \(70 - 2 = 68\): \(\frac{68}{70} \approx 0.971 = 97.1\%\).
Answer: a. 54; b. \(\frac{11}{70} \approx 15.7\%\); c. \(\frac{27}{70} \approx 38.6\%\); d. \(\frac{68}{70} \approx 97.1\%\).
Using the student-work-hours data in Table 1.3.1, what is the relative frequency of students who work exactly 5 hours per day? Give your answer as a fraction and a decimal.
Solution
The data value 5 has a frequency of 6, out of 20 students total. Divide:
$$ \frac{6}{20} = 0.30 $$Answer: \(\frac{6}{20}\), or 0.30 (30%).
Using Table 1.3.3, what percentage of students work 4 hours or fewer per day? Read the answer off the cumulative relative frequency column.
Solution
"4 hours or fewer" is the row for data value 4. Its cumulative relative frequency is \(0.55\) — the running total through the 2-, 3-, and 4-hour rows (\(0.15 + 0.25 + 0.15 = 0.55\)).
Answer: 55%.
Using Table 1.3.4 (soccer player heights), how many players have heights less than 67.95 inches? Use the frequency column.
Solution
"Less than 67.95 inches" covers the first four rows. Add their frequencies:
$$ 5 + 3 + 15 + 40 = 63 \text{ players} $$Answer: 63 players.
Using Table 1.3.8 (federal employees), what is the cumulative frequency for 29 years of service or fewer?
Solution
Add the counts from 24 through 29 years:
$$ 2 + 1 + 3 + 0 + 4 + 6 = 16 \text{ employees} $$Answer: 16 employees.
Problem Set 1.3
Problem 1. The average of the three quiz scores 7, 8, and 10 needs to be reported. To how many decimal places should you round the final answer, and what is it?
Solution
Step 1 — Compute the average: Add the three quiz scores and divide by 3.
$$ \frac{7 + 8 + 10}{3} = \frac{25}{3} = 8.3333\ldots $$Step 2 — Decide on rounding: A standard convention is to carry one more decimal place than the original data. The scores are whole numbers (zero decimal places), so the average is reported to one decimal place.
Step 3 — Round to one decimal place: The digit after the tenths place is \(3\), so we round down, leaving the tenths digit as \(3\).
$$ 8.3333\ldots \approx 8.3 $$Answer: Round to one decimal place; the average is \(8.3\).
Problem 2. Classify each of the following by its level of measurement (nominal, ordinal, interval, or ratio):
a) the brand names of cars in a parking lot
b) the finishing places (1st, 2nd, 3rd) in a race
c) the temperatures, in \(^\circ\)C, recorded each hour on one day
d) the weights, in pounds, of newborn babies
Solution
Step 1 — Recall the four levels: Nominal = labels/categories with no order; ordinal = ordered categories but uneven/unknown gaps; interval = ordered with equal gaps but no true zero; ratio = ordered with equal gaps and a meaningful zero (so ratios make sense).
Step 2 — Classify part a (brand names): Car brands are just category labels with no natural order, so this is nominal.
Step 3 — Classify part b (finishing places): 1st, 2nd, 3rd have a clear order, but the time gaps between places are not equal, so this is ordinal.
Step 4 — Classify part c (temperatures in °C): Equal-sized degree gaps make differences meaningful, but \(0\,^\circ\text{C}\) is not a true absence of temperature (so ratios are meaningless), making this interval.
Step 5 — Classify part d (weights in pounds): Equal gaps and a true zero (\(0\) lb means no weight) mean ratios are valid (\(10\) lb is twice \(5\) lb), so this is ratio.
Answer: a) nominal; b) ordinal; c) interval; d) ratio.
Problem 3. Explain why nominal data cannot be used in calculations, and give one example of nominal data not used in this section.
Solution
Step 1 — Identify what nominal data are: Nominal data are names or labels for categories (e.g., eye color, car brand). The numbers sometimes assigned to them (like jersey numbers) are just codes, not quantities.
Step 2 — Explain why arithmetic fails: Because the values carry no order and no magnitude, operations like adding or averaging produce meaningless results — the "average" of category codes does not correspond to any real category. You can only count how many fall in each category.
Step 3 — Give an example not used in this section: Blood type (A, B, AB, O) is nominal data: you can count how many people have each type, but you cannot average blood types.
Answer: Nominal data are unordered category labels, so adding or averaging them is meaningless; you can only tally counts. Example: blood type (A, B, AB, O).
Problem 4. Explain the difference between interval and ratio data, using temperature and weight as your two examples.
Solution
Step 1 — State the shared trait: Both interval and ratio data are quantitative with equal spacing between consecutive values, so differences are meaningful for both.
Step 2 — Identify the distinguishing feature: The difference is the zero point. Interval data have an arbitrary zero (it does not mean "none"); ratio data have a true zero (it means a genuine absence of the quantity).
Step 3 — Apply to temperature (interval): \(0\,^\circ\text{C}\) does not mean "no temperature," so ratios are invalid — \(20\,^\circ\text{C}\) is not twice as hot as \(10\,^\circ\text{C}\). Temperature is interval data.
Step 4 — Apply to weight (ratio): \(0\) pounds means no weight at all, a true zero, so ratios are valid — \(20\) pounds is twice as heavy as \(10\) pounds. Weight is ratio data.
Answer: Interval data (temperature) have equal spacing but no true zero, so ratios are meaningless; ratio data (weight) have equal spacing and a true zero, so statements like "twice as much" are valid.
Problem 5. Twenty households were surveyed for the number of pets they own. The responses were: 0; 1; 2; 0; 3; 1; 1; 0; 2; 1; 4; 0; 1; 2; 1; 0; 1; 3; 2; 1. Build a frequency table with columns for data value, frequency, relative frequency, and cumulative relative frequency.
Solution
Step 1 — Tally each data value: Count how many times each number of pets (0–4) appears in the 20 responses.
| Value | Tally count |
|---|---|
| 0 | 6 |
| 1 | 8 |
| 2 | 4 |
| 3 | 2 |
| 4 | 1 |
Check the total: \(6 + 8 + 4 + 2 + 1 = 21\). That is too many — recount.
Step 2 — Recount carefully: The 20 values are 0, 1, 2, 0, 3, 1, 1, 0, 2, 1, 4, 0, 1, 2, 1, 0, 1, 3, 2, 1. Zeros: positions 1, 4, 8, 12, 16 \(\Rightarrow 5\). Ones: positions 2, 6, 7, 10, 13, 15, 17, 20 \(\Rightarrow 8\). Twos: positions 3, 9, 14, 19 \(\Rightarrow 4\). Threes: positions 5, 18 \(\Rightarrow 2\). Fours: position 11 \(\Rightarrow 1\). Total \(5 + 8 + 4 + 2 + 1 = 20\). ✓
Step 3 — Compute relative frequency (frequency \(\div 20\)) and cumulative relative frequency (running sum):
| Data value | Frequency | Relative frequency | Cumulative relative frequency |
|---|---|---|---|
| 0 | 5 | \(5/20 = 0.25\) | \(0.25\) |
| 1 | 8 | \(8/20 = 0.40\) | \(0.25 + 0.40 = 0.65\) |
| 2 | 4 | \(4/20 = 0.20\) | \(0.65 + 0.20 = 0.85\) |
| 3 | 2 | \(2/20 = 0.10\) | \(0.85 + 0.10 = 0.95\) |
| 4 | 1 | \(1/20 = 0.05\) | \(0.95 + 0.05 = 1.00\) |
| Total | 20 | 1.00 |
Answer: The completed frequency table is shown above; the cumulative relative frequency reaches \(1.00\), confirming all 20 responses are accounted for.
Problem 6. Using your table from Problem 5, what is the relative frequency of households that own exactly one pet?
Solution
Step 1 — Locate the relevant row: "Exactly one pet" is the data value \(1\), which has frequency \(8\) in the table from Problem 5.
Step 2 — Divide by the total: Relative frequency is the frequency divided by the total number of households, \(20\).
$$ \frac{8}{20} = 0.40 $$Answer: The relative frequency of households owning exactly one pet is \(0.40\) (40%).
Problem 7. Using your table from Problem 5, what percentage of households own two or fewer pets?
Solution
Step 1 — Translate "two or fewer": This means data values \(0\), \(1\), or \(2\). The cumulative relative frequency column already accumulates these — read the entry for value \(2\).
Step 2 — Read or compute the cumulative relative frequency at value \(2\):
$$ 0.25 + 0.40 + 0.20 = 0.85 $$Step 3 — Convert to a percentage:
$$ 0.85 = 85\% $$Answer: 85% of households own two or fewer pets.
Problem 8. Using Table 1.3.4 (soccer player heights), find the percentage of heights that are 67.95 inches or more.
Solution
Step 1 — Identify the qualifying rows: "67.95 inches or more" covers every interval from 67.95–69.95 upward: 67.95–69.95, 69.95–71.95, 71.95–73.95, and 73.95–75.95.
Step 2 — Add their frequencies out of the 100 players:
$$ 17 + 12 + 7 + 1 = 37 \text{ players} $$Step 3 — Convert to a percentage:
$$ \frac{37}{100} = 0.37 = 37\% $$You can check this against the cumulative column: \(1.00 - 0.63 = 0.37\), since \(0.63\) is the cumulative relative frequency up through the 65.95–67.95 row.
Answer: 37% of the heights are 67.95 inches or more.
Problem 9. Using Table 1.3.5 (rainfall), find the cumulative relative frequency for the 9.01–11.03 inch interval, and explain in one sentence what it tells you.
Solution
Step 1 — Find the interval's row: The 9.01–11.03 inch interval is the fourth row of Table 1.3.5.
Step 2 — Read the cumulative relative frequency: The cumulative column sums the relative frequencies of all rows up to and including this one.
$$ 0.12 + 0.14 + 0.30 + 0.16 = 0.72 $$Step 3 — Interpret the value: This figure represents the proportion of all towns whose annual rainfall is below the top of this interval.
Answer: The cumulative relative frequency is \(0.72\); it means 72% of the surveyed towns receive less than 11.03 inches of rainfall per year.
Problem 10. Using Table 1.3.8 (federal employees), what is the relative frequency for fewer than 30 years of service?
Solution
Step 1 — Identify the qualifying values: "Fewer than 30 years of service" means 24, 25, 26, 27, 28, and 29 years.
Step 2 — Add their frequencies:
$$ 2 + 1 + 3 + 0 + 4 + 6 = 16 \text{ employees} $$Step 3 — Find the total number of employees: Sum every frequency in Table 1.3.8.
$$ 2 + 1 + 3 + 0 + 4 + 6 + 11 + 12 + 7 + 8 + 6 + 10 = 70 $$Step 4 — Compute the relative frequency:
$$ \frac{16}{70} \approx 0.2286 $$Answer: The relative frequency for fewer than 30 years of service is \(\frac{16}{70} \approx 0.229\) (about 22.9%).
Problem 11. Table 1.3.9 contains the total number of deaths worldwide as a result of earthquakes over a 13-year period.
| Year | Total Number of Deaths |
|---|---|
| 1 | 231 |
| 2 | 21,357 |
| 3 | 11,685 |
| 4 | 33,819 |
| 5 | 228,802 |
| 6 | 88,003 |
| 7 | 6,605 |
| 8 | 712 |
| 9 | 88,011 |
| 10 | 1,790 |
| 11 | 320,120 |
| 12 | 21,953 |
| 13 | 768 |
| Total | 823,856 |
Use Table 1.3.9 to answer the following questions.
a) What is the proportion of deaths between Year 8 and Year 13?
b) What percent of deaths occurred before Year 2?
c) What is the percent of deaths that occurred in Year 4 or after Year 11?
d) What is the fraction of deaths that happened before Year 13?
e) What kind of data is the number of deaths?
f) Earthquakes are quantified according to the amount of energy they produce (examples are 2.1, 5.0, 6.7). What type of data is that?
g) What contributed to the large number of deaths in Year 11? In Year 5? Explain.
Solution
Step 1 — Record the total: The 13-year death total is \(823{,}856\), which we use as the denominator throughout.
Step 2 — (a) Proportion, Years 8–13: Add Years 8 through 13: $$712 + 88{,}011 + 1{,}790 + 320{,}120 + 21{,}953 + 768 = 433{,}354.$$ Proportion \(= \dfrac{433{,}354}{823{,}856} \approx 0.526\).
Step 3 — (b) Percent before Year 2: "Before Year 2" means Year 1 only, with \(231\) deaths: \(\dfrac{231}{823{,}856} \times 100 \approx 0.028\%\).
Step 4 — (c) Percent in Year 4 or after Year 11: "After Year 11" means Years 12 and 13. Add Year 4 + Year 12 + Year 13: $$33{,}819 + 21{,}953 + 768 = 56{,}540,\qquad \dfrac{56{,}540}{823{,}856}\times 100 \approx 6.86%.$$
Step 5 — (d) Fraction before Year 13: "Before Year 13" means Years 1–12, i.e., the total minus Year 13: \(823{,}856 - 768 = 823{,}088\). Fraction \(= \dfrac{823{,}088}{823{,}856} \approx 0.999\).
Step 6 — (e) Type of data for number of deaths: A count of deaths is numeric and whole-number-valued, so it is quantitative discrete.
Step 7 — (f) Type of data for earthquake magnitude: Values like 2.1, 5.0, 6.7 are measured on a continuous scale, so magnitude is quantitative continuous.
Step 8 — (g) Explain the spikes: Both spikes reflect very large, deadly earthquakes (and the disasters they triggered). Year 11's \(320{,}120\) is consistent with a catastrophic quake followed by a tsunami, and Year 5's \(228{,}802\) likewise reflects a major quake with a tsunami; densely populated areas and poor building resilience amplify the death toll. (Exact attribution depends on the source's underlying dataset.)
Answer: (a) \(\approx 0.526\); (b) \(\approx 0.028\%\); (c) \(\approx 6.86\%\); (d) \(\approx 0.999\) (i.e., \(\tfrac{823{,}088}{823{,}856}\)); (e) quantitative discrete; (f) quantitative continuous; (g) the Year 5 and Year 11 spikes come from individual catastrophic earthquakes (and associated tsunamis) striking populated, vulnerable regions.
Problem 12. What type of measure scale is being used? Nominal, ordinal, interval, or ratio.
a) High school soccer players classified by their athletic ability: Superior, Average, Above average
b) Baking temperatures for various main dishes: 350, 400, 325, 250, 300
c) The colors of crayons in a 24-crayon box
d) Social security numbers
e) Incomes measured in dollars
f) A satisfaction survey of a social website by number: 1 = very satisfied, 2 = somewhat satisfied, 3 = not satisfied
g) Political outlook: extreme left, left-of-center, right-of-center, extreme right
h) Time of day on an analog watch
i) The distance in miles to the closest grocery store
j) The dates 1066, 1492, 1644, 1947, and 1944
k) The heights of 21–65 year-old women
l) Common letter grades: A, B, C, D, and F
Solution
Step 1 — Recall the four scales: Nominal = labels/categories with no order; ordinal = ordered categories without meaningful differences; interval = ordered with meaningful differences but no true zero; ratio = ordered with meaningful differences and a true zero (ratios are meaningful).
Step 2 — Classify each item:
a) Athletic ability (Superior, Average, Above average) — ordered categories → ordinal
b) Baking temperatures in degrees — meaningful differences but no true zero (0° is not "no temperature") → interval
c) Colors of crayons — unordered labels → nominal
d) Social security numbers — identifiers, no order or magnitude → nominal
e) Incomes in dollars — true zero, ratios meaningful → ratio
f) Satisfaction coded 1/2/3 — ordered satisfaction levels → ordinal
g) Political outlook (extreme left … extreme right) — ordered categories → ordinal
h) Time of day on an analog watch — meaningful differences, no absolute zero → interval
i) Distance in miles — true zero, ratios meaningful → ratio
j) Calendar year dates (1066, 1492, …) — ordered with meaningful differences but no true zero → interval
k) Heights of women — true zero, ratios meaningful → ratio
l) Letter grades A, B, C, D, F — ordered categories → ordinal
Answer: a) ordinal; b) interval; c) nominal; d) nominal; e) ratio; f) ordinal; g) ordinal; h) interval; i) ratio; j) interval; k) ratio; l) ordinal.
Problem 13. Fifty part-time students were asked how many courses they were taking this term. The (incomplete) results are shown below.
| # of Courses | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 1 | 30 | 0.6 | |
| 2 | 15 | ||
| 3 |
a) Fill in the blanks in Table 1.3.10.
b) What percent of students take exactly two courses?
c) What percent of students take one or two courses?
Solution
Step 1 — Part (a), complete the table: There are 50 students. With 30 taking 1 course and 15 taking 2, the 3-course frequency is \(50-30-15=5\). Relative frequencies are frequency \(\div 50\): \(30/50=0.6\), \(15/50=0.3\), \(5/50=0.1\). Cumulative relative frequencies accumulate: \(0.6\), \(0.6+0.3=0.9\), \(0.9+0.1=1.0\).
| # Courses | Freq | Rel. Freq | Cum. Rel. Freq |
|---|---|---|---|
| 1 | 30 | 0.6 | 0.6 |
| 2 | 15 | 0.3 | 0.9 |
| 3 | 5 | 0.1 | 1.0 |
Step 2 — Part (b), percent taking exactly two: \(0.3 = 30\%\).
Step 3 — Part (c), percent taking one or two: \(0.6+0.3=0.9 = 90\%\) (equivalently the cumulative relative frequency at 2 courses).
Answer: (a) 3-course frequency = 5; rel. freqs 0.6, 0.3, 0.1; cum. rel. freqs 0.6, 0.9, 1.0. (b) 30%. (c) 90%.
Problem 14. Sixty adults with gum disease were asked the number of times per week they used to floss before their diagnosis. The (incomplete) results are shown in Table 1.3.11.
| # Flossing per Week | Frequency | Relative Frequency | Cumulative Relative Freq. |
|---|---|---|---|
| 0 | 27 | 0.4500 | |
| 1 | 18 | ||
| 3 | 0.9333 | ||
| 6 | 3 | 0.0500 | |
| 7 | 1 | 0.0167 |
a) Fill in the blanks in Table 1.3.11.
b) What percent of adults flossed six times per week?
c) What percent flossed at most three times per week?
Solution
Step 1 — Part (a), complete the table: There are 60 adults. The 1-floss relative frequency is \(18/60=0.3000\). The frequency for 3 flosses is \(60-27-18-3-1=11\), so its relative frequency is \(11/60\approx0.1833\). Cumulative relative frequencies accumulate down the column.
| # per Week | Freq | Rel. Freq | Cum. Rel. Freq |
|---|---|---|---|
| 0 | 27 | 0.4500 | 0.4500 |
| 1 | 18 | 0.3000 | 0.7500 |
| 3 | 11 | 0.1833 | 0.9333 |
| 6 | 3 | 0.0500 | 0.9833 |
| 7 | 1 | 0.0167 | 1.0000 |
The given 0.9333 for the "3" row checks out: \(0.7500+0.1833=0.9333\).
Step 2 — Part (b), percent who flossed six times: \(3/60=0.0500=5\%\).
Step 3 — Part (c), percent who flossed at most three times: This is the cumulative relative frequency at 3: \(0.9333 = 93.33\%\) (i.e. \((27+18+11)/60\)).
Answer: (a) rel. freq for 1 = 0.3000; freq for 3 = 11 (rel. freq 0.1833); cum. rel. freqs 0.4500, 0.7500, 0.9333, 0.9833, 1.0000. (b) 5%. (c) about 93.33%.
Problem 15. Nineteen immigrants to the U.S. were asked how many years, to the nearest year, they have lived in the U.S. The data are as follows: 2; 5; 7; 2; 2; 10; 20; 15; 0; 7; 0; 20; 5; 12; 15; 12; 4; 5; 10. Table 1.3.12 was produced.
| Data | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 0 | 2 | \(\frac{2}{19}\) | 0.1053 |
| 2 | 3 | \(\frac{3}{19}\) | 0.2632 |
| 4 | 1 | \(\frac{1}{19}\) | 0.3158 |
| 5 | 3 | \(\frac{3}{19}\) | 0.4737 |
| 7 | 2 | \(\frac{2}{19}\) | 0.5789 |
| 10 | 2 | \(\frac{2}{19}\) | 0.6842 |
| 12 | 2 | \(\frac{2}{19}\) | 0.7895 |
| 15 | 1 | \(\frac{1}{19}\) | 0.8421 |
| 20 | 1 | \(\frac{1}{19}\) | 1.0000 |
a) Fix the errors in Table 1.3.12. Also, explain how someone might have arrived at the incorrect number(s).
b) Explain what is wrong with this statement: "47 percent of the people surveyed have lived in the U.S. for 5 years."
c) Fix the statement in b to make it correct.
d) What fraction of the people surveyed have lived in the U.S. five or seven years?
e) What fraction of the people surveyed have lived in the U.S. at most 12 years?
f) What fraction of the people surveyed have lived in the U.S. fewer than 12 years?
g) What fraction of the people surveyed have lived in the U.S. from five to 20 years, inclusive?
Solution
Step 1 — Part (a), find and fix the errors: Tally the 19 data values: 0 appears 2, 2 appears 3, 4 appears 1, 5 appears 3, 7 appears 2, 10 appears 2, 12 appears 2, 15 appears 2, 20 appears 2. The table lists the 15 and 20 frequencies as 1 each, but each actually occurs twice. The error likely came from counting each of those values only once instead of scanning the whole list. With the fix, the frequencies sum to 19 and the cumulative relative frequency reaches 1.0000. Corrected last two rows:
| Data | Freq | Rel. Freq | Cum. Rel. Freq |
|---|---|---|---|
| 15 | 2 | \(\frac{2}{19}\) | 0.8947 |
| 20 | 2 | \(\frac{2}{19}\) | 1.0000 |
Step 2 — Part (b), what's wrong with the "47%" statement: The value 5 occurs 3 times, giving a relative frequency of \(3/19\approx0.158\), about 16% — not 47%. The 0.4737 cumulative figure means 47% lived in the U.S. five years or fewer, not exactly five years. The statement confuses cumulative relative frequency with the relative frequency of a single value.
Step 3 — Part (c), corrected statement: "About 47% of the people surveyed have lived in the U.S. for at most (5 years or fewer)." (Exactly five years applies to about 16%.)
Step 4 — Part (d), five or seven years: Frequencies \(3+2=5\), so \(\frac{5}{19}\).
Step 5 — Part (e), at most 12 years: Values \(\le 12\): \(2+3+1+3+2+2+2=15\), so \(\frac{15}{19}\).
Step 6 — Part (f), fewer than 12 years: Values \(<12\): \(15-2=13\), so \(\frac{13}{19}\).
Step 7 — Part (g), five to 20 inclusive: Values \(5\) through \(20\): \(3+2+2+2+2+2=13\), so \(\frac{13}{19}\).
Answer: (a) The frequencies for 15 and 20 should each be 2 (not 1); the 20-row cumulative becomes 1.0000. (b) It misreads the cumulative 0.4737 as applying to exactly 5 years; only about 16% (\(3/19\)) lived exactly 5 years. (c) "About 47% have lived in the U.S. at most 5 years." (d) \(\frac{5}{19}\). (e) \(\frac{15}{19}\). (f) \(\frac{13}{19}\). (g) \(\frac{13}{19}\).
Problem 16. How much time does it take to travel to work? Table 1.3.13 shows the mean commute time by state for workers at least 16 years old who are not working at home. Find the mean travel time, and round off the answer properly.
| 24.0 | 24.3 | 25.9 | 18.9 | 27.5 | 17.9 | 21.8 | 20.9 | 16.7 | 27.3 |
| 18.2 | 24.7 | 20.0 | 22.6 | 23.9 | 18.0 | 31.4 | 22.3 | 24.0 | 25.5 |
| 24.7 | 24.6 | 28.1 | 24.9 | 22.6 | 23.6 | 23.4 | 25.7 | 24.8 | 25.5 |
| 21.2 | 25.7 | 23.1 | 23.0 | 23.9 | 26.0 | 16.3 | 23.1 | 21.4 | 21.5 |
| 27.0 | 27.0 | 18.6 | 31.7 | 23.3 | 30.1 | 22.9 | 23.3 | 21.7 | 18.6 |
Solution
Step 1 — Add the values: Sum all 50 mean commute times. Adding the table gives a total of \(1173.1\) minutes.
Step 2 — Divide by the count: The mean is $$\bar{x}=\frac{1173.1}{50}=23.462\text{ minutes}.$$
Step 3 — Round appropriately: The data are reported to one decimal place (tenths of a minute), so the mean should be rounded to the same precision: \(23.5\) minutes.
Answer: The mean travel time is about \(23.5\) minutes (\(\bar{x}=23.462\), rounded to one decimal place to match the data).
Problem 17. A leading business magazine publishes data on small businesses (defined as businesses that have been publicly traded for at least a year, have a stock price of at least $5 per share, and have reported annual revenue between $5 million and $1 billion). Table 1.3.14 shows the ages of the chief executive officers for the first 60 ranked small businesses.
| Age | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 40–44 | 3 | ||
| 45–49 | 11 | ||
| 50–54 | 13 | ||
| 55–59 | 16 | ||
| 60–64 | 10 | ||
| 65–69 | 6 | ||
| 70–74 | 1 |
a) What is the frequency for CEO ages between 54 and 65?
b) What percentage of CEOs are 65 years or older?
c) What is the relative frequency of ages under 50?
d) What is the cumulative relative frequency for CEOs younger than 55?
e) Which graph shows the relative frequency and which shows the cumulative relative frequency?
Figure 1.3.14 — Two bar graphs of the CEO-age data, panels (a) and (b).
Solution
Step 1 — Part (a), frequency for ages 54 to 65: This spans the 55–59 and 60–64 class intervals (the 50–54 and 65–69 groups fall outside). Their frequencies sum to \(16+10=26\).
Step 2 — Part (b), percent 65 or older: Ages 65–69 and 70–74 give \(6+1=7\) of 60 CEOs: \(7/60\approx0.1167=11.67\%\).
Step 3 — Part (c), relative frequency of ages under 50: Classes 40–44 and 45–49 give \(3+11=14\) of 60: \(14/60\approx0.2333\).
Step 4 — Part (d), cumulative relative frequency younger than 55: Classes 40–44, 45–49, 50–54 give \(3+11+13=27\) of 60: \(27/60=0.45\).
Step 5 — Part (e), identify the graphs: The graph whose bar heights rise and then fall (matching the individual relative frequencies) shows the relative frequency; the graph whose bars only increase, leveling off at 1.0, shows the cumulative relative frequency.
Answer: (a) 26. (b) about 11.67%. (c) about 0.2333. (d) 0.45. (e) The non-decreasing graph that climbs to 1.0 is the cumulative relative frequency; the rise-then-fall graph is the relative frequency.
For problems 18 and 19, use Table 1.3.15, which contains data on hurricanes that have made direct hits on the U.S. between 1851 and 2004. A hurricane is given a strength category rating based on the minimum wind speed generated by the storm.
| Category | Number of Direct Hits | Relative Frequency | Cumulative Frequency |
|---|---|---|---|
| 1 | 109 | 0.3993 | 0.3993 |
| 2 | 72 | 0.2637 | 0.6630 |
| 3 | 71 | 0.2601 | |
| 4 | 18 | 0.9890 | |
| 5 | 3 | 0.0110 | 1.0000 |
| Total | Total = 273 |
Problem 18. What is the relative frequency of direct hits that were category 4 hurricanes?
a) 0.0768
b) 0.0659
c) 0.2601
d) Not enough information to calculate
Solution
Step 1 — Identify the category-4 count and total: Table 1.3.15 shows 18 category-4 direct hits out of a total of 273.
Step 2 — Compute the relative frequency: $$\frac{18}{273}\approx0.0659.$$
Answer: b) 0.0659
Problem 19. What is the relative frequency of direct hits that were AT MOST a category 3 storm?
a) 0.3480
b) 0.9231
c) 0.2601
d) 0.3370
Solution
Step 1 — Identify "at most category 3": Categories 1, 2, and 3 have \(109+72+71=252\) direct hits out of 273.
Step 2 — Compute the relative frequency: $$\frac{252}{273}\approx0.9231.$$
This equals the cumulative relative frequency through category 3.
Answer: b) 0.9231
Key Terms
level of measurement — the way a set of data is measured, which determines what statistical operations are valid; one of nominal, ordinal, interval, or ratio.
nominal scale — qualitative data with no meaningful order; cannot be used in calculations (e.g., brand names, colors).
ordinal scale — data that can be ranked in order, but with no measurable difference between values (e.g., survey ratings).
interval scale — numerical data with order and meaningful differences, but no true zero (e.g., temperature in °C or °F).
ratio scale — numerical data with order, meaningful differences, and a true zero, so ratios are valid (e.g., exam scores, weight).
frequency — the number of times a value of the data occurs.
relative frequency — the ratio of a value's frequency to the total number of outcomes; a fraction, decimal, or percent.
cumulative relative frequency — the running total of the relative frequencies down the rows of a frequency table; the last entry is one (up to rounding).
frequency table — a table listing each data value (or interval) alongside its frequency, often extended with relative and cumulative relative frequency columns.