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Relative frequency from data

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5135556
A computer analyzes an English-language news article containing \(400\) letters. The letter E appears \(52\) times, N appears \(28\) times, and Q appears once. a) Find the relative frequencies of E, N, and Q. Give each result as a percent. b) A reference model for this activity uses \(12.7\%\) as the long-run proportion of E in English text. Compare your result from part a with this value and give one possible reason for the difference.

Hints

- Divide each letter count by the total number of letters. - How do you convert a decimal to a percent? - Why might one article differ from a long-run language model?

Solution

1. a) The relative frequency of E is \(\frac{52}{400} = 0.13 = 13\%\). 2. The relative frequency of N is \(\frac{28}{400} = 0.07 = 7\%\). 3. The relative frequency of Q is \(\frac{1}{400} = 0.0025 = 0.25\%\). 4. b) The observed E frequency of \(13\%\) is close to the model value of \(12.7\%\). A particular article can differ because it is a limited sample and its topic and word choices affect letter counts.

Answer

a) E: \(13\%\); N: \(7\%\); Q: \(0.25\%\) b) The observed \(13\%\) is slightly above \(12.7\%\). Sampling variation and the article’s vocabulary can explain the difference.
5135796
A survey of \(200\) students recorded their favorite activities: sports (\(80\)), music (\(50\)), gaming (\(40\)), and reading (\(30\)). Lucas says, “If I randomly select one student, the probability that reading is the student’s favorite activity is exactly \(25\%\) because there are four activity categories.” a) Explain why Lucas’s claim is incorrect. b) Find the actual probability that a randomly selected student chose reading. c) Find the probability that a randomly selected student chose sports or music.

Hints

- Use the number of students in each category, not just the number of categories. - Divide a category count by \(200\). - For “sports or music,” add the two category counts.

Solution

1. a) The categories do not contain equal numbers of students. The probability depends on the number of students in a category, not only on the number of categories. 2. b) \(P(\text{reading}) = \frac{30}{200} = 0.15 = 15\%\). 3. c) Sports and music are separate categories containing \(80 + 50 = 130\) students. Therefore, \(P(\text{sports or music}) = \frac{130}{200} = 0.65 = 65\%\).

Answer

a) The four categories are not equally frequent, so they do not each have probability \(25\%\). b) \(15\%\) c) \(65\%\)
5140746
A factory packs LED bulbs in boxes of \(20\). A quality check recorded the number of defective bulbs in each of \(200\) boxes. <table> <tr><th>Number of defective bulbs</th><th>0</th><th>1</th><th>2</th><th>3 or more</th></tr> <tr><td>Number of boxes</td><td>164</td><td>22</td><td>10</td><td>4</td></tr> </table> Find the relative frequency for each category. Give each result as a decimal and a percent.

Hints

- Confirm the total number of boxes by adding the category counts. - Relative frequency is category count divided by total count. - How can a fraction with denominator \(200\) be converted to a percent efficiently?

Solution

1. The total number of boxes is \(164 + 22 + 10 + 4 = 200\). 2. For \(0\) defective bulbs, the relative frequency is \(\frac{164}{200} = 0.82 = 82\%\). 3. For \(1\) defective bulb, the relative frequency is \(\frac{22}{200} = 0.11 = 11\%\). 4. For \(2\) defective bulbs, the relative frequency is \(\frac{10}{200} = 0.05 = 5\%\). 5. For \(3\) or more defective bulbs, the relative frequency is \(\frac{4}{200} = 0.02 = 2\%\).

Answer

\(0\) defective bulbs: \(0.82 = 82\%\) \(1\) defective bulb: \(0.11 = 11\%\) \(2\) defective bulbs: \(0.05 = 5\%\) \(3\) or more defective bulbs: \(0.02 = 2\%\)
5140756
Two archers, Ethan and Maya, each completed several sets of \(5\) shots. The table shows how many bullseyes they made in each set. <table> <tr><th>Bullseyes per set</th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th></tr> <tr><td>Ethan (number of sets)</td><td>2</td><td>4</td><td>6</td><td>12</td><td>10</td><td>6</td></tr> <tr><td>Maya (number of sets)</td><td>1</td><td>2</td><td>5</td><td>7</td><td>15</td><td>10</td></tr> </table> a) Find the relative frequency of the event “at least \(4\) bullseyes” for each archer. b) Compare the results. Which archer has the greater relative frequency for this event?

Hints

- Which table categories satisfy “at least \(4\)”? - Find the total number of sets for each archer. - Divide the favorable set count by the total set count.

Solution

1. Ethan completed \(2 + 4 + 6 + 12 + 10 + 6 = 40\) sets. 2. Ethan had at least \(4\) bullseyes in \(10 + 6 = 16\) sets, so his relative frequency is \(\frac{16}{40} = 0.4\). 3. Maya completed \(1 + 2 + 5 + 7 + 15 + 10 = 40\) sets. 4. Maya had at least \(4\) bullseyes in \(15 + 10 = 25\) sets, so her relative frequency is \(\frac{25}{40} = 0.625\). 5. Since \(0.625 > 0.4\), Maya has the greater relative frequency.

Answer

a) Ethan: \(0.4\); Maya: \(0.625\) b) Maya has the greater relative frequency.
5140766
A survey asked \(250\) households how many pets they own. - The relative frequency for “no pets” is \(0.36\). - Exactly \(90\) households have one pet. a) How many surveyed households have no pets? b) Find the relative frequency of households with exactly one pet. c) All remaining households have either \(2\) or \(3\) pets. What must the sum of the relative frequencies of those two categories be? Briefly explain.

Hints

- How can you find a count from a total and a relative frequency? - What must all relative frequencies in a complete distribution add to? - Find the portion that remains after the first two categories.

Solution

1. a) Multiply the total by the relative frequency: \(250 \times 0.36 = 90\) households. 2. b) The relative frequency for exactly one pet is \(\frac{90}{250} = 0.36\). 3. c) All relative frequencies must add to \(1\). Therefore, the remaining sum is \(1 - (0.36 + 0.36) = 1 - 0.72 = 0.28\).

Answer

a) \(90\) households b) \(0.36\) c) \(0.28\), because all category relative frequencies sum to \(1\).
5318246
A class surveyed students about their favorite fruit. Each student chose exactly one fruit. The bar graph shows the results. Find the relative frequency of each fruit as a percent.
Figure for problem 531824

Hints

- Read the number of votes for each fruit from the graph. - Add all four counts to find the total number of votes. - Divide each count by the total. - Convert each fraction to a percent.

Solution

1. The total number of votes is \(8+5+4+3=20\). 2. Apple: \(\frac{8}{20}=40\%\). 3. Banana: \(\frac{5}{20}=25\%\). 4. Strawberry: \(\frac{4}{20}=20\%\). 5. Cherry: \(\frac{3}{20}=15\%\).

Answer

Apple: \(40\%\) Banana: \(25\%\) Strawberry: \(20\%\) Cherry: \(15\%\)
5318386
A sixth-grade class surveyed students about their favorite sports. The bar graph shows the results. a) How many students participated in the survey? b) Find the relative frequency for soccer as a fraction in simplest form and as a percent.
Figure for problem 531838

Hints

- Read each bar height carefully. - Add all bar values to find the total number of students. - Relative frequency is the category count divided by the total. - Convert the fraction to a percent.

Solution

1. The graph shows soccer, \(10\); swimming, \(6\); gymnastics, \(4\); and basketball, \(5\). 2. The total is \(10+6+4+5=25\) students. 3. The relative frequency for soccer is \(\frac{10}{25}=\frac{2}{5}=40\%\).

Answer

a) \(25\) students b) \(\frac{2}{5}\), or \(40\%\)
5375066
In \(200\) trials of a two-stage experiment, the absolute frequencies are shown in the tree diagram. Find the relative frequency of event \(B\).
Figure for problem 537506

Hints

- Add the counts at all endpoints that belong to event \(B\). - Divide the total count for \(B\) by the number of trials.

Solution

1. Event \(B\) occurs in the two endpoint categories with counts \(72\) and \(20\), so its total frequency is \(72 + 20 = 92\). 2. The relative frequency of \(B\) is \(\frac{92}{200} = 0.46\).

Answer

\(0.46 = 46\%\)
5115066
Two sixth-grade classes were surveyed about their favorite type of pet. The results are summarized in the table. <table> <thead><tr><th>Class</th><th>Dog</th><th>Cat</th><th>Other</th><th>Total</th></tr></thead> <tbody> <tr><td>6A</td><td>8</td><td>6</td><td>6</td><td>20</td></tr> <tr><td>6B</td><td>10</td><td>5</td><td>10</td><td>25</td></tr> </tbody> </table> a) Find the relative frequency of each response category as a percent for both classes. b) In which class was the relative frequency of students who chose cats greater? Briefly justify your answer using your results from part a).

Hints

- Divide each category count by its class total. - Convert each fraction to a percent. - For part b), compare relative frequencies rather than only the counts.

Solution

1. For Class 6A, divide each count by the total of 20: dog, \(\frac{8}{20}=40\%\); cat, \(\frac{6}{20}=30\%\); other, \(\frac{6}{20}=30\%\). 2. For Class 6B, divide each count by the total of 25: dog, \(\frac{10}{25}=40\%\); cat, \(\frac{5}{25}=20\%\); other, \(\frac{10}{25}=40\%\). 3. The relative frequency for cats is \(30\%\) in Class 6A and \(20\%\) in Class 6B, so it is greater in Class 6A.

Answer

a) Class 6A: dog \(40\%\), cat \(30\%\), other \(30\%\) Class 6B: dog \(40\%\), cat \(20\%\), other \(40\%\) b) Class 6A, because \(30\%>20\%\).
5115076
A bicycle shop has \(50\) bicycles in stock. The owner started the table below but left some entries blank. <table> <thead><tr><th>Bicycle type</th><th>Count</th><th>Relative frequency (percent)</th></tr></thead> <tbody> <tr><td>Mountain bike</td><td>20</td><td>?</td></tr> <tr><td>City bike</td><td>15</td><td>?</td></tr> <tr><td>Electric bike</td><td>?</td><td>?</td></tr> </tbody> </table> a) Find the missing number of electric bikes. b) Complete the relative-frequency column by finding the percent for each bicycle type. c) Check that your percentages add to \(100\%\).

Hints

- Subtract the two known counts from the total. - Divide each count by \(50\), then write the result as a percent. - The relative frequencies of all categories should total \(100\%\).

Solution

1. The number of electric bikes is \(50-20-15=15\). 2. Mountain bikes have relative frequency \(\frac{20}{50}=40\%\). 3. City bikes have relative frequency \(\frac{15}{50}=30\%\). 4. Electric bikes have relative frequency \(\frac{15}{50}=30\%\). 5. The check is \(40\%+30\%+30\%=100\%\).

Answer

a) \(15\) electric bikes b) Mountain bike: \(40\%\); city bike: \(30\%\); electric bike: \(30\%\) c) \(40\%+30\%+30\%=100\%\)
5115596
Two sixth-grade classes participated in a donation drive. - In Class 6A, \(12\) of \(20\) students donated. - In Class 6B, \(15\) of \(30\) students donated. The Class 6B representative says, “Our class participated more because more students donated in our class than in Class 6A.” Use relative frequencies as percentages to evaluate the claim. Which class had the greater participation rate? Show your calculations.

Hints

- The class totals are different, so compare portions of each class rather than only the counts. - Divide the number who donated by the total number of students in each class. - Convert both ratios to percentages before comparing them.

Solution

1. Class 6A had a relative frequency of \(\frac{12}{20}=0.60=60\%\). 2. Class 6B had a relative frequency of \(\frac{15}{30}=0.50=50\%\). 3. Although more students donated in Class 6B, \(60\%>50\%\). Therefore, Class 6A had the greater participation rate, and the claim is incorrect.

Answer

The claim is incorrect. Class 6A had a participation rate of \(60\%\), while Class 6B had a participation rate of \(50\%\).
5118426
A survey asked \(150\) sixth-grade students to choose their favorite school subject. The results were: - Math: \(45\) - Science: \(30\) - English language arts: \(15\) - Physical education: \(60\) a) Find the relative frequency of each subject as a percent. b) Find the corresponding central angle for each category in a circle graph. c) A student says, “More than one third of the students chose physical education.” Is the student correct? Justify your answer using part a).

Hints

- Divide each count by \(150\) to find its relative frequency. - A full circle is \(360^\circ\). - Compare the physical-education percentage with one third written as a percent.

Solution

1. The relative frequencies are: math, \(\frac{45}{150}=30\%\); science, \(\frac{30}{150}=20\%\); English language arts, \(\frac{15}{150}=10\%\); physical education, \(\frac{60}{150}=40\%\). 2. The central angles are: math, \(360^\circ\times0.30=108^\circ\); science, \(360^\circ\times0.20=72^\circ\); English language arts, \(360^\circ\times0.10=36^\circ\); physical education, \(360^\circ\times0.40=144^\circ\). 3. One third is about \(33.3\%\). Since \(40\%>33.3\%\), the student is correct.

Answer

a) Math: \(30\%\); science: \(20\%\); English language arts: \(10\%\); physical education: \(40\%\) b) Math: \(108^\circ\); science: \(72^\circ\); English language arts: \(36^\circ\); physical education: \(144^\circ\) c) Yes. \(40\%\) is greater than one third, which is approximately \(33.3\%\).
5118436
Two classes compare how often students borrow books from the library. In Class 6A, \(15\) of \(25\) students regularly borrow books. In Class 6B, \(18\) of \(30\) students regularly borrow books. a) Find the relative frequency of regular library users in each class as a percent. b) Which class has the greater relative frequency? c) Suppose you display the results in a bar graph. Explain why percentages provide a fairer comparison than the numbers of students.

Hints

- Divide the number of regular library users by the class total. - Compare the resulting percentages. - Think about how different group sizes affect raw counts.

Solution

1. For Class 6A, \(\frac{15}{25}=0.60=60\%\). 2. For Class 6B, \(\frac{18}{30}=0.60=60\%\). 3. The relative frequencies are equal. 4. The classes have different total enrollments, so the counts \(15\) and \(18\) do not describe the same-sized groups. Percentages compare the same proportion of each class.

Answer

a) Class 6A: \(60\%\); Class 6B: \(60\%\) b) Neither class; the relative frequencies are equal. c) Percentages account for the different class sizes and therefore allow a fair comparison.
5142486
A survey asked \(40\) eighth-grade students how they travel to school. <table> <tr><td>Transportation</td><td>Bike</td><td>Bus</td><td>Walk</td><td>Car</td></tr> <tr><td>Number of students</td><td>10</td><td>16</td><td>10</td><td>4</td></tr> </table> a) Find the relative frequency of each transportation method as a percent. b) What is the probability that a randomly selected student from this class does not travel by car? Give the result as a fraction and a percent. c) A circle graph will display the data. What central angle should be used for the bus sector?

Hints

- Confirm the total number of students. - Relative frequency is category count divided by total count. - For a probability, compare the favorable count with the total count. - A full circle represents \(100\%\) and measures \(360^\circ\).

Solution

1. The total is \(10 + 16 + 10 + 4 = 40\) students. 2. a) Bike: \(\frac{10}{40} = 25\%\); bus: \(\frac{16}{40} = 40\%\); walk: \(\frac{10}{40} = 25\%\); car: \(\frac{4}{40} = 10\%\). 3. b) There are \(40 - 4 = 36\) students who do not travel by car. Therefore, \(P = \frac{36}{40} = \frac{9}{10} = 90\%\). 4. c) The bus proportion is \(0.4\), so the central angle is \(0.4 \times 360^\circ = 144^\circ\).

Answer

a) Bike: \(25\%\); bus: \(40\%\); walk: \(25\%\); car: \(10\%\) b) \(\frac{9}{10} = 90\%\) c) \(144^\circ\)
5142496
At a school, all \(120\) eighth-grade students choose exactly one elective course. - \(30\%\) choose French. - \(45\) choose engineering and technology. - The rest choose computer science. a) How many students choose French? b) Find the relative frequency for engineering and technology. c) What is the probability that a randomly selected student chose computer science?

Hints

- Convert the French percent to a count. - All relative frequencies must add to \(1\). - Find the number of computer science students before finding its probability.

Solution

1. a) The number choosing French is \(0.30 \times 120 = 36\). 2. b) The relative frequency for engineering and technology is \(\frac{45}{120} = \frac{3}{8} = 0.375 = 37.5\%\). 3. c) The number choosing computer science is \(120 - 36 - 45 = 39\). Therefore, \(P(\text{computer science}) = \frac{39}{120} = 0.325 = 32.5\%\).

Answer

a) \(36\) students b) \(0.375 = 37.5\%\) c) \(0.325 = 32.5\%\)
5142506
The results of a math test are shown in a circle graph. The central angles for the grade sectors are: - A: \(30^\circ\) - B: \(120^\circ\) - C: \(90^\circ\) - D: \(60^\circ\) - F: \(60^\circ\) a) Find the relative frequency of grade C. b) What is the probability that a randomly selected test earned an A or B? c) Exactly \(4\) students earned a D. Use this information to find the total number of students in the class.

Hints

- A full circle measures \(360^\circ\). - Divide a sector angle by \(360^\circ\) to find its relative frequency. - Add sector angles for an “or” event. - If one fraction of the class corresponds to \(4\) students, scale up to the whole class.

Solution

1. a) The grade-C sector is \(\frac{90^\circ}{360^\circ} = \frac{1}{4} = 25\%\) of the circle. 2. b) The A and B sectors total \(30^\circ + 120^\circ = 150^\circ\). Therefore, \(P(\text{A or B}) = \frac{150}{360} = \frac{5}{12} \approx 41.7\%\). 3. c) The D sector represents \(\frac{60}{360} = \frac{1}{6}\) of the class. If \(\frac{1}{6}\) of the class is \(4\) students, then the total is \(4 \times 6 = 24\).

Answer

a) \(25\%\) b) \(\frac{5}{12} \approx 41.7\%\) c) \(24\) students
5142656
A recreation program surveyed all \(80\) students about their favorite sport. Some results are shown as percentages. <table> <tr><td><b>Sport</b></td><td><b>Share (percent)</b></td><td><b>Number of students</b></td></tr> <tr><td>Soccer</td><td>\(25\%\)</td><td>?</td></tr> <tr><td>Basketball</td><td>\(20\%\)</td><td>?</td></tr> <tr><td>Tennis</td><td>?</td><td>\(8\)</td></tr> <tr><td>Other</td><td>?</td><td>?</td></tr> </table> a) Complete the table by finding the missing counts and percentages. b) In a bar graph with the sports on the x-axis, which category would have the tallest bar?

Hints

- The full group represents \(100\%\). - Use each known percent to find a part of \(80\). - All category percentages must add to \(100\%\).

Solution

1. Soccer: \(25\%\) of \(80\) is \(0.25\times80=20\) students. 2. Basketball: \(20\%\) of \(80\) is \(0.20\times80=16\) students. 3. Tennis: \(\frac{8}{80}=10\%\). 4. Other: \(100\%-25\%-20\%-10\%=45\%\). 5. The number in the Other category is \(0.45\times80=36\) students. 6. Other has the greatest count and percentage, so it would have the tallest bar.

Answer

a) Soccer: \(20\) students; basketball: \(16\) students; tennis: \(10\%\); other: \(45\%\) and \(36\) students b) Other
5350976
A sixth-grade class completed a survey about how students travel to school. The bar graph shows the results. a) How many students participated in the survey? b) Find the relative frequency of each transportation method as a percent. c) Which transportation method is used by the most students?
Figure for problem 535097

Hints

- Read each bar height from the y-axis. - Add the bar values to find the total. - Divide each count by the total, then convert to a percent. - The tallest bar identifies the most common method.

Solution

1. The graph shows bicycle, \(15\); walking, \(10\); bus or train, \(20\); and car, \(5\). 2. The total number of students is \(15+10+20+5=50\). 3. Bicycle: \(\frac{15}{50}=30\%\). 4. Walking: \(\frac{10}{50}=20\%\). 5. Bus or train: \(\frac{20}{50}=40\%\). 6. Car: \(\frac{5}{50}=10\%\). 7. Bus or train has the greatest count.

Answer

a) \(50\) students b) Bicycle: \(30\%\); walking: \(20\%\); bus or train: \(40\%\); car: \(10\%\) c) Bus or train
5351066
Visitors to a community youth center were surveyed about their favorite hobby. Each person chose exactly one response. The bar graph shows the results. a) Find the absolute frequency of the event “favorite hobby is sports or reading.” b) Find the relative frequency for gaming as a fraction in simplest form and as a percent.
Figure for problem 535106

Hints

- Read the value of each bar. - Add all bar values to find the total number surveyed. - For part a), add the counts for the two named hobbies. - Relative frequency is the category count divided by the total. - Convert the simplified fraction to a percent.

Solution

1. The graph shows sports, \(18\); gaming, \(12\); reading, \(10\); music, \(6\); and other, \(4\). 2. The total number surveyed is \(18+12+10+6+4=50\). 3. The absolute frequency for sports or reading is \(18+10=28\). 4. The relative frequency for gaming is \(\frac{12}{50}=\frac{6}{25}=24\%\).

Answer

a) \(28\) b) \(\frac{6}{25}\), or \(24\%\)
5351236
A survey asked \(40\) students to choose their preferred recess snack. The bar graph shows the relative frequencies. Which of the three distributions in the table matches the graph exactly? Justify your choice with calculations. <table> <tr><th>Distribution</th><th>Apple</th><th>Pretzel</th><th>Snack bar</th><th>Yogurt</th></tr> <tr><td>Distribution 1</td><td>\(10\)</td><td>\(16\)</td><td>\(8\)</td><td>\(6\)</td></tr> <tr><td>Distribution 2</td><td>\(8\)</td><td>\(20\)</td><td>\(4\)</td><td>\(8\)</td></tr> <tr><td>Distribution 3</td><td>\(12\)</td><td>\(12\)</td><td>\(10\)</td><td>\(6\)</td></tr> </table>
Figure for problem 535123

Hints

- Read the relative frequency of each snack from the graph. - Each row totals \(40\) students. - Divide each snack count by \(40\). - Compare the resulting decimals with the bar heights.

Solution

1. Distribution 1 has relative frequencies \(\frac{10}{40}=0.25\), \(\frac{16}{40}=0.40\), \(\frac{8}{40}=0.20\), and \(\frac{6}{40}=0.15\). 2. These values match the graph for Apple, Pretzel, Snack bar, and Yogurt. 3. Distribution 2 would give Pretzel a relative frequency of \(\frac{20}{40}=0.50\), which does not match the graph. 4. Distribution 3 would give Apple a relative frequency of \(\frac{12}{40}=0.30\), which does not match the graph. 5. Therefore, Distribution 1 is the only match.

Answer

Distribution 1. Its relative frequencies are Apple, \(0.25\); Pretzel, \(0.40\); Snack bar, \(0.20\); and Yogurt, \(0.15\), which match the graph.
5351376
A survey asked \(40\) sixth-grade students to choose their favorite hobby. The bar graph shows the results. Which statements are true? 1) More than one third of the students chose Sports. 2) Music and Reading together were more popular than Sports. 3) Gaming represents exactly \(25\%\) of all responses. 4) More than half of the class chose either Sports or Reading.
Figure for problem 535137

Hints

- Add the bar values to confirm the total. - Translate one third and one half into amounts out of \(40\). - Add category counts when a statement combines two hobbies. - Pay attention to words such as “more than” and “exactly.”

Solution

1. The graph totals \(14+10+10+6=40\) students. 2. Statement 1 is true because one third of \(40\) is about \(13.3\), and \(14>13.3\). 3. Statement 2 is true because Music and Reading total \(10+6=16\), and \(16>14\). 4. Statement 3 is true because \(\frac{10}{40}=\frac{1}{4}=25\%\). 5. Statement 4 is false because Sports and Reading total \(14+6=20\), which is exactly half of \(40\), not more than half.

Answer

Statements 1, 2, and 3 are true.

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