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Relative frequency from data

Learning goals

  • Computes relative frequencies from counts.
  • Expresses relative frequencies as fractions, decimals, or percents.
  • Interprets relative frequencies in context.
  • Uses conditional or marginal totals correctly when applicable.

Click problems to add them to your worksheet.

Relative frequency from data 1
In a class survey, \(3\) of \(12\) students chose chess. What fraction of the students chose chess? Simplify it.

Hints

- Use all surveyed students in the denominator. - Simplify the fraction after writing part over whole.

Solution

1. The relative frequency is the count choosing chess over the total: \(\frac{3}{12}\). 2. Simplify to \(\frac{1}{4}\).

Answer

\(\frac{1}{4}\).
Relative frequency from data 2
Eight out of \(20\) students walked to school. What percent walked?

Hints

- Divide the number who walked by the class total. - Convert the resulting part of one whole to a percent.

Solution

1. The relative frequency is \(\frac{8}{20}=0.4\). 2. Convert it to \(40\%\).

Answer

\(40\%\).
Relative frequency from data 3
In Class A, \(8\) of \(10\) students brought lunch. In Class B, \(8\) of \(20\) brought lunch. Which class had the greater share of students bringing lunch?

Hints

- A share compares the count with the total in the same class. - Why might the same count describe different-sized parts of two groups?

Solution

1. The shares are \(\frac{8}{10}=80\%\) for A and \(\frac{8}{20}=40\%\) for B. 2. Class A has the greater share, even though the counts are equal.

Answer

Class A.
Relative frequency from data 4
The table shows buttons that a class sorted by color. What is the relative frequency of red buttons? <table><tr><th>Color</th><th>Count</th></tr><tr><td>Blue</td><td>4</td></tr><tr><td>Red</td><td>3</td></tr><tr><td>Green</td><td>3</td></tr></table>

Hints

- Total all the button categories first. - Red buttons make up one part of that total.

Solution

1. There are \(4+3+3=10\) buttons. 2. The share that is red is \(\frac{3}{10}=0.3\).

Answer

\(\frac{3}{10}=0.3=30\%\).
Relative frequency from data 5
The graph shows the relative frequencies of students' favorite after-school clubs. Which club was chosen by half the students?
Figure for problem 559188

Hints

- Half of a whole is \(0.5\). - Find the bar that reaches this height on the vertical scale.

Solution

1. Half a group corresponds to a relative frequency of \(0.5\). 2. The bar for Art reaches \(0.5\).

Answer

Art.
Relative frequency from data 6
The circle graph shows students' club choices. What fraction of the students chose Chess?
Figure for problem 559189

Hints

- Each slice represents a share of the surveyed students. - Compare the Chess slice with the whole circle.

Solution

1. The Chess slice takes one fourth of the circle, so the same fraction of students chose it. 2. The observed relative frequency is \(\frac{1}{4}\).

Answer

\(\frac{1}{4}\) of the students.
Relative frequency from data 7
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.
Relative frequency from data 8
A survey of \(200\) students found their favorite activities: sports (\(80\)), music (\(50\)), gaming (\(40\)), and reading (\(30\)). Lucas says, “Four activities means each must make up \(25\%\) of the answers.” a) Explain the mistake. b) What percent of students chose reading? c) What percent chose sports or music?

Hints

- Does every activity have the same count? - Compare each relevant count with the total number of survey answers. - For the last part, combine the two non-overlapping groups first.

Solution

1. Four categories do not have equal relative frequencies unless they have equal counts. 2. Reading accounts for \(\frac{30}{200}\times100\%=15\%\) of the answers. 3. Sports or music accounts for \(\frac{80+50}{200}\times100\%=65\%\) of the answers.

Answer

a) The categories have different numbers of answers, so their shares are not all \(25\%\). b) \(15\%\) c) \(65\%\)
Relative frequency from data 9
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: \(0.02 = 2\%\)
Relative frequency from data 10
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.
Relative frequency from data 11
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 \(1\).
Relative frequency from data 12
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: \(15\%\)
Relative frequency from data 13
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 \(40\%\)
Relative frequency from data 14
A survey of \(75\) students has this relative-frequency distribution: - Option P: \(0.48\) - Option Q: \(0.24\) - Option R: \(0.28\) Find the count for each option. Then determine whether Option P was selected exactly twice as often as Option Q.

Hints

- Connect each share to the same total number of students. - Check the completed counts against the total. - Compare the two relevant counts multiplicatively, not by subtraction alone.

Solution

1. Option P count: \(0.48\times75=36\). 2. Option Q count: \(0.24\times75=18\). 3. Option R count: \(0.28\times75=21\). 4. Since \(36=2\times18\), Option P was selected exactly twice as often as Option Q. 5. The counts check because \(36+18+21=75\).

Answer

P: \(36\) students Q: \(18\) students R: \(21\) students Yes. Option P was selected exactly twice as often as Option Q.
Relative frequency from data 15
A report says, “\(36\) of \(45\) completed tasks were checked, so the relative frequency is \(\frac{45}{36}=1.25=125\%\).” Identify the error, give the correct relative frequency as a fraction, decimal, and percent, and explain why the reported percent is impossible in this context.

Hints

- Identify the part and the whole before writing the fraction. - A part-to-whole relative frequency has a predictable range of possible values. - Check whether the result is reasonable for a category contained inside the total.

Solution

1. Relative frequency is category count divided by total count, so the fraction should be \(\frac{36}{45}\), not \(\frac{45}{36}\). 2. Simplify: \(\frac{36}{45}=\frac{4}{5}=0.80=80\%\). 3. A checked task is part of the \(45\) total tasks, so its relative frequency cannot exceed \(100\%\).

Answer

The fraction was reversed. The correct relative frequency is \(\frac{36}{45}=\frac{4}{5}=0.80=80\%\).
Relative frequency from data 16
At Exhibit A, \(7\) of \(20\) visitors completed a challenge. At Exhibit B, \(9\) of \(25\) visitors completed it. Find each relative frequency as a decimal and percent. Which exhibit has the higher completion rate, and by how many percentage points?

Hints

- Compare part-to-whole ratios rather than the raw successful counts alone. - Express both rates in the same form. - Subtract the percentages to state the difference in percentage points.

Solution

1. Exhibit A: \(\frac{7}{20}=0.35=35\%\). 2. Exhibit B: \(\frac{9}{25}=0.36=36\%\). 3. Exhibit B has the higher relative frequency. 4. The difference is \(36\%-35\%=1\) percentage point.

Answer

Exhibit A: \(0.35=35\%\) Exhibit B: \(0.36=36\%\) Exhibit B is higher by \(1\) percentage point.
Relative frequency from data 17
A wildlife camera recorded \(9\) bird images, \(6\) deer images, \(5\) raccoon images, and \(4\) empty images. a) What is the relative frequency that an image contains an animal? b) What is the relative frequency that an image contains a mammal? Give each answer as a fraction and a percent rounded to the nearest tenth of a percent.

Hints

- Find the total number of recorded images first. - Combine only the categories that meet each description. - Use the same total as the denominator for both relative frequencies.

Solution

1. The camera recorded \(9+6+5+4=24\) images. 2. Animal images total \(9+6+5=20\), so the relative frequency is \(\frac{20}{24}=\frac{5}{6}\approx83.3\%\). 3. Mammal images total \(6+5=11\), so the relative frequency is \(\frac{11}{24}\approx45.8\%\).

Answer

a) \(\frac{5}{6}\approx83.3\%\) b) \(\frac{11}{24}\approx45.8\%\)
Relative frequency from data 18
Of \(72\) weather observations, \(27\) were sunny and \(18\) were cloudy. The rest were rainy. Find the count and relative frequency for each category. Give each relative frequency as a simplified fraction and decimal.

Hints

- Find the missing category count from the total. - Use the same whole as the denominator for every category. - Simplify each fraction and check that the shares form one whole.

Solution

1. Rainy observations total \(72-27-18=27\). 2. Sunny relative frequency is \(\frac{27}{72}=\frac{3}{8}=0.375\). 3. Cloudy relative frequency is \(\frac{18}{72}=\frac{1}{4}=0.25\). 4. Rainy relative frequency is \(\frac{27}{72}=\frac{3}{8}=0.375\). 5. The counts total \(72\), and the decimals total \(1\).

Answer

Sunny: \(27\), \(\frac{3}{8}=0.375\) Cloudy: \(18\), \(\frac{1}{4}=0.25\) Rainy: \(27\), \(\frac{3}{8}=0.375\)
Relative frequency from data 19
A team’s goals per game were summarized as follows. <table><tr><th>Goals</th><th>Number of games</th></tr><tr><td>\(0\)</td><td>\(8\)</td></tr><tr><td>\(1\)</td><td>\(12\)</td></tr><tr><td>\(2\)</td><td>\(6\)</td></tr><tr><td>\(3\)</td><td>\(4\)</td></tr></table> Find the relative frequency of games with at least \(2\) goals and the relative frequency of games with fewer than \(2\) goals. Give fractions and percents.

Hints

- Add all game frequencies for the denominator. - Combine the rows satisfying each inequality. - Use the complementary groups as a check.

Solution

1. The total number of games is \(8+12+6+4=30\). 2. At least \(2\) goals occurred in \(6+4=10\) games, so the relative frequency is \(\frac{10}{30}=\frac{1}{3}=33\frac{1}{3}\%\). 3. Fewer than \(2\) goals occurred in \(8+12=20\) games, so the relative frequency is \(\frac{20}{30}=\frac{2}{3}=66\frac{2}{3}\%\). 4. The two events are complements, so their relative frequencies total \(1\).

Answer

At least \(2\) goals: \(\frac{1}{3}=33\frac{1}{3}\%\) Fewer than \(2\) goals: \(\frac{2}{3}=66\frac{2}{3}\%\)
Relative frequency from data 20
A robot completed \(90\) route tests: \(27\) on Route A, \(36\) on Route B, \(18\) on Route C, and \(9\) on Route D. Find each route’s relative frequency. Then find the combined relative frequency for Routes A and C.

Hints

- Use the full number of tests as each denominator. - Convert every category to the same relative-frequency form. - Combine category shares only after checking that the categories do not overlap.

Solution

1. Route A: \(\frac{27}{90}=0.30=30\%\). 2. Route B: \(\frac{36}{90}=0.40=40\%\). 3. Route C: \(\frac{18}{90}=0.20=20\%\). 4. Route D: \(\frac{9}{90}=0.10=10\%\). 5. Routes A and C together have relative frequency \(0.30+0.20=0.50=50\%\).

Answer

Route A: \(30\%\) Route B: \(40\%\) Route C: \(20\%\) Route D: \(10\%\) Routes A and C combined: \(50\%\)
Relative frequency from data 21
Two reading groups each had a \(25\%\) relative frequency of students choosing an audiobook. Group A had \(48\) students, and Group B had \(84\) students. Find the count in each group and the difference between the counts. Explain why equal relative frequencies do not give equal counts.

Hints

- Apply the common relative frequency to each group’s own total. - Compare counts only after converting both shares. - Separate equal proportions from equal group sizes.

Solution

1. Group A’s count is \(0.25\times48=12\). 2. Group B’s count is \(0.25\times84=21\). 3. The difference is \(21-12=9\) students. 4. The same fraction of different group totals produces different counts.

Answer

Group A: \(12\) students Group B: \(21\) students Difference: \(9\) students
Relative frequency from data 22
A dot plot contains \(15\) observations. Exactly \(40\%\) of the dots are at \(6\), exactly \(\frac{1}{3}\) are at \(7\), and all remaining dots are at \(8\). Find the three stack heights and identify the mode.

Hints

- Convert each stated share into a number of dots. - Use the total to find the unassigned stack. - Compare the completed stack heights.

Solution

1. The stack at \(6\) has \(0.40\times15=6\) dots. 2. The stack at \(7\) has \(\frac{1}{3}\times15=5\) dots. 3. The remaining stack has \(15-6-5=4\) dots at \(8\). 4. The tallest stack is at \(6\), so the mode is \(6\).

Answer

Stack heights: \(6\) dots at value \(6\), \(5\) dots at value \(7\), and \(4\) dots at value \(8\). Mode: \(6\).
Relative frequency from data 23
Of \(250\) inspected devices, \(197\) passed immediately, \(38\) passed after a recheck, and \(15\) did not pass. Find the relative frequency of passing immediately, passing eventually, and not passing.

Hints

- Use all inspected devices as the denominator. - Combine both passing categories for the cumulative result. - Check the eventual outcome with its complement.

Solution

1. The immediate-pass relative frequency is \(\frac{197}{250}=0.788=78.8\%\). 2. Eventually passing includes immediate and recheck passes: \(197+38=235\). 3. The eventual-pass relative frequency is \(\frac{235}{250}=0.94=94\%\). 4. The not-pass relative frequency is \(\frac{15}{250}=0.06=6\%\). 5. Eventual pass and not pass are complementary: \(94\%+6\%=100\%\).

Answer

Passed immediately: \(78.8\%\) Passed eventually: \(94\%\) Did not pass: \(6\%\)
Relative frequency from data 24
In a class survey, \(6\) of \(24\) students chose basketball as their favorite sport. What was basketball's relative frequency, as a fraction? How many degrees should its slice have in a circle graph of the survey?

Hints

- Start with the basketball count over the total number surveyed. - A full circle stands for everyone who answered. - Use the same fraction of a full turn as the fraction of students.

Solution

1. The observed share choosing basketball is \(\frac{6}{24}=\frac{1}{4}\). 2. Its sector takes \(\frac{1}{4}\) of a full \(360^\circ\) circle. 3. The angle is \(360^\circ\div4=90^\circ\).

Answer

Relative frequency: \(\frac{1}{4}\); sector angle: \(90^\circ\).
Relative frequency from data 25
Two classes answered a favorite-color survey: <table><tr><th></th><th>Red</th><th>Blue</th><th>Total</th></tr><tr><td>Class A</td><td>3</td><td>9</td><td>12</td></tr><tr><td>Class B</td><td>5</td><td>3</td><td>8</td></tr></table> What percent of each class chose red? Which class had the greater red share?

Hints

- Compare red choices with each class's own total. - Do not use the total from both classes as the denominator. - Put both answers in percent form before comparing.

Solution

1. Class A: \(\frac{3}{12}=25\%\). 2. Class B: \(\frac{5}{8}=62.5\%\). 3. Class B has the higher relative frequency for red.

Answer

Class A: \(25\%\); Class B: \(62.5\%\). Class B has the greater red share.
Relative frequency from data 26
Give the relative frequency, as a percent, for \(1\), \(2\), and \(3\) pets in the dot plot.
Figure for problem 559192

Hints

- Count each stack of dots separately. - Every stack's relative frequency uses the same total number of students. - All percentages together should equal \(100\%\).

Solution

1. There are \(5+3+2=10\) dots in all. 2. One pet: \(\frac{5}{10}=50\%\). 3. Two pets: \(\frac{3}{10}=30\%\). 4. Three pets: \(\frac{2}{10}=20\%\).

Answer

\(1\) pet: \(50\%\); \(2\) pets: \(30\%\); \(3\) pets: \(20\%\).
Relative frequency from data 27
A park surveyed hikers about their favorite trail. Creek received \(12\) votes, Hill received \(8\), and Woods received \(10\). A sign says, “More than half of the hikers chose Creek.” Is that true? Find Creek's relative frequency to explain.

Hints

- Does “more than half” refer to the largest number of votes or to a fraction of all votes? - Find how many hikers voted altogether before judging the sign. - Compare Creek's part of the total with one half.

Solution

1. The total number of votes is \(12+8+10=30\). 2. Creek's relative frequency is \(\frac{12}{30}=\frac{2}{5}=40\%\). 3. Since \(40\%\) is less than half, the sign is wrong even though Creek received more votes than either other trail.

Answer

No. Creek received \(\frac{12}{30}=40\%\) of the votes, which is less than half.
Relative frequency from data 28
Students voted for a fruit for snack time. The results were: <table><tr><th>Fruit</th><th>Students</th></tr><tr><td>Apple</td><td>9</td></tr><tr><td>Banana</td><td>12</td></tr><tr><td>Orange</td><td>9</td></tr></table> A student says, “The relative frequency for Apple is \(9\).” Explain the mistake and find the correct relative frequency.

Hints

- Think about the difference between a count and a share of all answers. - Which total should the Apple count be compared with? - Express that part of the whole as a fraction or percent.

Solution

1. The number \(9\) is the count of students who picked Apple, not its share of all choices. 2. The total is \(9+12+9=30\) students. 3. The relative frequency for Apple is \(\frac{9}{30}=\frac{3}{10}=30\%\).

Answer

\(9\) is the count, not the relative frequency. The relative frequency is \(\frac{9}{30}=\frac{3}{10}=30\%\).
Relative frequency from data 29
Forty students chose their favorite after-school club. a) What fraction of the students chose Chess? Show how you found it from the angle. b) How many students chose Chess?
Figure for problem 559205

Hints

- A full circle stands for all the students. - Compare the marked angle with the angle of a full circle. - Use the resulting fraction of the \(40\) students.

Solution

1. A whole circle is \(360^\circ\), so Chess represents \(\frac{72}{360}=\frac{1}{5}\) of the surveyed students. 2. One fifth of \(40\) students is \(40\div5=8\) students.

Answer

a) \(\frac{1}{5}\) of the students. b) \(8\) students.
Relative frequency from data 30
Volunteers counted butterflies at a nature center. They saw \(9\) monarch butterflies, which were \(\frac{1}{4}\) of all the butterflies they counted. a) How many butterflies did the volunteers count altogether? b) How many were not monarch butterflies?

Hints

- Which number is a count, and which is a fraction of the whole group? - Should the total number of butterflies be smaller or larger than the monarch count? - Check your total by comparing the observed monarch count with the stated relative frequency.

Solution

1. The \(9\) monarchs are one fourth of the total, so the total is \(9\times4=36\) butterflies. 2. Of those \(36\), the number that were not monarchs is \(36-9=27\).

Answer

a) \(36\) butterflies. b) \(27\) butterflies.
Relative frequency from data 31
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 \(30\%>20\%\).
Relative frequency from data 32
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) \(40\%+30\%+30\%=100\%\)
Relative frequency from data 33
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 \(50\%\).
Relative frequency from data 34
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) Kagiso says, “More than one third of the students chose physical education.” Is Kagiso 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 exactly \(33\frac{1}{3}\%\). Since \(40\%>33\frac{1}{3}\%\), Kagiso 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 exactly \(33\frac{1}{3}\%\).
Relative frequency from data 35
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.
Relative frequency from data 36
A survey asked \(40\) eighth-grade students how they travel to school. <table><tr><th>Travel method</th><th>Bike</th><th>Bus</th><th>Walk</th><th>Car</th></tr><tr><td>Students</td><td>10</td><td>16</td><td>10</td><td>4</td></tr></table> a) Find the relative frequency of each method as a percent. b) What fraction and percent of those surveyed do not travel by car? c) A circle graph will show the results. What central angle should the bus sector have?

Hints

- Use the number of surveyed students as the denominator each time. - “Not by car” includes all the other listed travel methods. - For a circle graph, match a fraction of the total to the same fraction of a full turn.

Solution

1. There are \(40\) students in all. 2. Bike: \(\frac{10}{40}=25\%\); bus: \(\frac{16}{40}=40\%\); walk: \(\frac{10}{40}=25\%\); car: \(\frac{4}{40}=10\%\). 3. Not by car: \(40-4=36\) students, so \(\frac{36}{40}=\frac{9}{10}=90\%\). 4. Bus covers \(40\%\) of a circle: \(0.4\times360^\circ=144^\circ\).

Answer

a) Bike: \(25\%\); bus: \(40\%\); walk: \(25\%\); car: \(10\%\) b) \(\frac{9}{10}=90\%\) c) \(144^\circ\)
Relative frequency from data 37
All \(120\) eighth-grade students at a school choose one elective course. - \(30\%\) choose French. - \(45\) choose engineering and technology. - The rest choose computer science. a) How many choose French? b) Find the relative frequency of engineering and technology. c) Find the relative frequency of computer science.

Hints

- First turn the French percentage into a number of students. - A relative frequency compares one course's count with the whole grade. - All the course counts must add up to the total.

Solution

1. French: \(0.30\times120=36\) students. 2. Engineering and technology: \(\frac{45}{120}=0.375=37.5\%\). 3. Computer science: \(120-36-45=39\) students, so \(\frac{39}{120}=0.325=32.5\%\). 4. Check: \(30\%+37.5\%+32.5\%=100\%\).

Answer

a) \(36\) students b) \(0.375=37.5\%\) c) \(0.325=32.5\%\)
Relative frequency from data 38
A class made a circle graph of its math test grades. The sectors have these central angles: - A: \(30^\circ\) - B: \(120^\circ\) - C: \(90^\circ\) - D: \(60^\circ\) - F: \(60^\circ\) a) What percent of tests earned a C? b) What fraction of tests earned an A or B? Give the percent to the nearest tenth. c) Exactly \(4\) students earned a D. How many students took the test?

Hints

- A full circle is \(360^\circ\). - A sector's share is its angle divided by the full angle. - Combine the angles for the two separate grade categories before finding their share. - Use the D share to work back to the total number of students.

Solution

1. Grade C: \(\frac{90}{360}=\frac{1}{4}=25\%\). 2. Grades A or B: \(30^\circ+120^\circ=150^\circ\), so \(\frac{150}{360}=\frac{5}{12}\approx41.7\%\). 3. Grade D covers \(\frac{60}{360}=\frac{1}{6}\) of the class. If this is \(4\) students, the total is \(4\times6=24\).

Answer

a) \(25\%\) b) \(\frac{5}{12}\approx41.7\%\) c) \(24\) students
Relative frequency from data 39
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
Relative frequency from data 40
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
Relative frequency from data 41
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 frequency, or count, for “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 count represented by each bar. - Add all bar values to find the total number surveyed. - Relative frequency is the category count divided by the total.

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 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 \(24\%\)
Relative frequency from data 42
A survey asked \(40\) students to choose their preferred recess snack. The bar graph shows the relative frequencies. Which distribution in the table matches the graph exactly? Justify the match by checking enough categories to rule out both other rows. <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>\(10\)</td><td>\(14\)</td><td>\(10\)</td><td>\(6\)</td></tr><tr><td>Distribution 3</td><td>\(8\)</td><td>\(16\)</td><td>\(8\)</td><td>\(8\)</td></tr></table>
Figure for problem 535123

Hints

- Each incorrect row has some categories that do match the graph, so one category is not enough. - Convert counts to relative frequencies using the common total of \(40\). - Check categories that distinguish each remaining candidate from the graph.

Solution

1. The graph shows relative frequencies \(0.25,0.40,0.20,0.15\) for Apple, Pretzel, Snack bar, and Yogurt. 2. Distribution 1 gives \(10\div40=0.25\), \(16\div40=0.40\), \(8\div40=0.20\), and \(6\div40=0.15\), so it matches all four bars. 3. Distribution 2 matches Apple and Yogurt but gives Pretzel \(14\div40=0.35\) and Snack bar \(10\div40=0.25\), so it does not match exactly. 4. Distribution 3 matches Pretzel and Snack bar but gives Apple \(8\div40=0.20\) and Yogurt \(8\div40=0.20\), so it does not match exactly. 5. Therefore Distribution 1 is the unique complete match.

Answer

Distribution 1. Its relative frequencies are Apple \(0.25\), Pretzel \(0.40\), Snack bar \(0.20\), and Yogurt \(0.15\).
Relative frequency from data 43
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.
Relative frequency from data 44
In Group A, \(80\%\) of \(10\) people solve a problem. In Group B, \(40\%\) of \(30\) people solve it. Mia averages the two percentages and gets \(60\%\). Is \(60\%\) the percent of all people who solved the problem?

Hints

- The two percentages describe groups of different sizes. - Convert each percentage to a number of people first. - Find the overall percent using all participants.

Solution

1. In Group A, \(0.8 \times 10 = 8\) people solved the problem. 2. In Group B, \(0.4 \times 30 = 12\) people solved the problem. 3. Altogether, \(8 + 12 = 20\) of \(10 + 30 = 40\) people solved it. 4. The overall percent is \(20 \div 40 = 50\%\), not \(60\%\).

Answer

No. Altogether, \(20\) of \(40\) people solved the problem, so the overall percent is \(50\%\).
Relative frequency from data 45
On the first day of a museum program, \(12\) of \(30\) visitors chose the rooftop tour. On the second day, an unknown number of the \(20\) visitors chose it. Across both days, the relative frequency of choosing the rooftop tour was \(0.40\). How many second-day visitors chose the rooftop tour?

Hints

- Combine the two group sizes before using the overall relative frequency. - Work backward from the overall share to the overall category count. - Remove the known first-day count from the combined category count.

Solution

1. The two days had \(30+20=50\) visitors in all. 2. A relative frequency of \(0.40\) means \(0.40\times50=20\) visitors chose the rooftop tour across both days. 3. The second-day count is \(20-12=8\) visitors.

Answer

\(8\) second-day visitors chose the rooftop tour.
Relative frequency from data 46
A field camera recorded the surface under \(80\) animal tracks. The relative frequency for sand was \(0.275\), and the relative frequency for mud was \(0.35\). Every track was classified as sand, mud, or gravel. Find the count and relative frequency for each surface.

Hints

- The relative frequencies of all exclusive categories must combine to one whole. - Use the total number of observations to connect a share with a count. - Check the completed counts against the stated total.

Solution

1. Sand count: \(0.275\times80=22\). 2. Mud count: \(0.35\times80=28\). 3. Gravel relative frequency: \(1-0.275-0.35=0.375\). 4. Gravel count: \(0.375\times80=30\). 5. The counts check because \(22+28+30=80\).

Answer

Sand: \(22\) tracks, relative frequency \(0.275\) Mud: \(28\) tracks, relative frequency \(0.35\) Gravel: \(30\) tracks, relative frequency \(0.375\)
Relative frequency from data 47
A mural contains \(80\) tiles. The relative frequency of warm-color tiles is \(0.45\), the relative frequency of cool-color tiles is \(0.325\), and all remaining tiles are neutral. Among the neutral tiles, \(10\) are gray and the rest are white. Find the count and relative frequency for warm, cool, gray, and white tiles.

Hints

- First find the share and count of the unsplit category. - Use the stated subgroup count to find the remaining subgroup. - Divide each subgroup count by the original total, not by the category subtotal.

Solution

1. Warm-color count: \(0.45\times80=36\). 2. Cool-color count: \(0.325\times80=26\). 3. Neutral relative frequency: \(1-0.45-0.325=0.225\), so the neutral count is \(0.225\times80=18\). 4. White count: \(18-10=8\). 5. Gray relative frequency: \(\frac{10}{80}=0.125\). White relative frequency: \(\frac{8}{80}=0.10\).

Answer

Warm: \(36\) tiles, relative frequency \(0.45\) Cool: \(26\) tiles, relative frequency \(0.325\) Gray: \(10\) tiles, relative frequency \(0.125\) White: \(8\) tiles, relative frequency \(0.10\)
Relative frequency from data 48
A survey received \(72\) forms. Six forms were blank. Of the completed forms, \(22\) selected “evening session.” a) Find the relative frequency of “evening session” among all returned forms. b) Find its relative frequency among completed forms only. c) Explain why the two relative frequencies differ.

Hints

- Decide exactly which responses belong in each denominator. - Keep the category count fixed unless the question changes which category is counted. - Compare the two part-to-whole relationships rather than only their numerators.

Solution

1. Among all returned forms, the relative frequency is \(\frac{22}{72}=\frac{11}{36}\approx0.306\). 2. The number of completed forms is \(72-6=66\). 3. Among completed forms, the relative frequency is \(\frac{22}{66}=\frac{1}{3}\approx0.333\). 4. The numerator is the same in both calculations, but the second denominator excludes the blank forms.

Answer

a) \(\frac{11}{36}\approx0.306\) b) \(\frac{1}{3}\approx0.333\) c) The denominators differ because the second calculation excludes the \(6\) blank forms.
Relative frequency from data 49
In a vote with \(37\) valid ballots, Candidate A received \(13\) votes, Candidate B received \(12\), and Candidate C received \(12\). a) Find each candidate’s relative frequency as a percent to the nearest tenth. b) Round each percent to the nearest whole percent and add the rounded values. c) Explain why the rounded total is not exactly \(100\%\).

Hints

- Use the same total number of ballots for every category. - Keep extra decimal places until each percentage is rounded as requested. - Distinguish the exact shares from their rounded displays.

Solution

1. Candidate A: \(\frac{13}{37}\times100\%\approx35.1\%\). 2. Candidate B: \(\frac{12}{37}\times100\%\approx32.4\%\). 3. Candidate C: \(\frac{12}{37}\times100\%\approx32.4\%\). 4. Rounded to whole percents, the values are \(35\%\), \(32\%\), and \(32\%\), which total \(99\%\). 5. Each category was rounded separately, so the small rounding changes accumulate.

Answer

a) A: \(35.1\%\), B: \(32.4\%\), C: \(32.4\%\) b) \(35\%+32\%+32\%=99\%\) c) Separate rounding changes the category values slightly, so their rounded total need not equal exactly \(100\%\).
Relative frequency from data 50
In a collection survey, \(9\) responses represent a relative frequency of \(0.375\). a) How many responses were collected in all? b) A different category received \(5\) responses. Find its relative frequency as a fraction, as a decimal to the nearest thousandth, and as a percent to the nearest tenth.

Hints

- Treat the known count as a stated fraction of the unknown total. - Work backward from the relative frequency before using the second category. - Use the recovered total as the denominator for every category.

Solution

1. The total satisfies \(\frac{9}{n}=0.375=\frac{3}{8}\). 2. Since \(9\) is \(\frac{3}{8}\) of the total, the total is \(9\div\frac{3}{8}=24\). 3. The other category’s relative frequency is \(\frac{5}{24}\). 4. As a decimal, \(\frac{5}{24}\approx0.208\), and as a percent it is approximately \(20.8\%\).

Answer

a) \(24\) responses b) \(\frac{5}{24}\), approximately \(0.208\), or approximately \(20.8\%\)
Relative frequency from data 51
In the first \(30\) responses to a survey, \(18\) people selected Option A. Ten more responses arrived, and \(4\) of those selected Option A. Find the relative frequency of Option A before and after the new responses. Did the count and the relative frequency change in the same direction? Explain.

Hints

- Update the category count and the total separately. - Compare the new subgroup’s share with the original overall share. - A larger count does not always mean a larger proportion.

Solution

1. Initially, the relative frequency is \(\frac{18}{30}=0.60=60\%\). 2. After the new responses, the Option A count is \(18+4=22\), and the total is \(30+10=40\). 3. The new relative frequency is \(\frac{22}{40}=0.55=55\%\). 4. The count increased from \(18\) to \(22\), but the relative frequency decreased because only \(4\) of the \(10\) new responses selected Option A, a rate below the original \(60\%\).

Answer

Before: \(0.60=60\%\) After: \(0.55=55\%\) The count increased, but the relative frequency decreased.
Relative frequency from data 52
A survey of \(40\) visitors produced this table. One relative frequency is incorrect. <table> <tr><th>Choice</th><th>Count</th><th>Reported relative frequency</th></tr> <tr><td>A</td><td>\(12\)</td><td>\(0.30\)</td></tr> <tr><td>B</td><td>\(10\)</td><td>\(0.25\)</td></tr> <tr><td>C</td><td>\(8\)</td><td>\(0.20\)</td></tr> <tr><td>D</td><td>\(10\)</td><td>\(0.20\)</td></tr> </table> Identify and correct the error. Give two different checks that confirm your correction.

Hints

- Compare each category count with the same overall total. - Check the completed frequency column against one whole. - Use both the count total and the relative-frequency total as independent checks.

Solution

1. Choice D has relative frequency \(\frac{10}{40}=0.25\), not \(0.20\). 2. Count check: \(12+10+8+10=40\), so all responses are represented. 3. Relative-frequency check: \(0.30+0.25+0.20+0.25=1.00\). 4. Both checks confirm that the corrected value for D is \(0.25\).

Answer

Choice D is incorrect. Its relative frequency should be \(0.25\). The counts total \(40\), and the corrected relative frequencies total \(1.00\).
Relative frequency from data 53
An archive initially has \(64\) labeled photographs, and \(8\) show a lighthouse. Later, \(64\) more photographs are added, but none show a lighthouse. Find the lighthouse relative frequency before and after the addition. Explain why the count stays the same while the relative frequency changes.

Hints

- Track the category count and the overall total separately. - Recalculate the part-to-whole ratio after the collection changes. - Compare how the numerator and denominator change.

Solution

1. Initially, the relative frequency is \(\frac{8}{64}=0.125=12.5\%\). 2. After the addition, the lighthouse count remains \(8\), while the total becomes \(64+64=128\). 3. The new relative frequency is \(\frac{8}{128}=0.0625=6.25\%\). 4. The category count is unchanged, but the whole collection doubled, so the category’s share was cut in half.

Answer

Before: \(0.125=12.5\%\) After: \(0.0625=6.25\%\) The count remains \(8\), but the total doubles, so the relative frequency is halved.
Relative frequency from data 54
Fifty students could select every workshop that interested them. Music was selected by \(28\) students, sports by \(32\), and coding by \(20\). Find the relative frequency of each selection. Add the three relative frequencies and explain why the total can be greater than \(1\).

Hints

- Use the same total number of students as the denominator for each selection. - Check whether the response categories are mutually exclusive. - A sum of one is required only when each observation belongs to exactly one category.

Solution

1. Music relative frequency: \(\frac{28}{50}=0.56\). 2. Sports relative frequency: \(\frac{32}{50}=0.64\). 3. Coding relative frequency: \(\frac{20}{50}=0.40\). 4. The sum is \(0.56+0.64+0.40=1.60\). 5. Students could select more than one workshop, so the categories overlap and are not parts of one exclusive whole.

Answer

Music: \(0.56\) Sports: \(0.64\) Coding: \(0.40\) Total: \(1.60\) The total exceeds \(1\) because some students are counted in more than one category.
Relative frequency from data 55
A collection has \(40\) labeled objects: \(18\) in category A, \(12\) in category B, and \(10\) in category C. Two objects are reclassified from B to C. a) Find the relative frequencies of A, B, and C before and after the change. b) Explain why the combined relative frequency of B and C does not change.

Hints

- Keep the total fixed while updating the affected category counts. - Calculate individual shares before and after the reclassification. - Track the combined count of the two categories as a separate quantity.

Solution

1. Before the change, the relative frequencies are \(\frac{18}{40}=0.45\), \(\frac{12}{40}=0.30\), and \(\frac{10}{40}=0.25\). 2. After the change, the counts are \(18, 10, 12\), so the relative frequencies are \(0.45, 0.25, 0.30\). 3. Before the change, B and C together have relative frequency \(\frac{12+10}{40}=\frac{22}{40}=0.55\). 4. After the change, B and C together still contain the same \(22\) objects, so their combined relative frequency remains \(0.55\).

Answer

Before: A \(0.45\), B \(0.30\), C \(0.25\) After: A \(0.45\), B \(0.25\), C \(0.30\) The combined B-and-C relative frequency remains \(0.55\) because objects moved only between those two categories.
Relative frequency from data 56
A category contains \(14\) observations and has relative frequency \(\frac{7}{15}\). a) How many observations are in the full data set? b) Six new observations are added, and none belong to the category. Find the new relative frequency as a fraction, as a decimal to the nearest thousandth, and as a percent to the nearest tenth.

Hints

- Work backward from the stated part-to-whole fraction. - Update only the denominator when the added observations are outside the category. - Convert the simplified fraction after finding the new share.

Solution

1. Since \(14\) is \(\frac{7}{15}\) of the total, the original total is \(14\div\frac{7}{15}=30\). 2. The new total is \(30+6=36\), while the category count remains \(14\). 3. The new relative frequency is \(\frac{14}{36}=\frac{7}{18}\). 4. As a decimal, \(\frac{7}{18}\approx0.389\), and as a percent it is approximately \(38.9\%\).

Answer

a) \(30\) observations b) \(\frac{7}{18}\), approximately \(0.389\), or approximately \(38.9\%\)
Relative frequency from data 57
Two equal-size groups of \(40\) students answered the same question. In Group A, the selected category had relative frequency \(0.35\). In Group B, it had relative frequency \(0.55\). Find the category count in each group and its relative frequency in the combined group of \(80\) students. Explain why the combined relative frequency is the average of the two group frequencies in this case.

Hints

- Convert each group’s share to a count using the same group size. - Combine both category counts and both group totals. - Notice why equal group sizes make the two rates equally weighted.

Solution

1. Group A count: \(0.35\times40=14\). 2. Group B count: \(0.55\times40=22\). 3. The combined count is \(14+22=36\), so the combined relative frequency is \(\frac{36}{80}=0.45\). 4. Because the groups have equal size, each group contributes equal weight, and \((0.35+0.55)\div2=0.45\).

Answer

Group A: \(14\) students Group B: \(22\) students Combined relative frequency: \(0.45=45\%\)
Relative frequency from data 58
In a survey, \(25\%\) of students chose blue, and \(15\%\) chose yellow. The remaining students chose red or green. The relative frequency for red was twice the relative frequency for green. Find the relative frequency of every color. If \(80\) students were surveyed, find each color's count.

Hints

- First find the share left after the two stated categories. - Represent the relationship between the remaining two categories with equal parts. - Check that the four shares and the four counts each total one whole group.

Solution

1. Blue and yellow account for \(25\%+15\%=40\%\), leaving \(60\%\) for red and green. 2. Red and green are in a \(2\)-to-\(1\) relationship, so the remaining \(60\%\) is divided into \(3\) equal parts. 3. One part is \(60\%\div3=20\%\). Green is \(20\%\), and red is \(40\%\). 4. For \(80\) students, the counts are blue \(0.25\times80=20\), yellow \(0.15\times80=12\), red \(0.40\times80=32\), and green \(0.20\times80=16\).

Answer

Blue: \(25\%\), \(20\) students Yellow: \(15\%\), \(12\) students Red: \(40\%\), \(32\) students Green: \(20\%\), \(16\) students
Relative frequency from data 59
The histogram shows how long students spent on an art activity. Convert its bar heights to relative frequencies. What stays the same, and what changes, if it is redrawn as a relative-frequency histogram?
Figure for problem 541361

Hints

- Add all four bar heights to get the total. - Use that same total for every interval's fraction. - Compare which intervals have taller and shorter bars before and after the conversion.

Solution

1. There are \(3+5+8+4=20\) observations. 2. The relative frequencies, from left to right, are \(\frac{3}{20}=0.15\), \(\frac{5}{20}=0.25\), \(\frac{8}{20}=0.40\), and \(\frac{4}{20}=0.20\). 3. The time intervals and shape stay the same. The vertical axis changes from counts to shares of the total.

Answer

The four relative frequencies are \(0.15,\ 0.25,\ 0.40,\ 0.20\). The shape and intervals stay the same; the vertical scale changes to relative frequency.
Relative frequency from data 60
In a school survey, \(12\) of the \(80\) students were left-handed. Of those \(12\) students, \(5\) chose blue as their favorite color. Find the relative frequency of choosing blue among the left-handed students and the relative frequency of being both left-handed and choosing blue among all surveyed students. Explain why the denominators differ.

Hints

- Identify the whole group named in each question. - Keep the same numerator while changing the reference group. - State what each denominator counts.

Solution

1. Within the left-handed group, \(5\) of \(12\) chose blue, so the relative frequency is \(\frac{5}{12}\approx0.417\), or about \(41.7\%\). 2. In the whole survey, \(5\) of \(80\) students were both left-handed and chose blue, so the relative frequency is \(\frac{5}{80}=\frac{1}{16}=0.0625=6.25\%\). 3. The first question uses the subgroup of \(12\) as its whole; the second uses all \(80\) students as its whole.

Answer

Among left-handed students: \(\frac{5}{12}\approx41.7\%\). Among all students: \(\frac{1}{16}=6.25\%\).
Relative frequency from data 61
During the first \(30\) shuttle arrivals, the relative frequency of arriving on time was \(0.70\). During the next \(20\) arrivals, \(14\) were on time. Find the combined relative frequency of on-time arrivals. Explain why it is unchanged.

Hints

- Convert the first relative frequency into a count. - Combine the successful counts and the total counts separately. - Compare the rate in each part before explaining the unchanged result.

Solution

1. In the first group, \(0.70\times30=21\) arrivals were on time. 2. The combined number of on-time arrivals is \(21+14=35\). 3. The combined total is \(30+20=50\). 4. The combined relative frequency is \(\frac{35}{50}=0.70\). 5. The second group also had relative frequency \(\frac{14}{20}=0.70\), so combining groups with the same rate preserves that rate.

Answer

The combined relative frequency is \(0.70=70\%\).
Relative frequency from data 62
A histogram summarizes \(40\) observations. The middle interval contains \(35\%\) of all observations. The two outermost intervals have frequencies \(6\) and \(4\). Find the frequency of the middle interval. Then find the combined relative frequency of the two outermost intervals.

Hints

- Convert the stated percentage of the total into a count for the middle interval. - Combine the two requested outer frequencies before forming a part-to-whole ratio. - Check that a relative frequency lies between \(0\) and \(1\).

Solution

1. The middle interval contains \(35\%\) of \(40\) observations: \(0.35 \times 40=14\). 2. The two outermost intervals contain \(6+4=10\) observations. 3. Their combined relative frequency is \(\frac{10}{40}=0.25=25\%\).

Answer

Middle-interval frequency: \(14\) Combined relative frequency of the two outermost intervals: \(0.25=25\%\)
Relative frequency from data 63
A scanner report included \(120\) items, and \(18\) were marked for review. Later, workers learned that \(20\) test items had been included by mistake; \(8\) of those test items were marked for review. Find the original relative frequency marked for review and the corrected relative frequency after removing all test items.

Hints

- Remove the excluded records from both the total and the marked count. - Form each part-to-whole ratio using the matching version of the data. - Compare the rates only after both corrections are made.

Solution

1. The original relative frequency is \(\frac{18}{120}=0.15=15\%\). 2. Removing the test items leaves \(120-20=100\) actual items. 3. Removing the \(8\) marked test items leaves \(18-8=10\) actual items marked for review. 4. The corrected relative frequency is \(\frac{10}{100}=0.10=10\%\).

Answer

Original relative frequency: \(15\%\) Corrected relative frequency: \(10\%\)
Relative frequency from data 64
In a survey with three exclusive choices, Choice A had relative frequency \(0.45\), Choice B had relative frequency \(0.35\), and \(8\) people chose Choice C. Find the total number surveyed and the counts for Choices A and B.

Hints

- Find the missing category’s share first. - Use its known count to recover the whole group. - Convert the other relative frequencies to counts and check the total.

Solution

1. Choice C has relative frequency \(1-0.45-0.35=0.20\). 2. If \(8\) people represent \(0.20\) of the total, the total is \(8\div0.20=40\). 3. Choice A’s count is \(0.45\times40=18\). 4. Choice B’s count is \(0.35\times40=14\). 5. The counts check: \(18+14+8=40\).

Answer

Total surveyed: \(40\) Choice A: \(18\) Choice B: \(14\)
Relative frequency from data 65
A relative-frequency histogram shows how many minutes students practiced a song. What would the interval counts be for a group of \(10\) students? What would they be for a group of \(20\)? Why can't the group size be found from this histogram alone?
Figure for problem 541429

Hints

- Read each bar height using the vertical relative-frequency scale. - Multiply each share by the stated group size. - Would doubling every count change the relative frequencies?

Solution

1. The three bar heights are \(0.20\), \(0.50\), and \(0.30\). 2. With \(10\) students the interval counts are \(2,5,3\). 3. With \(20\) students the interval counts are \(4,10,6\). 4. Both groups have the same relative frequencies; shares alone do not give the total number of observations.

Answer

For \(10\) students: \(2,5,3\). For \(20\) students: \(4,10,6\). The group size cannot be found from the relative-frequency bars alone.
Relative frequency from data 66
Histograms A and B show students' reading times in minutes. Which group has the greater share of students who read for \(10\) to less than \(20\) minutes? Give both shares.
Figure for problem 541455

Hints

- Find the bar for the requested time interval in each display. - Compare each count with its own group's total. - Taller bars can mean more students but not necessarily a larger share.

Solution

1. Histogram A has \(10\) of \(20\) students in the middle interval: \(\frac{10}{20}=50\%\). 2. Histogram B has \(16\) of \(40\) students in that interval: \(\frac{16}{40}=40\%\). 3. Histogram A has the greater share, even though its middle bar is shorter.

Answer

Histogram A: \(50\%\); Histogram B: \(40\%\). A has the greater share.
Relative frequency from data 67
A report says an outcome had relative frequency \(0.33\) in \(30\) trials. Could \(0.33\) be the exact relative frequency? If it was rounded to the nearest hundredth, what whole-number count produced it? State the unrounded fraction.

Hints

- An exact relative frequency in \(30\) trials must come from a whole-number count out of \(30\). - Ask whether the displayed decimal could equal one of those possible fractions exactly. - If the value was rounded, compare nearby possible counts out of \(30\) with the stated hundredths value.

Solution

1. An exact relative frequency in \(30\) trials must have the form \(\frac{k}{30}\) for a whole-number count \(k\). 2. The product \(0.33\times30=9.9\) is not a whole number, so \(0.33\) cannot be exact. 3. A count of \(10\) gives \(\frac{10}{30}=\frac{1}{3}=0.333\ldots\), which rounds to \(0.33\) to the nearest hundredth.

Answer

\(0.33\) cannot be exact. A count of \(10\) outcomes produced the rounded value, with unrounded relative frequency \(\frac{1}{3}=0.333\ldots\).
Relative frequency from data 68
Three exclusive categories have relative frequencies \(\frac{1}{4}\), \(\frac{3}{10}\), and \(\frac{9}{20}\). What is the smallest possible whole-number total of observations? Find the corresponding count in each category and explain why no smaller total works.

Hints

- A category count must be a whole number. - Look for a total compatible with every fraction’s denominator. - Check that the resulting counts add back to the total.

Solution

1. The fractions have denominators \(4, 10, 20\). 2. A total must make each product of total and relative frequency a whole-number count. 3. The smallest common multiple of \(4, 10, 20\) is \(20\). 4. The counts are \(\frac{1}{4}\times20=5\), \(\frac{3}{10}\times20=6\), and \(\frac{9}{20}\times20=9\). 5. The counts total \(5+6+9=20\), and any smaller total would fail to be divisible by all required denominators after simplification.

Answer

The smallest possible total is \(20\). The category counts are \(5, 6, 9\).

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