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Relative frequency from data

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Difficulty: 1 – 5

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 \(40\%\)

5375067

In \(200\) trials of a two-stage experiment, the frequencies are shown in the tree diagram. Find the relative frequency of event \(B\).

Figure for problem 537506

Hints

- Identify every endpoint in the compound experiment that belongs to event \(B\). - Add those endpoint counts before forming a part-to-whole ratio. - Divide by the total 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

\(0.46=46\%\)

5413216

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.

5413326

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 \(\frac{36}{45}=\frac{4}{5}=0.80=80\%\).

5413406

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.

5413606

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) \(\frac{11}{24}\approx45.8\%\)

5414007

In a sample of \(60\) library visitors, \(18\) used the self-checkout station. Find the relative frequency of self-checkout use. Assuming the sample rate is representative, estimate how many of a similar group of \(300\) visitors would use self-checkout. Explain why the second result is an estimate.

Hints

- Form the sample's part-to-whole ratio first. - Apply that same proportion to the larger group. - Distinguish using a sample rate from observing the larger group directly.

Solution

1. The sample relative frequency is \(\frac{18}{60}=0.30=30\%\). 2. Applying the same rate to \(300\) visitors gives \(0.30\times300=90\). 3. The result is an estimate because a new group's observed relative frequency may differ from the sample rate.

Answer

Relative frequency: \(0.30=30\%\) Estimated users among \(300\): about \(90\)

5414116

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: \(27\), \(\frac{3}{8}=0.375\)

5414286

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: \(\frac{2}{3}=66\frac{2}{3}\%\)

5414636

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: \(50\%\)

5414806

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

5414856

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: \(6\).

5414926

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: \(6\%\)

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 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 \(24\%\)

5351236

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 \(0.15\).

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.

5387976

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 \(50\%\).

5412696

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.

5412736

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 \(0.375\)

5412786

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 \(0.10\)

5412826

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.

5412876

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 \(100\%\).

5412916

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 \(20.8\%\)

5412966

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.

5413006

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 \(1.00\).

5413036

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.

5413086

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.

5413176

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.

5413286

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 \(38.9\%\)

5413366

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: \(0.45=45\%\)

5413506

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

5413616

The frequency histogram shows four equal-width intervals. Convert the bar heights to relative frequencies. Explain what stays the same and what changes when the display is redrawn as a relative-frequency histogram.

Figure for problem 541361

Hints

- Find the total represented by all bars. - Compare each interval’s count with that same total. - Think about whether dividing every bar height by one common number changes the shape.

Solution

1. The total frequency is \(3+5+8+4=20\). 2. The relative-frequency heights are \(\frac{3}{20}=0.15\), \(\frac{5}{20}=0.25\), \(\frac{8}{20}=0.40\), and \(\frac{4}{20}=0.20\). 3. The intervals and the relative pattern of bar heights stay the same. 4. The vertical scale changes from numbers of observations to parts of the total.

Answer

Relative-frequency heights: \(0.15, 0.25, 0.40, 0.20\). The histogram keeps the same intervals and shape, but its y-axis changes from frequency to relative frequency.

5413736

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: \(\frac{1}{16}=6.25\%\).

5413786

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 \(0.70=70\%\).

5413856

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: \(0.25=25\%\)

5413886

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: \(10\%\)

5414196

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: \(14\)

5414296

The relative-frequency histogram has three equal-width intervals. Give one possible set of interval counts if the histogram represents \(10\) observations and another if it represents \(20\) observations. Explain why the histogram alone does not reveal the sample size.

Figure for problem 541429

Hints

- Multiply each bar height by the proposed total. - Check that each set of counts adds to its sample size. - Consider what information is removed when counts are divided by one common total.

Solution

1. For \(10\) observations, the counts are \(0.20\times10=2\), \(0.50\times10=5\), and \(0.30\times10=3\). 2. For \(20\) observations, the counts are \(4,10,6\). 3. Both count sets have the same proportions, so they produce the same relative-frequency bar heights. 4. A relative-frequency scale shows shares of a total but not the total itself.

Answer

For \(10\) observations: \(2,5,3\). For \(20\) observations: \(4,10,6\). The sample size cannot be determined from relative frequencies alone.

5414556

Histogram A represents \(20\) observations, and Histogram B represents \(40\) observations. Both use the same three intervals. Which histogram has the greater relative concentration in the middle interval? Give both middle-interval relative frequencies.

Figure for problem 541455

Hints

- Compare each middle bar with its own histogram total. - Do not rank concentration by raw bar height when sample sizes differ. - Express both shares in the same form.

Solution

1. Histogram A’s middle relative frequency is \(\frac{10}{20}=0.50=50\%\). 2. Histogram B’s middle relative frequency is \(\frac{16}{40}=0.40=40\%\). 3. Although B has more observations in the middle interval, A has the greater proportion there.

Answer

Histogram A has the greater middle-interval concentration. A: \(50\%\) B: \(40\%\)

5414716

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

- A trial count must be a whole number. - Multiply the displayed decimal by the sample size to test exactness. - Check nearby whole-number counts and round their fractions.

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 \(\frac{1}{3}=0.333\ldots\).

5413456

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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