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Relative frequency in two-way tables

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5501208
The table summarizes \(120\) commuters. <table><tr><th></th><th>Morning</th><th>Evening</th><th>Total</th></tr><tr><th>Uses the app</th><td>\(27\)</td><td>\(18\)</td><td>\(45\)</td></tr><tr><th>Does not use the app</th><td>\(33\)</td><td>\(42\)</td><td>\(75\)</td></tr><tr><th>Total</th><td>\(60\)</td><td>\(60\)</td><td>\(120\)</td></tr></table> What overall relative frequency represents commuters who use the app and travel in the evening?

Hints

- Locate the intersection of both stated categories. - For an overall relative frequency, use the grand total as the denominator.

Solution

1. The relevant cell count is \(18\), and an overall relative frequency uses the grand total of \(120\). 2. \(\frac{18}{120}=0.15=15\%\).

Answer

\(15\%\).
5501228
Complete the row relative-frequency table for workshop choice. <table><tr><th></th><th>Robotics</th><th>Art</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(35\%\)</td><td></td><td>\(100\%\)</td></tr><tr><th>Grade \(8\)</th><td></td><td>\(58\%\)</td><td>\(100\%\)</td></tr></table>

Hints

- Use the defining total for each grade row. - Find each missing category as the complement to \(100\%\).

Solution

1. Each grade row must total \(100\%\). 2. Grade \(7\), Art: \(100\%-35\%=65\%\). 3. Grade \(8\), Robotics: \(100\%-58\%=42\%\).

Answer

<table><tr><th></th><th>Robotics</th><th>Art</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(35\%\)</td><td>\(65\%\)</td><td>\(100\%\)</td></tr><tr><th>Grade \(8\)</th><td>\(42\%\)</td><td>\(58\%\)</td><td>\(100\%\)</td></tr></table>
5501238
Complete the column relative-frequency table. <table><tr><th></th><th>Online</th><th>In person</th></tr><tr><th>Completed</th><td>\(64\%\)</td><td></td></tr><tr><th>Not completed</th><td></td><td>\(45\%\)</td></tr><tr><th>Total</th><td>\(100\%\)</td><td>\(100\%\)</td></tr></table>

Hints

- Work vertically because the table uses column relative frequencies. - Each column is a complete conditional distribution.

Solution

1. In the Online column, not completed is \(100\%-64\%=36\%\). 2. In the In person column, completed is \(100\%-45\%=55\%\). 3. Each column totals \(100\%\).

Answer

<table><tr><th></th><th>Online</th><th>In person</th></tr><tr><th>Completed</th><td>\(64\%\)</td><td>\(55\%\)</td></tr><tr><th>Not completed</th><td>\(36\%\)</td><td>\(45\%\)</td></tr><tr><th>Total</th><td>\(100\%\)</td><td>\(100\%\)</td></tr></table>
5501488
The table shows travel choices for two grades. <table><tr><th></th><th>Bus</th><th>Walk</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(12\)</td><td>\(8\)</td><td>\(20\)</td></tr><tr><th>Grade \(8\)</th><td>\(18\)</td><td>\(12\)</td><td>\(30\)</td></tr><tr><th>Total</th><td>\(30\)</td><td>\(20\)</td><td>\(50\)</td></tr></table> What percent of Grade \(7\) students ride the bus?

Hints

- Use the total for the Grade \(7\) row as the denominator. - Divide the Grade \(7\) bus count by the number of Grade \(7\) students.

Solution

1. The question is about the Grade \(7\) group, whose row total is \(20\). 2. Of those students, \(12\) ride the bus. 3. The row relative frequency is \(\frac{12}{20}=60\%\).

Answer

\(60\%\).
5153498
The table shows whether \(30\) students regularly play a sport and whether they play a musical instrument. <table><tr><th></th><th>Plays an instrument</th><th>Does not play an instrument</th><th>Total</th></tr><tr><th>Plays a sport</th><td>\(6\)</td><td>\(14\)</td><td>\(20\)</td></tr><tr><th>Does not play a sport</th><td>\(4\)</td><td>\(6\)</td><td>\(10\)</td></tr><tr><th>Total</th><td>\(10\)</td><td>\(20\)</td><td>\(30\)</td></tr></table> What fraction and percent of the students participate in at least one of the two activities?

Hints

- “At least one” includes students in either activity or both activities. - Add the three qualifying interior cells. - Divide by all \(30\) students and convert the result to a percent.

Solution

1. Students in at least one activity are in the cells for both activities, sport only, and instrument only. 2. The count is \(6+14+4=24\). 3. The fraction is \(\frac{24}{30}=\frac{4}{5}\). 4. Since \(\frac{4}{5}=0.80\), the percentage is \(80\%\).

Answer

\(\frac{4}{5}=80\%\).
5386788
A two-way table classifies \(200\) transit trips by whether the trip requires a transfer and whether it arrives on time. <table><tr><th></th><th>On time</th><th>Late</th><th>Total</th></tr><tr><th>Transfer</th><td>\(72\)</td><td>\(28\)</td><td>\(100\)</td></tr><tr><th>No transfer</th><td>\(84\)</td><td>\(16\)</td><td>\(100\)</td></tr><tr><th>Total</th><td>\(156\)</td><td>\(44\)</td><td>\(200\)</td></tr></table> Find the overall percentage of trips that: a) require a transfer and arrive late; b) require no transfer and arrive on time; c) require a transfer or arrive on time, or both.

Hints

- For parts a and b, locate the single cell where both stated categories meet. - An overall percentage uses all \(200\) trips as the denominator. - For part c, include each qualifying cell once; do not count the transfer-and-on-time cell twice.

Solution

1. The cell for a transfer and a late arrival contains \(28\) trips, so the overall percentage is \(\frac{28}{200}=14\%\). 2. The cell for no transfer and an on-time arrival contains \(84\) trips, so the overall percentage is \(\frac{84}{200}=42\%\). 3. A trip that requires a transfer or arrives on time is in one of three cells: transfer and on time, transfer and late, or no transfer and on time. 4. Those cells contain \(72+28+84=184\) trips, so the percentage is \(\frac{184}{200}=92\%\).

Answer

a) \(14\%\). b) \(42\%\). c) \(92\%\).
5501188
The table shows choir participation. <table><tr><th></th><th>Participates in choir</th><th>Does not participate in choir</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(18\)</td><td>\(42\)</td><td>\(60\)</td></tr><tr><th>Grade \(8\)</th><td>\(30\)</td><td>\(30\)</td><td>\(60\)</td></tr><tr><th>Total</th><td>\(48\)</td><td>\(72\)</td><td>\(120\)</td></tr></table> Construct the row relative-frequency table. Write each entry as a percent.

Hints

- Use each grade row total as the denominator. - Check that the percentages across each row add to \(100\%\).

Solution

1. Divide each Grade \(7\) count by its row total of \(60\): \(\frac{18}{60}=30\%\) and \(\frac{42}{60}=70\%\). 2. Divide each Grade \(8\) count by its row total of \(60\): \(\frac{30}{60}=50\%\) and \(\frac{30}{60}=50\%\). 3. Each row totals \(100\%\).

Answer

<table><tr><th></th><th>Participates in choir</th><th>Does not participate in choir</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(30\%\)</td><td>\(70\%\)</td><td>\(100\%\)</td></tr><tr><th>Grade \(8\)</th><td>\(50\%\)</td><td>\(50\%\)</td><td>\(100\%\)</td></tr></table>
5501198
Use the table to construct a column relative-frequency table. <table><tr><th></th><th>Tutoring</th><th>No tutoring</th><th>Total</th></tr><tr><th>Passed</th><td>\(36\)</td><td>\(24\)</td><td>\(60\)</td></tr><tr><th>Did not pass</th><td>\(14\)</td><td>\(26\)</td><td>\(40\)</td></tr><tr><th>Total</th><td>\(50\)</td><td>\(50\)</td><td>\(100\)</td></tr></table>

Hints

- Use the total at the bottom of each column as the denominator. - Check that the percentages in each column add vertically to \(100\%\).

Solution

1. In the Tutoring column, divide by \(50\): \(\frac{36}{50}=72\%\) passed and \(\frac{14}{50}=28\%\) did not pass. 2. In the No tutoring column, divide by \(50\): \(\frac{24}{50}=48\%\) passed and \(\frac{26}{50}=52\%\) did not pass. 3. Each column totals \(100\%\).

Answer

<table><tr><th></th><th>Tutoring</th><th>No tutoring</th></tr><tr><th>Passed</th><td>\(72\%\)</td><td>\(48\%\)</td></tr><tr><th>Did not pass</th><td>\(28\%\)</td><td>\(52\%\)</td></tr><tr><th>Total</th><td>\(100\%\)</td><td>\(100\%\)</td></tr></table>
5501248
A row relative-frequency table shows that \(62.5\%\) of \(48\) Grade \(7\) students ride the bus to school. How many ride the bus, and how many use another travel method?

Hints

- Apply the row relative frequency to the size of the Grade \(7\) row. - Use the row total to find the complementary count.

Solution

1. Convert \(62.5\%\) to \(0.625\). 2. The bus-rider count is \(0.625\cdot48=30\). 3. The remaining count is \(48-30=18\).

Answer

\(30\) ride the bus, and \(18\) use another travel method.
5501298
Use the table to find both quantities: a) The overall percent of employees who use the shuttle. b) The percent of night-shift employees who use the shuttle. <table><tr><th></th><th>Uses the shuttle</th><th>Does not use the shuttle</th><th>Total</th></tr><tr><th>Day shift</th><td>\(80\)</td><td>\(120\)</td><td>\(200\)</td></tr><tr><th>Night shift</th><td>\(45\)</td><td>\(5\)</td><td>\(50\)</td></tr><tr><th>Total</th><td>\(125\)</td><td>\(125\)</td><td>\(250\)</td></tr></table>

Hints

- Identify whether each question refers to the full population or one subgroup. - Write the relevant denominator before calculating each percentage.

Solution

1. Overall shuttle use is \(\frac{125}{250}=50\%\). 2. Night-shift shuttle use is \(\frac{45}{50}=90\%\). 3. The denominators differ because part a) uses all employees, while part b) uses only night-shift employees.

Answer

a) \(50\%\) b) \(90\%\)
5501318
An overall relative-frequency table summarizes \(200\) students by grade level and lunch choice. Convert it to a count table. <table><tr><th></th><th>Cafeteria</th><th>Packed lunch</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(0.18\)</td><td>\(0.27\)</td><td>\(0.45\)</td></tr><tr><th>Grade \(8\)</th><td>\(0.32\)</td><td>\(0.23\)</td><td>\(0.55\)</td></tr><tr><th>Total</th><td>\(0.50\)</td><td>\(0.50\)</td><td>\(1.00\)</td></tr></table>

Hints

- Overall relative frequencies use the grand total as the denominator. - Multiply each displayed cell value by \(200\). - Add rows and columns after converting the four interior cells.

Solution

1. Multiply each interior overall relative frequency by the grand total of \(200\). 2. Grade \(7\) counts are \(0.18\cdot200=36\) cafeteria meals and \(0.27\cdot200=54\) packed lunches. 3. Grade \(8\) counts are \(0.32\cdot200=64\) cafeteria meals and \(0.23\cdot200=46\) packed lunches. 4. The row totals are \(90\) and \(110\); both lunch-choice columns total \(100\), and the grand total is \(200\).

Answer

<table><tr><th></th><th>Cafeteria</th><th>Packed lunch</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(36\)</td><td>\(54\)</td><td>\(90\)</td></tr><tr><th>Grade \(8\)</th><td>\(64\)</td><td>\(46\)</td><td>\(110\)</td></tr><tr><th>Total</th><td>\(100\)</td><td>\(100\)</td><td>\(200\)</td></tr></table>
5501328
The display is labeled a row relative-frequency table for workshop choices. Find the error. <table><tr><th></th><th>Robotics</th><th>Art</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(48\%\)</td><td>\(57\%\)</td><td>\(100\%\)</td></tr><tr><th>Grade \(8\)</th><td>\(35\%\)</td><td>\(65\%\)</td><td>\(100\%\)</td></tr></table>

Hints

- Use the table label to decide which direction must total \(100\%\). - Check the two grade rows separately.

Solution

1. Each grade row in a row relative-frequency table must total \(100\%\). 2. The Grade \(7\) entries total \(48\%+57\%=105\%\). 3. Therefore, at least one Grade \(7\) entry is incorrect; the Grade \(8\) row is consistent.

Answer

The Grade \(7\) row is invalid because its entries total \(105\%\), not \(100\%\).
5501398
Use the table to find the difference in on-time completion rates, in percentage points. <table><tr><th></th><th>Completed on time</th><th>Late</th><th>Total</th></tr><tr><th>Uses a planner</th><td>\(51\)</td><td>\(9\)</td><td>\(60\)</td></tr><tr><th>Does not use a planner</th><td>\(42\)</td><td>\(18\)</td><td>\(60\)</td></tr><tr><th>Total</th><td>\(93\)</td><td>\(27\)</td><td>\(120\)</td></tr></table>

Hints

- Compute each within-group rate using the same outcome. - Subtract the percentages and report the result as percentage points.

Solution

1. The planner-user on-time rate is \(\frac{51}{60}=85\%\). 2. The nonuser on-time rate is \(\frac{42}{60}=70\%\). 3. The difference is \(85\%-70\%=15\) percentage points.

Answer

\(15\) percentage points.
5501498
Use the table to answer: “What percent of bus riders are Grade \(8\) students?” <table><tr><th></th><th>Bus</th><th>Walk</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(12\)</td><td>\(8\)</td><td>\(20\)</td></tr><tr><th>Grade \(8\)</th><td>\(18\)</td><td>\(22\)</td><td>\(40\)</td></tr><tr><th>Total</th><td>\(30\)</td><td>\(30\)</td><td>\(60\)</td></tr></table> Explain why the denominator is not \(40\), the Grade \(8\) total.

Hints

- Identify the group named immediately after the word “of.” - Use the total of the Bus column, then find how many members of that column are in Grade \(8\). - Check that your fraction has the requested group in its denominator.

Solution

1. The phrase “of bus riders” restricts the group to the \(30\) students in the Bus column. 2. Of those bus riders, \(18\) are Grade \(8\) students. 3. The required percentage is \(\frac{18}{30}=60\%\). 4. The Grade \(8\) row total answers questions about all Grade \(8\) students, not questions about all bus riders.

Answer

\(60\%\). The denominator is the \(30\) bus riders because the question asks for the percent of that group who are in Grade \(8\).
5147378
A summer camp has \(150\) campers, each of whom is either a middle school student or a high school student. Of the campers, \(90\) play soccer. Middle school students make up \(40\%\) of the soccer players, and \(30\) middle school campers do not play soccer. a) Create a complete two-way table of counts for school level and soccer participation. b) What percentage of the high school campers play soccer? Round to the nearest tenth of a percent.

Hints

- Place the directly stated counts and percentage-based count in the table first. - Use row, column, and grand totals to complete the remaining counts. - For part b, divide the high school soccer count by the total number of high school campers.

Solution

1. The number of middle school soccer players is \(0.40\cdot90=36\). 2. The number of high school soccer players is \(90-36=54\). 3. The total number of middle school campers is \(36+30=66\), so the total number of high school campers is \(150-66=84\). 4. The number of high school campers who do not play soccer is \(84-54=30\). 5. The requested conditional percentage is \(\frac{54}{84}\cdot100\%\approx64.3\%\).

Answer

a) <table><tr><th></th><th>Plays soccer</th><th>Does not play soccer</th><th>Total</th></tr><tr><th>Middle school</th><td>\(36\)</td><td>\(30\)</td><td>\(66\)</td></tr><tr><th>High school</th><td>\(54\)</td><td>\(30\)</td><td>\(84\)</td></tr><tr><th>Total</th><td>\(90\)</td><td>\(60\)</td><td>\(150\)</td></tr></table> b) Approximately \(64.3\%\) of the high school campers play soccer.
5147388
An electronic component is produced by two different machines. Of the \(200\) components inspected, \(120\) were made by Machine A, and the rest were made by Machine B. Quality-control results show that \(5\%\) of the components from Machine A and \(10\%\) of the components from Machine B are defective. a) Organize the data in a complete two-way table of counts. b) What percentage of all inspected components are defective?

Hints

- First find how many inspected components came from Machine B. - Find the number of defective components produced by each machine separately. - Add the defective counts before finding the overall percentage.

Solution

1. Machine B produced \(200-120=80\) of the inspected components. 2. Machine A produced \(120\cdot0.05=6\) defective components, so it produced \(120-6=114\) nondefective components. 3. Machine B produced \(80\cdot0.10=8\) defective components, so it produced \(80-8=72\) nondefective components. 4. In all, \(6+8=14\) components are defective and \(114+72=186\) are nondefective. 5. The overall defective percentage is \(\frac{14}{200}\cdot100\%=7\%\).

Answer

a) <table><tr><th></th><th>Defective</th><th>Nondefective</th><th>Total</th></tr><tr><th>Machine A</th><td>\(6\)</td><td>\(114\)</td><td>\(120\)</td></tr><tr><th>Machine B</th><td>\(8\)</td><td>\(72\)</td><td>\(80\)</td></tr><tr><th>Total</th><td>\(14\)</td><td>\(186\)</td><td>\(200\)</td></tr></table> b) \(7\%\) of all inspected components are defective.
5147448
The incomplete table shows overall relative frequencies for students who play a sport and students who play a musical instrument. <table><tr><th></th><th>Plays an instrument</th><th>Does not play an instrument</th><th>Total</th></tr><tr><th>Plays a sport</th><td></td><td></td><td>\(0.65\)</td></tr><tr><th>Does not play a sport</th><td>\(0.15\)</td><td></td><td></td></tr><tr><th>Total</th><td>\(0.40\)</td><td></td><td>\(1.00\)</td></tr></table> a) Complete the table. b) What percent of students play neither a sport nor an instrument? c) What percent play at least one of the two?

Hints

- Every row or column of overall relative frequencies must add to its stated total. - Complete margins before using them to find interior cells. - “At least one” includes every cell except the “neither” cell.

Solution

1. The missing row total is \(1.00-0.65=0.35\), and the missing column total is \(1.00-0.40=0.60\). 2. The upper-left cell is \(0.40-0.15=0.25\). 3. The other cell in the sport row is \(0.65-0.25=0.40\). 4. The remaining interior cell is \(0.35-0.15=0.20\). 5. The relative frequency for at least one activity is \(1.00-0.20=0.80\).

Answer

a) <table><tr><th></th><th>Plays an instrument</th><th>Does not play an instrument</th><th>Total</th></tr><tr><th>Plays a sport</th><td>\(0.25\)</td><td>\(0.40\)</td><td>\(0.65\)</td></tr><tr><th>Does not play a sport</th><td>\(0.15\)</td><td>\(0.20\)</td><td>\(0.35\)</td></tr><tr><th>Total</th><td>\(0.40\)</td><td>\(0.60\)</td><td>\(1.00\)</td></tr></table> b) \(20\%\). c) \(80\%\).
5147458
A school surveyed \(250\) students about whether they like algebra and whether they like band. Of the students, \(150\) like algebra, \(100\) like band, and \(60\) like both. a) Create a complete two-way table of counts. b) Are students more likely to like exactly one of the two subjects or neither subject? Support your answer with percentages. c) What percent like at least one of the two subjects?

Hints

- “Exactly one” uses the two off-diagonal interior cells. - Find “neither” after completing the other three interior cells. - Divide each comparison count by all \(250\) students.

Solution

1. The number who like algebra but not band is \(150-60=90\). 2. The number who like band but not algebra is \(100-60=40\). 3. The number who like at least one subject is \(60+90+40=190\), so \(250-190=60\) like neither. 4. Exactly one subject is liked by \(90+40=130\) students, so the percent is \(\frac{130}{250}=0.52=52\%\). 5. The percent who like neither is \(\frac{60}{250}=0.24=24\%\). Therefore, exactly one subject is more likely. 6. The percent who like at least one subject is \(\frac{190}{250}=0.76=76\%\).

Answer

a) <table><tr><th></th><th>Likes band</th><th>Does not like band</th><th>Total</th></tr><tr><th>Likes algebra</th><td>\(60\)</td><td>\(90\)</td><td>\(150\)</td></tr><tr><th>Does not like algebra</th><td>\(40\)</td><td>\(60\)</td><td>\(100\)</td></tr><tr><th>Total</th><td>\(100\)</td><td>\(150\)</td><td>\(250\)</td></tr></table> b) Exactly one subject: \(52\%\); neither subject: \(24\%\). Exactly one is more likely. c) \(76\%\).
5147498
The table shows participation in soccer and track and field for \(120\) students. <table><tr><th></th><th>Plays soccer</th><th>Does not play soccer</th><th>Total</th></tr><tr><th>Participates in track</th><td>\(15\)</td><td>\(25\)</td><td>\(40\)</td></tr><tr><th>Does not participate in track</th><td>\(35\)</td><td>\(45\)</td><td>\(80\)</td></tr><tr><th>Total</th><td>\(50\)</td><td>\(70\)</td><td>\(120\)</td></tr></table> a) Convert the four interior counts to overall relative frequencies as percentages. b) What percent of track participants also play soccer? c) What percent of students who do not participate in track play soccer? d) Based on the two percentages, does soccer participation appear to be associated with track participation? Explain.

Hints

- For overall relative frequencies, divide each cell by all \(120\) students. - For each “among” question, use the named row total as the denominator. - Compare the two within-row percentages rather than the raw soccer counts.

Solution

1. Divide each interior count by all \(120\) students. The overall relative frequencies are \(\frac{15}{120}=12.5\%\), \(\frac{25}{120}\approx20.8\%\), \(\frac{35}{120}\approx29.2\%\), and \(\frac{45}{120}=37.5\%\). 2. Among track participants, the soccer rate is \(\frac{15}{40}=37.5\%\). 3. Among students who do not participate in track, the soccer rate is \(\frac{35}{80}=43.75\%\approx43.8\%\). 4. The rates differ by about \(6.3\) percentage points. This suggests a small association in this sample, with soccer participation slightly less common among track participants.

Answer

a) <table><tr><th></th><th>Plays soccer</th><th>Does not play soccer</th></tr><tr><th>Participates in track</th><td>\(12.5\%\)</td><td>\(20.8\%\)</td></tr><tr><th>Does not participate in track</th><td>\(29.2\%\)</td><td>\(37.5\%\)</td></tr></table> b) \(37.5\%\). c) Approximately \(43.8\%\). d) The sample suggests a small association because the two soccer rates are not equal; the rate is slightly lower among track participants.
5147768
A survey of \(120\) students asked whether they play a musical instrument. Of the \(64\) eighth-grade students, \(28\) play an instrument. Altogether, \(40\) students play an instrument. a) Create a complete two-way table of counts for grade level and whether a student plays an instrument. b) How many ninth-grade students do not play an instrument? c) What percent of the ninth-grade students play an instrument? Round to the nearest tenth of a percent.

Hints

- First find the number of ninth-grade students. - Use the column total for students who play an instrument. - For part c, use the ninth-grade total as the denominator.

Solution

1. The number of ninth-grade students is \(120-64=56\). 2. The number of eighth-grade students who do not play an instrument is \(64-28=36\). 3. The number of ninth-grade students who play an instrument is \(40-28=12\). 4. The number of ninth-grade students who do not play an instrument is \(56-12=44\). 5. The percentage of ninth-grade students who play an instrument is \(\frac{12}{56}\cdot100\%\approx21.4\%\).

Answer

a) <table><tr><th></th><th>Plays an instrument</th><th>Does not play an instrument</th><th>Total</th></tr><tr><th>Eighth grade</th><td>\(28\)</td><td>\(36\)</td><td>\(64\)</td></tr><tr><th>Ninth grade</th><td>\(12\)</td><td>\(44\)</td><td>\(56\)</td></tr><tr><th>Total</th><td>\(40\)</td><td>\(80\)</td><td>\(120\)</td></tr></table> b) \(44\) ninth-grade students do not play an instrument. c) Approximately \(21.4\%\) of the ninth-grade students play an instrument.
5147778
A survey of \(500\) students asked about their reading habits. Ninth-grade students make up \(45\%\) of those surveyed. Of the eighth-grade students, \(60\%\) regularly read books. Among the ninth-grade students, \(120\) do not regularly read books. a) Create a complete two-way table of counts. b) What percent of all surveyed students regularly read books?

Hints

- Convert each percentage to a count before completing the table. - Pay attention to whether a percentage refers to all students or to one grade level. - Add the regular-reader counts from both grade levels.

Solution

1. The number of ninth-grade students is \(0.45\cdot500=225\), so the number of eighth-grade students is \(500-225=275\). 2. The number of eighth-grade students who regularly read books is \(0.60\cdot275=165\), so \(275-165=110\) do not. 3. Of the \(225\) ninth-grade students, \(225-120=105\) regularly read books. 4. In all, \(165+105=270\) students regularly read books. 5. The overall percentage is \(\frac{270}{500}\cdot100\%=54\%\).

Answer

a) <table><tr><th></th><th>Regularly reads books</th><th>Does not regularly read books</th><th>Total</th></tr><tr><th>Eighth grade</th><td>\(165\)</td><td>\(110\)</td><td>\(275\)</td></tr><tr><th>Ninth grade</th><td>\(105\)</td><td>\(120\)</td><td>\(225\)</td></tr><tr><th>Total</th><td>\(270\)</td><td>\(230\)</td><td>\(500\)</td></tr></table> b) \(54\%\) of all surveyed students regularly read books.
5147788
A company with \(800\) employees studies participation in its fitness program by work shift. Night-shift employees make up \(25\%\) of the workforce. Of the night-shift employees, \(20\%\) participate in the fitness program. Among the day-shift employees, \(320\) do not participate. a) Complete a two-way table of counts. b) Compare the fitness-program participation rates for night-shift and day-shift employees. Which group has the higher rate? Support your answer with calculations.

Hints

- First divide the workforce into night-shift and day-shift employees. - Complete each row using its row total. - Compare within-group percentages, not just the two participant counts.

Solution

1. The number of night-shift employees is \(0.25\cdot800=200\), so the number of day-shift employees is \(800-200=600\). 2. The number of night-shift participants is \(0.20\cdot200=40\), so \(200-40=160\) night-shift employees do not participate. 3. Since \(320\) day-shift employees do not participate, \(600-320=280\) do participate. 4. The night-shift participation rate is \(20\%\). 5. The day-shift participation rate is \(\frac{280}{600}\cdot100\%\approx46.7\%\). 6. The day-shift participation rate is higher.

Answer

a) <table><tr><th></th><th>Participates</th><th>Does not participate</th><th>Total</th></tr><tr><th>Night shift</th><td>\(40\)</td><td>\(160\)</td><td>\(200\)</td></tr><tr><th>Day shift</th><td>\(280\)</td><td>\(320\)</td><td>\(600\)</td></tr><tr><th>Total</th><td>\(320\)</td><td>\(480\)</td><td>\(800\)</td></tr></table> b) The night-shift participation rate is \(20\%\), and the day-shift rate is approximately \(46.7\%\). The day-shift rate is higher.
5153598
A company inspects \(500\) components for a material defect and a surface defect. Of the components, \(25\) have a material defect, \(15\) have a surface defect, and \(5\) have both defects. a) Create a complete two-way table of overall relative frequencies. b) What percent of components have exactly one defect? c) What percent have neither defect?

Hints

- Complete the count table before dividing each entry by \(500\). - “Exactly one” uses the two single-defect cells. - “Neither” is the cell with no defect of either type.

Solution

1. The count with a material defect but no surface defect is \(25-5=20\). 2. The count with a surface defect but no material defect is \(15-5=10\). 3. The count with neither defect is \(500-(5+20+10)=465\). 4. Divide each count by \(500\): \(\frac{5}{500}=0.01\), \(\frac{20}{500}=0.04\), \(\frac{10}{500}=0.02\), and \(\frac{465}{500}=0.93\). 5. Exactly one defect has relative frequency \(0.04+0.02=0.06=6\%\). 6. Neither defect has relative frequency \(0.93=93\%\).

Answer

a) <table><tr><th></th><th>Surface defect</th><th>No surface defect</th><th>Total</th></tr><tr><th>Material defect</th><td>\(0.01\)</td><td>\(0.04\)</td><td>\(0.05\)</td></tr><tr><th>No material defect</th><td>\(0.02\)</td><td>\(0.93\)</td><td>\(0.95\)</td></tr><tr><th>Total</th><td>\(0.03\)</td><td>\(0.97\)</td><td>\(1.00\)</td></tr></table> b) \(6\%\). c) \(93\%\).
5153608
A school cafeteria surveyed \(200\) students about food preferences. Of the students, \(80\) prefer a vegetarian option, \(150\) like pizza, and \(190\) are in at least one of those two groups. a) How many students both prefer a vegetarian option and like pizza? b) Create a complete two-way table of counts. c) Xavier claims, “A majority of students who prefer a vegetarian option like pizza.” Determine whether the claim is correct and justify your answer.

Hints

- The two category totals double-count students who are in both groups. - Complete the other cells by subtracting from category totals and the grand total. - For the claim, use only students who prefer a vegetarian option as the denominator.

Solution

1. Adding the two category totals counts students in both groups twice. The overlap is \(80+150-190=40\). 2. The number who prefer a vegetarian option but do not like pizza is \(80-40=40\). 3. The number who like pizza but do not prefer a vegetarian option is \(150-40=110\). 4. The number in neither group is \(200-190=10\). 5. Among the \(80\) students who prefer a vegetarian option, \(40\) like pizza. The proportion is \(\frac{40}{80}=0.50=50\%\). 6. A majority must be more than \(50\%\), so the claim is not correct.

Answer

a) \(40\) students. b) <table><tr><th></th><th>Likes pizza</th><th>Does not like pizza</th><th>Total</th></tr><tr><th>Prefers a vegetarian option</th><td>\(40\)</td><td>\(40\)</td><td>\(80\)</td></tr><tr><th>Does not prefer a vegetarian option</th><td>\(110\)</td><td>\(10\)</td><td>\(120\)</td></tr><tr><th>Total</th><td>\(150\)</td><td>\(50\)</td><td>\(200\)</td></tr></table> c) The claim is not correct. Exactly \(50\%\) of the students who prefer a vegetarian option like pizza, and a majority must be greater than \(50\%\).
5155728
A school surveyed \(200\) students in Grades 9 and 10 about their cafeteria meal choice. There are \(90\) Grade 9 students and \(110\) Grade 10 students. Of the Grade 9 students, \(60\%\) chose the vegetarian meal. Of the Grade 10 students, \(40\%\) chose the vegetarian meal. a) Create a complete two-way table of counts. b) What percent of all surveyed students are Grade 9 students who chose the vegetarian meal? What percent are Grade 10 students who chose the vegetarian meal? c) What percent of all surveyed students chose the vegetarian meal?

Hints

- Convert each within-grade percentage to a count before completing the table. - In parts b and c, the phrase “of all surveyed students” makes \(200\) the denominator. - Add the two vegetarian-meal counts before finding the overall percentage.

Solution

1. The number of Grade 9 students who chose the vegetarian meal is \(0.60\cdot90=54\), so \(90-54=36\) chose another meal. 2. The number of Grade 10 students who chose the vegetarian meal is \(0.40\cdot110=44\), so \(110-44=66\) chose another meal. 3. In all, \(54+44=98\) students chose the vegetarian meal, and \(36+66=102\) chose another meal. 4. As percentages of all \(200\) surveyed students, the two vegetarian-meal cells are \(\frac{54}{200}=27\%\) and \(\frac{44}{200}=22\%\). 5. The overall vegetarian-meal percentage is \(\frac{98}{200}=49\%\).

Answer

a) <table><tr><th></th><th>Vegetarian meal</th><th>Other meal</th><th>Total</th></tr><tr><th>Grade 9</th><td>\(54\)</td><td>\(36\)</td><td>\(90\)</td></tr><tr><th>Grade 10</th><td>\(44\)</td><td>\(66\)</td><td>\(110\)</td></tr><tr><th>Total</th><td>\(98\)</td><td>\(102\)</td><td>\(200\)</td></tr></table> b) Grade 9 and vegetarian: \(27\%\); Grade 10 and vegetarian: \(22\%\). c) \(49\%\).
5501258
The table compares schools of different sizes. <table><tr><th></th><th>Offers robotics</th><th>Does not offer robotics</th><th>Total</th></tr><tr><th>Small school</th><td>\(18\)</td><td>\(12\)</td><td>\(30\)</td></tr><tr><th>Large school</th><td>\(72\)</td><td>\(48\)</td><td>\(120\)</td></tr><tr><th>Total</th><td>\(90\)</td><td>\(60\)</td><td>\(150\)</td></tr></table> Lila says the large-school group is more likely to offer robotics because \(72>18\). Evaluate the claim using relative frequencies.

Hints

- Compare proportions within each school-size group rather than the numerators alone. - Different row totals can produce very different raw counts for the same rate.

Solution

1. The small-school robotics-offering rate is \(\frac{18}{30}=60\%\). 2. The large-school robotics-offering rate is \(\frac{72}{120}=60\%\). 3. The raw counts differ because the group sizes differ; the conditional rates are equal.

Answer

The claim is incorrect. Both groups have a \(60\%\) robotics-offering rate.
5501268
The table shows transit delays. <table><tr><th></th><th>Delayed</th><th>On time</th><th>Total</th></tr><tr><th>Route A</th><td>\(18\)</td><td>\(42\)</td><td>\(60\)</td></tr><tr><th>Route B</th><td>\(24\)</td><td>\(96\)</td><td>\(120\)</td></tr><tr><th>Total</th><td>\(42\)</td><td>\(138\)</td><td>\(180\)</td></tr></table> Which route has the higher delay rate, even though it has fewer delayed trips?

Hints

- Use each route's total number of trips as the denominator. - A smaller count can represent a larger proportion in a smaller group.

Solution

1. Route A's delay rate is \(\frac{18}{60}=30\%\). 2. Route B's delay rate is \(\frac{24}{120}=20\%\). 3. Route A has fewer delayed trips but the higher delay rate.

Answer

Route A, with a \(30\%\) delay rate compared with Route B's \(20\%\).
5501278
The two pie charts show course choice within two session groups. Panel a) represents morning participants, and panel b) represents evening participants. a) Which session group has the higher relative frequency of choosing the advanced course, and by how many percentage points? b) Describe the possible association between session time and course choice shown by the charts. Do not claim causation. c) Explain why the two session groups do not need to contain the same number of people for this comparison to be valid.
Figure for problem 550127

Hints

- Compare the Advanced slice in panel a) with the Advanced slice in panel b). - A percentage-point difference compares two percentages by subtraction. - Each pie chart uses its own session group as the whole, so focus on proportions rather than numbers of people.

Solution

1. The advanced-course relative frequency is \(60\%\) for morning participants and \(35\%\) for evening participants. 2. The difference is \(60\%-35\%=25\) percentage points, so the morning group has the higher advanced-course rate. 3. The different within-group percentages suggest an association between session time and course choice in these data, but they do not establish causation. 4. Each pie chart represents \(100\%\) of its own session group. Relative frequencies therefore compare proportions within the groups rather than raw counts, so equal group sizes are unnecessary.

Answer

a) Morning participants, by \(25\) percentage points. b) Morning participants choose the advanced course at a higher relative frequency than evening participants, suggesting a possible association between session time and course choice in these data. c) Each chart shows within-group relative frequencies that sum to \(100\%\), so the comparison uses proportions rather than raw group sizes.
5501288
The table summarizes one week of school travel. Suppose the same pattern is repeated for three identical weeks, so every count is multiplied by \(3\). <table><tr><th></th><th>Bus</th><th>Walk</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(12\)</td><td>\(8\)</td><td>\(20\)</td></tr><tr><th>Grade \(8\)</th><td>\(21\)</td><td>\(9\)</td><td>\(30\)</td></tr><tr><th>Total</th><td>\(33\)</td><td>\(17\)</td><td>\(50\)</td></tr></table> What happens to the row relative frequencies? Explain.

Hints

- Compare a cell-to-row-total fraction before and after scaling. - The same factor appears in both numerator and denominator.

Solution

1. The Grade \(7\) bus rate is \(\frac{12}{20}=60\%\); after scaling, it is \(\frac{36}{60}=60\%\). 2. The Grade \(8\) bus rate is \(\frac{21}{30}=70\%\); after scaling, it is \(\frac{63}{90}=70\%\). 3. Multiplying every cell and its row total by the same factor leaves every row relative frequency unchanged.

Answer

All row relative frequencies remain unchanged.
5501308
The table shows recycling behavior in three neighborhoods. <table><tr><th></th><th>Recycled</th><th>Did not recycle</th><th>Total</th></tr><tr><th>North</th><td>\(45\)</td><td>\(15\)</td><td>\(60\)</td></tr><tr><th>Central</th><td>\(32\)</td><td>\(8\)</td><td>\(40\)</td></tr><tr><th>South</th><td>\(28\)</td><td>\(12\)</td><td>\(40\)</td></tr><tr><th>Total</th><td>\(105\)</td><td>\(35\)</td><td>\(140\)</td></tr></table> Rank the neighborhoods from highest to lowest recycling relative frequency.

Hints

- Calculate a within-neighborhood percentage for each row. - Rank the rates rather than the recycling counts.

Solution

1. North: \(\frac{45}{60}=75\%\). 2. Central: \(\frac{32}{40}=80\%\). 3. South: \(\frac{28}{40}=70\%\). 4. The ranking from highest to lowest is Central, North, South.

Answer

Central \((80\%)\), North \((75\%)\), South \((70\%)\).
5501358
In Grade \(8\), \(24\) students chose the robotics workshop. The row relative frequency for robotics is \(60\%\). How many Grade \(8\) students were surveyed, and how many chose the art workshop?

Hints

- Treat the known count as \(60\%\) of the unknown grade total. - After finding the row total, subtract the robotics count to find the complement.

Solution

1. Let the Grade \(8\) row total be \(n\). Then \(\frac{24}{n}=0.60\). 2. Solving gives \(n=24\div0.60=40\). 3. The art-workshop count is \(40-24=16\).

Answer

Grade \(8\) has \(40\) surveyed students, and \(16\) chose the art workshop.
5501368
Which table supports the claim “Grade \(7\) students are more likely than Grade \(8\) students to choose robotics”? Justify your choice by comparing the robotics row relative frequencies. a) <table><tr><th></th><th>Robotics</th><th>Art</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(18\)</td><td>\(12\)</td><td>\(30\)</td></tr><tr><th>Grade \(8\)</th><td>\(24\)</td><td>\(36\)</td><td>\(60\)</td></tr><tr><th>Total</th><td>\(42\)</td><td>\(48\)</td><td>\(90\)</td></tr></table> b) <table><tr><th></th><th>Robotics</th><th>Art</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(21\)</td><td>\(9\)</td><td>\(30\)</td></tr><tr><th>Grade \(8\)</th><td>\(50\)</td><td>\(10\)</td><td>\(60\)</td></tr><tr><th>Total</th><td>\(71\)</td><td>\(19\)</td><td>\(90\)</td></tr></table>

Hints

- “More likely” requires comparing proportions within the two grade groups. - Use each grade total as the denominator for that grade's robotics relative frequency. - Do not decide from the robotics counts alone when the grade totals are different.

Solution

1. In table a), the Grade \(7\) robotics relative frequency is \(\frac{18}{30}=60\%\). 2. In table a), the Grade \(8\) robotics relative frequency is \(\frac{24}{60}=40\%\), so Grade \(7\) has the higher robotics rate. 3. In table b), the Grade \(7\) robotics relative frequency is \(\frac{21}{30}=70\%\). 4. In table b), the Grade \(8\) robotics relative frequency is \(\frac{50}{60}\approx83.3\%\), so Grade \(7\) does not have the higher robotics rate. 5. Therefore, table a) supports the claim. Notice that its Grade \(7\) robotics count is smaller than its Grade \(8\) robotics count, so raw counts alone would give the wrong conclusion.

Answer

Table a). The robotics row relative frequencies are \(60\%\) for Grade \(7\) and \(40\%\) for Grade \(8\). In table b), they are \(70\%\) and approximately \(83.3\%\), respectively.
5501378
Which table shows the stronger association between grade level and workshop choice? For each table, compare the Grade \(7\) and Grade \(8\) robotics row relative frequencies and use the percentage-point gap to justify your answer. a) <table><tr><th></th><th>Robotics</th><th>Art</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(24\)</td><td>\(16\)</td><td>\(40\)</td></tr><tr><th>Grade \(8\)</th><td>\(40\)</td><td>\(40\)</td><td>\(80\)</td></tr><tr><th>Total</th><td>\(64\)</td><td>\(56\)</td><td>\(120\)</td></tr></table> b) <table><tr><th></th><th>Robotics</th><th>Art</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(28\)</td><td>\(12\)</td><td>\(40\)</td></tr><tr><th>Grade \(8\)</th><td>\(32\)</td><td>\(48\)</td><td>\(80\)</td></tr><tr><th>Total</th><td>\(60\)</td><td>\(60\)</td><td>\(120\)</td></tr></table>

Hints

- Association strength here depends on how different the conditional distributions are. - Convert the robotics count in each grade row to a percentage of that row total. - Compare percentage-point gaps only after using the correct denominator for each grade.

Solution

1. In table a), the robotics row relative frequencies are \(\frac{24}{40}=60\%\) for Grade \(7\) and \(\frac{40}{80}=50\%\) for Grade \(8\). 2. The percentage-point gap in table a) is \(60\%-50\%=10\) percentage points. 3. In table b), the robotics row relative frequencies are \(\frac{28}{40}=70\%\) for Grade \(7\) and \(\frac{32}{80}=40\%\) for Grade \(8\). 4. The percentage-point gap in table b) is \(70\%-40\%=30\) percentage points. 5. Table b) shows the stronger association because its conditional distributions differ more. The raw robotics-count gap is actually larger in table a), so comparing counts would give the wrong ranking.

Answer

Table b). Table a) has robotics rates of \(60\%\) and \(50\%\), a \(10\)-percentage-point gap. Table b) has robotics rates of \(70\%\) and \(40\%\), a \(30\)-percentage-point gap.
5501388
A company has \(80\) day-shift employees, and \(55\%\) of them use the shuttle. The night shift has \(40\) employees. Altogether, \(70\) employees use the shuttle. Find the shuttle-use relative frequency for night-shift employees.

Hints

- Convert the day-shift percentage into a count. - Use the overall shuttle-user total to find the night-shift count. - Divide by the night-shift total to obtain the conditional relative frequency.

Solution

1. Day-shift shuttle users: \(0.55\cdot80=44\). 2. Night-shift shuttle users: \(70-44=26\). 3. The night-shift shuttle-use rate is \(\frac{26}{40}=65\%\).

Answer

\(65\%\).
5501428
A school wants to know whether use of a review guide is associated with meeting a standard. Write a two-sentence conclusion supported by row relative frequencies. Do not claim causation. <table><tr><th></th><th>Met standard</th><th>Did not meet</th><th>Total</th></tr><tr><th>Uses the review guide</th><td>\(51\)</td><td>\(24\)</td><td>\(75\)</td></tr><tr><th>Does not use the review guide</th><td>\(36\)</td><td>\(44\)</td><td>\(80\)</td></tr><tr><th>Total</th><td>\(87\)</td><td>\(68\)</td><td>\(155\)</td></tr></table>

Hints

- Use each group’s own row total as the denominator. - Compare the two within-group percentages rather than the raw counts. - Use association language rather than language claiming that the guide caused the outcome.

Solution

1. Among guide users, the row relative frequency for meeting the standard is \(\frac{51}{75}=0.68=68\%\). 2. Among nonusers, the corresponding row relative frequency is \(\frac{36}{80}=0.45=45\%\). 3. The \(23\)-percentage-point difference suggests an association, but the table does not establish that guide use caused the difference.

Answer

Students who used the review guide met the standard at a rate of \(68\%\), compared with \(45\%\) for students who did not use it. This difference suggests an association between guide use and meeting the standard, but it does not establish causation.
5513838
A community survey classifies \(150\) adults by work arrangement and whether they bike to work at least once per week. <table><tr><th></th><th>Hybrid work</th><th>On-site work</th><th>Total</th></tr><tr><th>Bikes to work</th><td>\(24\)</td><td>\(18\)</td><td>\(42\)</td></tr><tr><th>Does not bike to work</th><td>\(36\)</td><td>\(72\)</td><td>\(108\)</td></tr><tr><th>Total</th><td>\(60\)</td><td>\(90\)</td><td>\(150\)</td></tr></table> a) Construct a column relative-frequency table. Write each entry as a percent. b) Use the column relative frequencies to describe a possible association between work arrangement and biking to work. Do not claim causation.

Hints

- Because the task asks for column relative frequencies, use each work-arrangement column total as its denominator. - Compare the same biking category across the two columns after converting to percentages. - Describe what the different conditional rates suggest, but do not say one category caused the other.

Solution

1. For the Hybrid work column, divide by \(60\): \(\frac{24}{60}=40\%\) bike to work and \(\frac{36}{60}=60\%\) do not. 2. For the On-site work column, divide by \(90\): \(\frac{18}{90}=20\%\) bike to work and \(\frac{72}{90}=80\%\) do not. 3. The biking rate is higher among hybrid workers, \(40\%\) compared with \(20\%\) among on-site workers. 4. This difference suggests an association between work arrangement and biking to work in this survey, but it does not establish causation.

Answer

a) <table><tr><th></th><th>Hybrid work</th><th>On-site work</th></tr><tr><th>Bikes to work</th><td>\(40\%\)</td><td>\(20\%\)</td></tr><tr><th>Does not bike to work</th><td>\(60\%\)</td><td>\(80\%\)</td></tr><tr><th>Total</th><td>\(100\%\)</td><td>\(100\%\)</td></tr></table> b) Hybrid workers bike to work at a higher rate, \(40\%\) versus \(20\%\) for on-site workers. This suggests a possible association in this survey, not a causal relationship.
5501338
The displayed table has grade level as rows and travel method as columns. <table><tr><th></th><th>Bus</th><th>Walk</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(30\)</td><td>\(20\)</td><td>\(50\)</td></tr><tr><th>Grade \(8\)</th><td>\(10\)</td><td>\(40\)</td><td>\(50\)</td></tr><tr><th>Total</th><td>\(40\)</td><td>\(60\)</td><td>\(100\)</td></tr></table> Suppose the table is transposed so Bus and Walk become rows and the grades become columns. Will the original row relative frequencies become the new row relative frequencies? Explain.

Hints

- Track which category totals serve as denominators before and after transposing. - The cell counts stay the same, but the conditioning direction changes.

Solution

1. The original row relative frequencies divide by the grade totals of \(50\). 2. After transposing, the new row totals are the travel totals of \(40\) and \(60\). 3. Because the conditioning totals change, the original row relative frequencies become column relative frequencies in the transposed table, not row relative frequencies.

Answer

No. Transposing swaps the roles of row and column relative frequencies.
5501348
A small training session has \(20\) participants and a \(70\%\) success rate. A large session has \(80\) participants and a \(40\%\) success rate. Find the overall success relative frequency. Explain why it is not the simple average of \(70\%\) and \(40\%\).

Hints

- Convert each session's rate to a success count before combining sessions. - Add the success counts and participant totals separately. - The larger session must contribute more weight to the overall rate.

Solution

1. Small-session successes: \(0.70\cdot20=14\). 2. Large-session successes: \(0.40\cdot80=32\). 3. Altogether, \(14+32=46\) of \(100\) participants succeed, so the overall rate is \(46\%\). 4. The simple average is inappropriate because it gives equal weight to sessions with different numbers of participants.

Answer

\(46\%\). It is not \(55\%\) because the two sessions have different sizes.
5501408
The table currently shows different robotics-choice rates. <table><tr><th></th><th>Robotics</th><th>Art</th><th>Total</th></tr><tr><th>Grade \(7\)</th><td>\(18\)</td><td>\(12\)</td><td>\(30\)</td></tr><tr><th>Grade \(8\)</th><td>\(20\)</td><td>\(30\)</td><td>\(50\)</td></tr><tr><th>Total</th><td>\(38\)</td><td>\(42\)</td><td>\(80\)</td></tr></table> How many Grade \(8\) responses would need to change from Art to Robotics for Grade \(8\) to have the same robotics rate as Grade \(7\)?

Hints

- Find the target conditional rate from the Grade \(7\) row. - Apply that rate to the unchanged Grade \(8\) row total. - Compare the target robotics count with the current count.

Solution

1. The Grade \(7\) robotics rate is \(\frac{18}{30}=60\%\). 2. Sixty percent of the \(50\) Grade \(8\) responses is \(0.60\cdot50=30\) robotics choices. 3. Grade \(8\) currently has \(20\) robotics choices, so \(30-20=10\) responses must change.

Answer

\(10\) Grade \(8\) responses.
5501418
Two programs report success separately for beginners and experienced participants. Beginners: <table><tr><th></th><th>Success</th><th>Not successful</th><th>Total</th></tr><tr><th>Program A</th><td>\(9\)</td><td>\(1\)</td><td>\(10\)</td></tr><tr><th>Program B</th><td>\(40\)</td><td>\(10\)</td><td>\(50\)</td></tr></table> Experienced participants: <table><tr><th></th><th>Success</th><th>Not successful</th><th>Total</th></tr><tr><th>Program A</th><td>\(24\)</td><td>\(36\)</td><td>\(60\)</td></tr><tr><th>Program B</th><td>\(6\)</td><td>\(14\)</td><td>\(20\)</td></tr></table> a) Which program has the higher success rate within each experience level? b) Combine the two levels. Which program has the higher overall success rate? c) Explain how the overall comparison can reverse.

Hints

- Compute success rates separately within the beginner and experienced tables. - Add each program's successes and totals across the two tables for the overall rates. - Compare how each program's participants are distributed between the higher- and lower-success subgroups.

Solution

1. Among beginners, Program A's success rate is \(\frac{9}{10}=90\%\). 2. Among beginners, Program B's success rate is \(\frac{40}{50}=80\%\), so Program A is higher. 3. Among experienced participants, Program A's success rate is \(\frac{24}{60}=40\%\). 4. Among experienced participants, Program B's success rate is \(\frac{6}{20}=30\%\), so Program A is again higher. 5. Overall, Program A has \(9+24=33\) successes among \(10+60=70\) participants, so its rate is \(\frac{33}{70}\approx47.1\%\). 6. Overall, Program B has \(40+6=46\) successes among \(50+20=70\) participants, so its rate is \(\frac{46}{70}\approx65.7\%\). 7. Program B has many more beginners, the subgroup with higher success rates in both programs. The different subgroup mixes reverse the overall comparison.

Answer

a) Program A in both experience levels: \(90\%\) versus \(80\%\) for beginners, and \(40\%\) versus \(30\%\) for experienced participants. b) Program B overall: about \(65.7\%\) versus about \(47.1\%\) for Program A. c) Program B contains a much larger proportion of beginners, the higher-success subgroup, so combining the differently mixed groups reverses the comparison.

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