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Tables for one categorical variable

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53964212
A survey of \(96\) students gave the table below. <table><thead><tr><th>Response</th><th>Frequency</th></tr></thead><tbody><tr><td>Strongly agree</td><td>\(31\)</td></tr><tr><td>Agree</td><td>\(24\)</td></tr><tr><td>Neutral</td><td>\(19\)</td></tr><tr><td>Disagree</td><td>?</td></tr></tbody></table> Find the missing frequency.

Hints

- Use the fact that every observational unit belongs to exactly one listed category. - Subtract the sum of the known category counts from the stated sample size.

Solution

1. The known frequencies sum to \(74\). 2. The missing frequency is \(96-74=22\).

Answer

The missing frequency is \(22\).
53965212
Two relative frequency tables are proposed for the same categorical variable. <table><thead><tr><th>Category</th><th>Table A</th><th>Table B</th></tr></thead><tbody><tr><td>Alpha</td><td>\(0.18\)</td><td>\(0.18\)</td></tr><tr><td>Beta</td><td>\(0.27\)</td><td>\(0.27\)</td></tr><tr><td>Gamma</td><td>\(0.31\)</td><td>\(0.36\)</td></tr><tr><td>Delta</td><td>\(0.24\)</td><td>\(0.24\)</td></tr></tbody></table> Which table could represent a complete distribution, and why?

Hints

- Check the required total for all categories together. - A complete relative-frequency column must total \(1\), allowing only small rounding error.

Solution

1. The values in Table A sum to \(1.00\). 2. The values in Table B sum to \(1.05\). 3. A complete relative frequency distribution must sum to \(1\), so only Table A is valid.

Answer

Table A, because its relative frequencies sum to \(1\).
53965512
In a categorical frequency table, one category has relative frequency \(0.125\). The total sample size is \(64\). Verify that the corresponding frequency is an integer and find it.

Hints

- Relate the category’s share to the full sample size. - Check that the resulting category frequency is a whole number.

Solution

1. Multiply the relative frequency by the total: \(0.125\cdot 64=8\). 2. The result is an integer, as required for a frequency count.

Answer

The frequency is \(8\).
54866212
A frequency table lists Category A \(40\), Category B \(40\), and Category C \(20\). Farzana reports A as the mode because it appears first in the table. Evaluate Farzana's report and state how the distribution's mode should be described.

Hints

- Compare the largest frequencies rather than the order of the rows. - Check whether the maximum frequency occurs once or more than once. - Use language that preserves a tie instead of selecting one category arbitrarily.

Solution

1. A mode is any category with the greatest frequency. 2. Categories A and B both have the greatest frequency, \(40\). 3. Table order does not break a frequency tie, so the distribution has two modal categories.

Answer

Farzana's report is incorrect. Categories A and B are tied as modes; the distribution is bimodal with respect to category frequency.
55620912
A library surveyed \(20\) teens about the workshop they most wanted next month. The responses were: Coding, Art, Writing, Coding, Robotics, Art, Coding, Writing, Art, Robotics, Coding, Writing, Coding, Art, Robotics, Writing, Coding, Art, Robotics, Writing. Construct a frequency and relative-frequency table with the categories in alphabetical order. Report each relative frequency as a percentage.

Hints

- Tally one category at a time so every response is counted exactly once. - Use the total number of responses as the denominator for every relative frequency. - Check that the final frequencies and percentages have the correct totals.

Solution

1. Count each category: Art \(5\), Coding \(6\), Robotics \(4\), and Writing \(5\). 2. Divide each frequency by \(20\): Art \(25\%\), Coding \(30\%\), Robotics \(20\%\), and Writing \(25\%\). 3. The frequencies total \(20\), and the relative frequencies total \(100\%\), which checks the table.

Answer

<table><thead><tr><th>Workshop</th><th>Frequency</th><th>Relative frequency</th></tr></thead><tbody><tr><td>Art</td><td>\(5\)</td><td>\(25\%\)</td></tr><tr><td>Coding</td><td>\(6\)</td><td>\(30\%\)</td></tr><tr><td>Robotics</td><td>\(4\)</td><td>\(20\%\)</td></tr><tr><td>Writing</td><td>\(5\)</td><td>\(25\%\)</td></tr></tbody></table>
53964012
The preferred music genres of \(12\) students are: Jazz, Rock, Jazz, Classical, Rock, Jazz, Hip-hop, Rock, Classical, Jazz, Hip-hop, Jazz. Determine the frequency and relative frequency for each genre, and present the results in a table. Round relative frequencies to three decimals.

Hints

- Organize identical responses before counting. - Check that the frequencies total the sample size and the relative frequencies total approximately one.

Solution

1. Count each category. 2. Divide each count by \(12\). 3. The table is <table><thead><tr><th>Genre</th><th>Frequency</th><th>Relative frequency</th></tr></thead><tbody><tr><td>Classical</td><td>\(2\)</td><td>\(0.167\)</td></tr><tr><td>Hip-hop</td><td>\(2\)</td><td>\(0.167\)</td></tr><tr><td>Jazz</td><td>\(5\)</td><td>\(0.417\)</td></tr><tr><td>Rock</td><td>\(3\)</td><td>\(0.250\)</td></tr></tbody></table>

Answer

The completed table is <table><thead><tr><th>Genre</th><th>Frequency</th><th>Relative frequency</th></tr></thead><tbody><tr><td>Classical</td><td>\(2\)</td><td>\(0.167\)</td></tr><tr><td>Hip-hop</td><td>\(2\)</td><td>\(0.167\)</td></tr><tr><td>Jazz</td><td>\(5\)</td><td>\(0.417\)</td></tr><tr><td>Rock</td><td>\(3\)</td><td>\(0.250\)</td></tr></tbody></table>
53964112
A student activities survey produced this frequency table. <table><thead><tr><th>Club</th><th>Frequency</th></tr></thead><tbody><tr><td>Ceramics</td><td>\(18\)</td></tr><tr><td>Photography</td><td>\(27\)</td></tr><tr><td>Robotics</td><td>\(21\)</td></tr><tr><td>Theater</td><td>\(14\)</td></tr></tbody></table> What proportion and percentage of respondents selected Photography?

Hints

- Identify the whole represented by the table before comparing one category with it. - Express the Photography count relative to that whole, then state the same comparison as a percent.

Solution

1. The total is \(80\). 2. The proportion is \(\frac{27}{80}=0.3375\). 3. The percentage is \(33.75\%\).

Answer

The proportion is \(0.3375\), or \(33.75\%\).
53964312
The relative frequency table describes the primary way students travel to school. <table><thead><tr><th>Method</th><th>Relative frequency</th></tr></thead><tbody><tr><td>Walk</td><td>\(0.18\)</td></tr><tr><td>Bike</td><td>\(0.12\)</td></tr><tr><td>Bus</td><td>\(0.46\)</td></tr><tr><td>Car</td><td>?</td></tr></tbody></table> Find the missing relative frequency and interpret it in context.

Hints

- Use the required total for a complete relative frequency distribution. - Subtract the listed shares from the whole, then express the remaining share in context.

Solution

1. Relative frequencies total \(1\). 2. The missing value is \(1-0.18-0.12-0.46=0.24\). 3. Thus, \(24\%\) of the students primarily travel by car.

Answer

The missing relative frequency is \(0.24\); \(24\%\) primarily travel by car.
53964412
In a relative frequency table for \(240\) survey responses, the category “Evening” has relative frequency \(0.275\). What frequency should appear for that category?

Hints

- Relate the category’s share to the total number of responses. - Use \(\text{frequency}=\text{relative frequency}\cdot\text{total}\).

Solution

1. Convert the relative frequency to a count: \(0.275\cdot 240=66\).

Answer

The frequency is \(66\).
53964612
A frequency table shows how 80 customers received an order: 36 chose curbside pickup, 24 chose in-store pickup, and 20 chose delivery. A manager notes that the ratio of curbside pickup to in-store pickup is \(3:2\) and concludes that \(60\%\) of all customers chose curbside pickup. Is that conclusion correct? Find the actual relative frequency of curbside pickup and explain the denominator error.

Hints

- Distinguish a ratio between two categories from a relative frequency out of the whole table. - Identify which observations belong in the denominator for “all customers.” - Check whether every category in the table has been included before converting to a percentage.

Solution

1. The ratio \(36:24=3:2\), so curbside is \(\frac{3}{3+2}=60\%\) only when comparing those two categories by themselves. 2. The full table contains \(80\) customers, including the \(20\) delivery customers. 3. The curbside relative frequency in the full table is \(\frac{36}{80}=0.45=45\%\). 4. The manager used the subtotal \(36+24=60\) as the denominator instead of the total \(80\).

Answer

No. The actual curbside relative frequency is \(45\%\). The \(60\%\) figure uses only the curbside and in-store subtotal as its denominator and ignores the delivery category.
53964712
Shanice made this table for \(100\) responses. <table><thead><tr><th>Color</th><th>Frequency</th><th>Relative frequency</th></tr></thead><tbody><tr><td>Red</td><td>\(22\)</td><td>\(0.22\)</td></tr><tr><td>Blue</td><td>\(38\)</td><td>\(0.38\)</td></tr><tr><td>Green</td><td>\(25\)</td><td>\(0.30\)</td></tr><tr><td>Yellow</td><td>\(15\)</td><td>\(0.15\)</td></tr></tbody></table> Identify and correct the error.

Hints

- Compare each row's count with the common total number of responses. - Use the completed relative-frequency column as a consistency check on your correction.

Solution

1. For Green, the relative frequency should be \(\frac{25}{100}=0.25\). 2. With that correction, the relative frequencies sum to \(1\).

Answer

The Green relative frequency should be \(0.25\), not \(0.30\).
53964812
The table classifies \(100\) train trips by delay category. <table><thead><tr><th>Delay category</th><th>Frequency</th></tr></thead><tbody><tr><td>No delay</td><td>\(54\)</td></tr><tr><td>Under \(10\) minutes</td><td>\(31\)</td></tr><tr><td>At least \(10\) minutes</td><td>\(15\)</td></tr></tbody></table> What percentage of trips had any delay? Justify using the table.

Hints

- Identify all categories that satisfy the description before combining them. - The phrase “any delay” includes every nonzero-delay category.

Solution

1. Trips with any delay are in the last two categories. 2. Their count is \(31+15=46\). 3. Out of \(100\) trips, this is \(46\%\).

Answer

\(46\%\) of the trips had a delay.
53964912
A sample of students produced this table. <table><thead><tr><th>Bottle status</th><th>Frequency</th></tr></thead><tbody><tr><td>Reusable bottle</td><td>\(84\)</td></tr><tr><td>Disposable bottle</td><td>\(39\)</td></tr><tr><td>No bottle</td><td>\(17\)</td></tr></tbody></table> Ndapewa claims, “More than three-fifths brought a reusable bottle.” Is the claim supported?

Hints

- Translate the verbal benchmark into the same form as the table's information. - Compare the reusable-bottle proportion with the benchmark using equality as well as greater-than.

Solution

1. The reusable-bottle proportion is \(\frac{84}{140}=0.60\). 2. Three-fifths is \(0.60\). 3. The values are equal, so the proportion is not more than three-fifths.

Answer

No. The proportion is exactly \(0.60=\frac35\), not more than three-fifths.
53965012
The category codes in a quality check are: \(A, C, B, A, D, A, B, C, A, B, A, D, C, A, B, A\). Which category is most frequent, and what fraction of the observations does it represent?

Hints

- Count systematically, then compare each count. - A tally table can prevent missed or duplicated category codes.

Solution

1. Category \(A\) occurs \(7\) times, more than any other category. 2. Its fraction is \(\frac{7}{16}\).

Answer

Category \(A\); it represents \(\frac{7}{16}\) of the observations.
53965112
A survey of \(250\) deliveries gives the percentage table below. <table><thead><tr><th>Delivery window</th><th>Percentage</th></tr></thead><tbody><tr><td>Morning</td><td>\(32\%\)</td></tr><tr><td>Afternoon</td><td>\(44\%\)</td></tr><tr><td>Evening</td><td>\(18\%\)</td></tr><tr><td>Overnight</td><td>\(6\%\)</td></tr></tbody></table> Convert the table to frequencies.

Hints

- Apply the same part-to-whole relationship to each category. - Check that the resulting frequencies total the sample size.

Solution

1. Multiply each percentage by \(250\). 2. The frequencies are Morning \(80\), Afternoon \(110\), Evening \(45\), and Overnight \(15\).

Answer

Morning: \(80\) Afternoon: \(110\) Evening: \(45\) Overnight: \(15\)
53965412
A table reports that the relative frequency for “Accepted” is \(0.375\). Give the equivalent percentage and an equivalent ratio of accepted observations to all observations.

Hints

- Represent the same part-to-whole relationship in different forms. - Convert the decimal to a percent and to a simplified fraction before writing the ratio.

Solution

1. Convert to a percentage: \(0.375=37.5\%\). 2. Convert to a fraction: \(0.375=\frac{375}{1000}=\frac{3}{8}\). 3. The accepted-to-total ratio is \(3:8\).

Answer

\(37.5\%\) and the ratio \(3:8\).
53965712
A poll of \(400\) people gives this relative frequency table. <table><thead><tr><th>Choice</th><th>Relative frequency</th></tr></thead><tbody><tr><td>Option P</td><td>\(42\%\)</td></tr><tr><td>Option Q</td><td>\(35\%\)</td></tr><tr><td>Option R</td><td>\(23\%\)</td></tr></tbody></table> How many more people selected Option P than Option Q?

Hints

- Find the difference between the two relative frequencies. - Because both options use the same sample, apply that share difference to the common total.

Solution

1. The relative-frequency difference is \(0.42-0.35=0.07\). 2. The count difference is \(0.07\cdot 400=28\).

Answer

\(28\) more people selected Option P.
53965912
School A has \(120\) band students, and \(48\) play a brass instrument. School B has \(80\) band students, and \(36\) play a brass instrument. Which school has the larger relative frequency of brass players?

Hints

- Compare proportions rather than counts because the totals differ. - Compute each school’s brass count as a fraction of its own band total.

Solution

1. School A has relative frequency \(\frac{48}{120}=0.40\). 2. School B has relative frequency \(\frac{36}{80}=0.45\). 3. School B has the larger relative frequency.

Answer

School B, with relative frequency \(0.45\) compared with \(0.40\) for School A.
54864712
Two samples produced these frequency tables for the same response categories. <table><thead><tr><th>Response</th><th>Sample A</th><th>Sample B</th></tr></thead><tbody><tr><td>Option 1</td><td>\(12\)</td><td>\(20\)</td></tr><tr><td>Option 2</td><td>\(18\)</td><td>\(30\)</td></tr><tr><td>Option 3</td><td>\(30\)</td><td>\(50\)</td></tr></tbody></table> Construct the relative frequency column for each sample. Do the samples have the same categorical distribution? Explain why their frequencies can differ while their relative frequencies match.

Hints

- Find each sample total separately. - Divide each category count by its own sample total. - Compare proportions rather than raw counts when sample sizes differ.

Solution

1. Sample A has total \(60\), giving relative frequencies \(0.20\), \(0.30\), and \(0.50\). 2. Sample B has total \(100\), giving relative frequencies \(0.20\), \(0.30\), and \(0.50\). 3. The relative frequency distributions are identical, while the raw counts differ because Sample B contains more observations.

Answer

<table><thead><tr><th>Response</th><th>Sample A relative frequency</th><th>Sample B relative frequency</th></tr></thead><tbody><tr><td>Option 1</td><td>\(0.20\)</td><td>\(0.20\)</td></tr><tr><td>Option 2</td><td>\(0.30\)</td><td>\(0.30\)</td></tr><tr><td>Option 3</td><td>\(0.50\)</td><td>\(0.50\)</td></tr></tbody></table> Yes. The samples have the same relative distribution despite different sample sizes.
54864912
A categorical table has frequencies A \(18\), B \(27\), and C \(15\). Every observation in the data set is copied twice, producing three identical copies of the original data. Give the new frequencies and explain what happens to the relative frequencies and the modal category.

Hints

- Determine the multiplication factor applied to every count. - Compare what happens to each numerator and to the total denominator. - Check whether multiplying all counts changes which one is largest.

Solution

1. Three copies multiply each frequency by \(3\), giving A \(54\), B \(81\), and C \(45\). 2. Both each category count and the total are multiplied by the same factor, so every relative frequency is unchanged. 3. Category B remains the mode because multiplying all frequencies by the same positive factor preserves their order.

Answer

New frequencies: A \(54\), B \(81\), C \(45\). The relative frequencies do not change, and B remains the modal category.
54865812
A questionnaire offers four mutually exclusive responses: Yes, No, Unsure, and Not applicable. In one sample, the frequencies are Yes \(46\), No \(34\), Unsure \(20\), and Not applicable \(0\). Construct the complete relative frequency table and explain why the zero-frequency category may still be useful to display.

Hints

- Include every category that was part of the response design. - A zero count still produces a valid relative frequency. - Distinguish an unavailable category from an available category selected by no one.

Solution

1. The total is \(46+34+20+0=100\). 2. The relative frequencies are \(0.46\), \(0.34\), \(0.20\), and \(0.00\). 3. Keeping Not applicable documents every permitted response and makes clear that the category was available but not selected.

Answer

<table><thead><tr><th>Response</th><th>Frequency</th><th>Relative frequency</th></tr></thead><tbody><tr><td>Yes</td><td>\(46\)</td><td>\(0.46\)</td></tr><tr><td>No</td><td>\(34\)</td><td>\(0.34\)</td></tr><tr><td>Unsure</td><td>\(20\)</td><td>\(0.20\)</td></tr><tr><td>Not applicable</td><td>\(0\)</td><td>\(0.00\)</td></tr></tbody></table> The zero row preserves the full set of allowed responses.
53964512
A frequency table shows \(42\) observations in the “Hybrid” category. The corresponding relative frequency is \(0.35\). Find the total sample size.

Hints

- Work backward from a part-to-whole relationship. - Represent the unknown total in \(\frac{\text{category frequency}}{\text{total}}=\text{relative frequency}\). - Check that the recovered total makes the given count the stated share.

Solution

1. Let \(n\) be the total. Then \(\frac{42}{n}=0.35\). 2. Solve: \(n=\frac{42}{0.35}=120\).

Answer

The sample size is \(120\).
53965312
A listening-preference table initially has these frequencies. <table><thead><tr><th>Preference</th><th>Frequency</th></tr></thead><tbody><tr><td>Podcast</td><td>\(45\)</td></tr><tr><td>Music</td><td>\(75\)</td></tr><tr><td>Audiobook</td><td>\(30\)</td></tr></tbody></table> Ten additional respondents all select Podcast. What is the new relative frequency for Podcast?

Hints

- Recover the original sample total from all table frequencies. - Add the new responses to both the target category and the overall total. - Form the updated part-to-whole ratio and use the old share as a reasonableness check.

Solution

1. The new Podcast frequency is \(45+10=55\). 2. The new total is \(150+10=160\). 3. The new relative frequency is \(\frac{55}{160}=\frac{11}{32}=0.34375\).

Answer

The new relative frequency is \(0.34375\), or \(34.375\%\).
53965612
The frequencies of three categories are in the ratio \(5:3:2\), and the total frequency is \(200\). Find the frequency and relative frequency of each category.

Hints

- Interpret the ratio as a collection of equal parts. - Divide the total by the sum of the ratio parts to find the count represented by one part. - Convert each category count to a share and verify that the shares total \(1\).

Solution

1. The ratio has \(5+3+2=10\) equal parts. 2. Each part represents \(\frac{200}{10}=20\) observations. 3. The frequencies are \(100\), \(60\), and \(40\). 4. The relative frequencies are \(0.50\), \(0.30\), and \(0.20\).

Answer

Frequencies: \(100, 60, 40\). Relative frequencies: \(0.50, 0.30, 0.20\).
53965812
A frequency table has category counts \(A=28\), \(B=22\), \(C=30\), and \(D=20\). If category \(D\) is removed from the analysis, what are the new relative frequencies for \(A\), \(B\), and \(C\)?

Hints

- Recalculate the total after removing the category. - Removing a category changes the denominator for every remaining relative frequency. - Use the remaining counts as numerators and verify that the new shares total \(1\).

Solution

1. The remaining total is \(28+22+30=80\). 2. The new relative frequencies are \(\frac{28}{80}=0.35\), \(\frac{22}{80}=0.275\), and \(\frac{30}{80}=0.375\).

Answer

\(A=0.35\) \(B=0.275\) \(C=0.375\)
54863712
A survey of \(80\) students reports the preferred rehearsal space using percentages rounded to the nearest whole percent. <table><thead><tr><th>Space</th><th>Reported percentage</th></tr></thead><tbody><tr><td>Auditorium</td><td>\(26\%\)</td></tr><tr><td>Music room</td><td>\(24\%\)</td></tr><tr><td>Classroom</td><td>\(19\%\)</td></tr><tr><td>Outdoor stage</td><td>\(31\%\)</td></tr></tbody></table> Recover the frequency for each category. Explain why rounding does not create ambiguity in this case.

Hints

- Determine how much one student changes a category’s percentage. - Find the integer count whose exact percentage lies in the rounding interval for each reported value. - Verify that the recovered counts total the stated sample size.

Solution

1. For \(80\) students, each observation changes a percentage by \(\frac{100\%}{80}=1.25\%\). 2. The unique whole-number counts that round to the reported percentages are Auditorium \(21\), Music room \(19\), Classroom \(15\), and Outdoor stage \(25\). 3. The counts total \(21+19+15+25=80\), and their exact percentages are \(26.25\%\), \(23.75\%\), \(18.75\%\), and \(31.25\%\), which round as reported.

Answer

Auditorium: \(21\) Music room: \(19\) Classroom: \(15\) Outdoor stage: \(25\) Each reported percentage corresponds to exactly one possible integer count for a total of \(80\).
54863812
An electronics collection event produced this table. <table><thead><tr><th>Disposition</th><th>Frequency</th></tr></thead><tbody><tr><td>Reused</td><td>\(22\)</td></tr><tr><td>Recycled</td><td>\(41\)</td></tr><tr><td>Landfilled</td><td>\(17\)</td></tr><tr><td>Outcome unknown</td><td>\(10\)</td></tr></tbody></table> Reconstruct the table using the categories Diverted, Landfilled, and Outcome unknown, where Diverted combines Reused and Recycled. Give each new frequency and relative frequency, rounded to three decimals.

Hints

- Combine only the source categories named in the new definition. - Keep the original total as the denominator for every new relative frequency. - Check that the new counts and rounded relative frequencies each form a complete distribution.

Solution

1. The total frequency is \(22+41+17+10=90\). 2. Diverted has frequency \(22+41=63\), Landfilled has frequency \(17\), and Outcome unknown has frequency \(10\). 3. The relative frequencies are \(\frac{63}{90}=0.700\), \(\frac{17}{90}\approx0.189\), and \(\frac{10}{90}\approx0.111\).

Answer

<table><thead><tr><th>Category</th><th>Frequency</th><th>Relative frequency</th></tr></thead><tbody><tr><td>Diverted</td><td>\(63\)</td><td>\(0.700\)</td></tr><tr><td>Landfilled</td><td>\(17\)</td><td>\(0.189\)</td></tr><tr><td>Outcome unknown</td><td>\(10\)</td><td>\(0.111\)</td></tr></tbody></table>
54863912
A public report summarizes \(200\) cases but suppresses any category frequency below \(5\). One category is displayed as “fewer than \(5\).” Give every possible frequency for that category and the corresponding interval of possible relative frequencies.

Hints

- Translate the suppression phrase into allowable integer counts. - Divide each endpoint count by the full sample size. - Preserve the fact that only certain increments are possible.

Solution

1. “Fewer than \(5\)” permits the integer frequencies \(0,1,2,3,4\). 2. Dividing by \(200\) gives possible relative frequencies \(0,0.005,0.010,0.015,0.020\). 3. Therefore, the relative frequency lies from \(0\) through \(0.020\), inclusive, although its exact value is not disclosed.

Answer

Possible frequencies: \(0,1,2,3,4\). Possible relative-frequency interval: \([0,0.020]\), with values occurring in increments of \(0.005\).
54864012
A commuting survey originally reported these frequencies: Bus \(40\), Rail \(35\), Bike \(15\), and Walk \(10\). A data audit finds that \(6\) Rail responses were mistakenly coded as Bus and \(4\) Walk responses were mistakenly coded as Bike. Construct the corrected frequency and relative frequency table.

Hints

- A correction transfers observations between categories without changing the total. - Update both the category that loses records and the category that gains them. - Recalculate the relative frequencies from the corrected counts.

Solution

1. Move \(6\) observations from Bus to Rail, giving Bus \(34\) and Rail \(41\). 2. Move \(4\) observations from Bike to Walk, giving Bike \(11\) and Walk \(14\). 3. The total remains \(100\), so the corrected relative frequencies are \(0.34\), \(0.41\), \(0.11\), and \(0.14\).

Answer

<table><thead><tr><th>Method</th><th>Frequency</th><th>Relative frequency</th></tr></thead><tbody><tr><td>Bus</td><td>\(34\)</td><td>\(0.34\)</td></tr><tr><td>Rail</td><td>\(41\)</td><td>\(0.41\)</td></tr><tr><td>Bike</td><td>\(11\)</td><td>\(0.11\)</td></tr><tr><td>Walk</td><td>\(14\)</td><td>\(0.14\)</td></tr></tbody></table>
54864112
Two support centers recorded customers’ preferred contact methods. <table><thead><tr><th>Method</th><th>Center A</th><th>Center B</th></tr></thead><tbody><tr><td>Text</td><td>\(54\)</td><td>\(28\)</td></tr><tr><td>Email</td><td>\(42\)</td><td>\(36\)</td></tr><tr><td>Phone</td><td>\(24\)</td><td>\(16\)</td></tr></tbody></table> Construct the combined relative frequency table. Explain why simply averaging the two centers’ relative frequencies would not generally give the combined distribution.

Hints

- Add matching category counts before forming any combined proportions. - Use the total number of responses from both centers as the common denominator. - Compare the two center totals when deciding whether an unweighted average is valid.

Solution

1. The combined frequencies are Text \(82\), Email \(78\), and Phone \(40\), for a total of \(200\). 2. The combined relative frequencies are \(\frac{82}{200}=0.41\), \(\frac{78}{200}=0.39\), and \(\frac{40}{200}=0.20\). 3. Center A has \(120\) responses and Center B has \(80\), so their relative frequencies must be weighted by different sample sizes rather than averaged equally.

Answer

<table><thead><tr><th>Method</th><th>Combined frequency</th><th>Combined relative frequency</th></tr></thead><tbody><tr><td>Text</td><td>\(82\)</td><td>\(0.41\)</td></tr><tr><td>Email</td><td>\(78\)</td><td>\(0.39\)</td></tr><tr><td>Phone</td><td>\(40\)</td><td>\(0.20\)</td></tr></tbody></table> Equal averaging is inappropriate because the centers have unequal sample sizes.
54864212
A survey of \(150\) residents asks which emergency-notification methods they would use, allowing multiple selections. <table><thead><tr><th>Method</th><th>Number selecting</th></tr></thead><tbody><tr><td>Email alerts</td><td>\(92\)</td></tr><tr><td>Text alerts</td><td>\(71\)</td></tr><tr><td>App alerts</td><td>\(46\)</td></tr><tr><td>Phone calls</td><td>\(33\)</td></tr></tbody></table> Find the selection rate for each method, rounded to one decimal percent. Explain why the rates do not need to sum to \(100\%\), and identify any method selected by a majority of respondents.

Hints

- Use the number of respondents, not the sum of selections, as each denominator. - Check whether the response options are mutually exclusive. - Compare each individual selection rate with the majority benchmark.

Solution

1. Divide each count by \(150\): Email \(\frac{92}{150}\approx61.3\%\), Text \(\frac{71}{150}\approx47.3\%\), App \(\frac{46}{150}\approx30.7\%\), and Phone \(\frac{33}{150}=22.0\%\). 2. Respondents could select several methods, so one person may contribute to multiple counts and the rates are not parts of one mutually exclusive distribution. 3. Email alerts were selected by more than \(50\%\), so that method alone was selected by a majority.

Answer

Email alerts: \(61.3\%\) Text alerts: \(47.3\%\) App alerts: \(30.7\%\) Phone calls: \(22.0\%\) The rates may exceed \(100\%\) in total because selections overlap. Email alerts were selected by a majority.
54864312
A survey invitation produced this table. <table><thead><tr><th>Outcome</th><th>Frequency</th></tr></thead><tbody><tr><td>Completed survey</td><td>\(132\)</td></tr><tr><td>Declined</td><td>\(48\)</td></tr><tr><td>No response</td><td>\(20\)</td></tr></tbody></table> Find the relative frequency of Completed survey using a) all invited people represented in the table and b) only people who gave a definite response. Explain why the two values differ.

Hints

- Identify the population included in each requested denominator. - Keep the numerator fixed while changing only the eligible total. - State explicitly which category is excluded from the second calculation.

Solution

1. The full table total is \(132+48+20=200\), so the all-invited relative frequency is \(\frac{132}{200}=0.66\). 2. The definite-response total is \(132+48=180\), so the valid-response relative frequency is \(\frac{132}{180}\approx0.733\). 3. The values differ because the second calculation excludes the \(20\) nonresponses from its denominator.

Answer

a) Using all represented invitees: \(0.66\), or \(66\%\). b) Using definite responses only: \(0.733\), or about \(73.3\%\). The denominators differ because nonresponses are excluded in b).
54864412
A published table combines two rare response categories into “Other,” with frequency \(40\). An accompanying note says that \(35\%\) of the “Other” responses were write-in answers and the rest were blank-but-valid responses. Recover the two original frequencies and state their relative frequencies within the “Other” category.

Hints

- Treat the combined category as the total for the note’s percentage. - Find one subcategory count from its share of that total. - Use the remainder for the second subcategory and verify the two shares sum to \(1\).

Solution

1. The write-in frequency is \(0.35\cdot40=14\). 2. The blank-but-valid frequency is \(40-14=26\). 3. Within “Other,” the relative frequencies are \(\frac{14}{40}=0.35\) and \(\frac{26}{40}=0.65\).

Answer

Write-in answers: frequency \(14\), relative frequency within “Other” \(0.35\). Blank-but-valid responses: frequency \(26\), relative frequency within “Other” \(0.65\).
54864612
An ordinal variable has \(120\) observations. Its cumulative relative frequency table is shown below. <table><thead><tr><th>Level</th><th>Cumulative relative frequency</th></tr></thead><tbody><tr><td>Level 1</td><td>\(0.10\)</td></tr><tr><td>Level 2</td><td>\(0.35\)</td></tr><tr><td>Level 3</td><td>\(0.75\)</td></tr><tr><td>Level 4</td><td>\(1.00\)</td></tr></tbody></table> Recover the ordinary frequency and relative frequency for each level.

Hints

- Convert cumulative proportions to cumulative counts first. - Each ordinary category count is the increase from the previous cumulative total. - Check that the recovered relative frequencies sum to \(1\).

Solution

1. The cumulative frequencies are \(12,42,90,120\). 2. Subtract consecutive cumulative frequencies to get ordinary frequencies \(12,30,48,30\). 3. Dividing by \(120\) gives relative frequencies \(0.10,0.25,0.40,0.25\).

Answer

<table><thead><tr><th>Level</th><th>Frequency</th><th>Relative frequency</th></tr></thead><tbody><tr><td>Level 1</td><td>\(12\)</td><td>\(0.10\)</td></tr><tr><td>Level 2</td><td>\(30\)</td><td>\(0.25\)</td></tr><tr><td>Level 3</td><td>\(48\)</td><td>\(0.40\)</td></tr><tr><td>Level 4</td><td>\(30\)</td><td>\(0.25\)</td></tr></tbody></table>
54864812
In a categorical table, Category A has frequency \(31\) and Category B has frequency \(19\). The relative frequency of A exceeds the relative frequency of B by \(0.15\). Find the total sample size and the relative frequencies of A and B.

Hints

- Express the difference between the two relative frequencies using their count difference. - Use the common unknown denominator for both categories. - Check the recovered total by recomputing each proportion.

Solution

1. The frequency difference is \(31-19=12\). 2. If the sample size is \(n\), then \(\frac{12}{n}=0.15\), so \(n=80\). 3. The relative frequencies are \(\frac{31}{80}=0.3875\) and \(\frac{19}{80}=0.2375\).

Answer

Sample size: \(80\). Category A relative frequency: \(0.3875\). Category B relative frequency: \(0.2375\).
54865012
A transportation table contains two labels that refer to the same category. <table><thead><tr><th>Recorded label</th><th>Frequency</th></tr></thead><tbody><tr><td>E-bike</td><td>\(18\)</td></tr><tr><td>Electric bicycle</td><td>\(27\)</td></tr><tr><td>Standard bicycle</td><td>\(25\)</td></tr><tr><td>Walk</td><td>\(30\)</td></tr></tbody></table> Merge the duplicate labels, give the corrected relative frequency table, and identify the modal category.

Hints

- Identify labels that describe the same response before calculating proportions. - Preserve the original total while combining their counts. - Compare corrected frequencies to determine the mode.

Solution

1. Merge E-bike and Electric bicycle to obtain frequency \(18+27=45\). 2. The total is \(100\), so the corrected relative frequencies are Electric bicycle \(0.45\), Standard bicycle \(0.25\), and Walk \(0.30\). 3. Electric bicycle has the greatest frequency, so it is the modal category.

Answer

<table><thead><tr><th>Category</th><th>Frequency</th><th>Relative frequency</th></tr></thead><tbody><tr><td>Electric bicycle</td><td>\(45\)</td><td>\(0.45\)</td></tr><tr><td>Standard bicycle</td><td>\(25\)</td><td>\(0.25\)</td></tr><tr><td>Walk</td><td>\(30\)</td><td>\(0.30\)</td></tr></tbody></table> The modal category is Electric bicycle.
54865112
A report states that its sample size is \(240\), but it gives this table. <table><thead><tr><th>Zone</th><th>Frequency</th><th>Relative frequency</th></tr></thead><tbody><tr><td>North</td><td>\(70\)</td><td>\(0.28\)</td></tr><tr><td>South</td><td>\(65\)</td><td>\(0.26\)</td></tr><tr><td>East</td><td>\(60\)</td><td>\(0.24\)</td></tr><tr><td>West</td><td>\(55\)</td><td>\(0.22\)</td></tr></tbody></table> Determine whether the row entries or the stated sample size are inconsistent, and give the corrected sample size.

Hints

- Check the frequency total independently of the stated sample size. - Test each relative frequency against a common denominator. - Decide which part of the report makes all entries consistent.

Solution

1. The frequencies sum to \(70+65+60+55=250\). 2. Each listed relative frequency matches its count divided by \(250\): \(0.28\), \(0.26\), \(0.24\), and \(0.22\). 3. The relative frequencies also sum to \(1\), so the row entries are internally consistent and the stated sample size \(240\) is incorrect.

Answer

The corrected sample size is \(250\). The frequencies and relative frequencies are mutually consistent; the reported total of \(240\) is the error.
54865312
A proposed age-group table uses the categories Under \(18\), \(18\)–\(24\), \(24\)–\(30\), and Over \(30\). Assume age is recorded in completed whole years. Explain why these categories do not define a valid one-variable frequency table. Rewrite the categories so every possible recorded age belongs to exactly one group.

Hints

- Test each boundary value against the two neighboring categories. - A valid frequency table needs groups that are mutually exclusive and exhaustive. - Use the fact that age is recorded in completed whole years when rewriting the endpoints.

Solution

1. Age \(24\) belongs to both \(18\)–\(24\) and \(24\)–\(30\), so the original categories are not mutually exclusive. 2. With age recorded in completed whole years, age \(30\) is already included in \(24\)–\(30\), while Over \(30\) begins at \(31\). 3. One valid revision is Under \(18\), \(18\)–\(23\), \(24\)–\(30\), and \(31\) or older. These groups do not overlap and together cover all possible recorded ages.

Answer

The original categories overlap at age \(24\). One valid set is Under \(18\), \(18\)–\(23\), \(24\)–\(30\), and \(31\) or older.
54865412
An ordered opinion variable has this table. <table><thead><tr><th>Response</th><th>Frequency</th></tr></thead><tbody><tr><td>Strongly oppose</td><td>\(14\)</td></tr><tr><td>Oppose</td><td>\(21\)</td></tr><tr><td>Neutral</td><td>\(25\)</td></tr><tr><td>Support</td><td>\(30\)</td></tr><tr><td>Strongly support</td><td>\(10\)</td></tr></tbody></table> Find the percentage that selected Support or Strongly support, and identify the median response category.

Hints

- Combine every category included in the stated threshold. - Preserve the response order when accumulating frequencies. - Locate the middle ordered observations using the cumulative totals.

Solution

1. The total is \(100\), and Support or Strongly support has frequency \(30+10=40\), so the percentage is \(40\%\). 2. The cumulative frequencies are \(14\), \(35\), \(60\), \(90\), and \(100\). 3. The \(50\)th and \(51\)st observations fall in Neutral, so Neutral is the median category.

Answer

Support or Strongly support: \(40\%\). Median response category: Neutral.
54865512
The ratio of observations in Category K to observations not in Category K is \(7:13\). The sample contains \(180\) observations. Find Category K’s frequency and relative frequency.

Hints

- Interpret the ratio as category versus remainder, not category versus total. - Find how many equal parts make up the full sample. - Convert the resulting category count to a proportion of the total.

Solution

1. The ratio represents \(7+13=20\) equal parts. 2. Each part represents \(\frac{180}{20}=9\) observations. 3. Category K has frequency \(7\cdot9=63\), and its relative frequency is \(\frac{63}{180}=0.35\).

Answer

Category K has frequency \(63\) and relative frequency \(0.35\), or \(35\%\).
54865612
A response table has frequencies Option A \(40\), Option B \(35\), Option C \(25\), and Unknown \(50\). Identify the modal category in the complete table and the modal known-response category. Explain why the two answers serve different reporting purposes.

Hints

- First compare all category frequencies exactly as displayed. - Then restrict attention to categories that represent known preferences. - Interpret what each mode says about the data rather than treating one as universally correct.

Solution

1. In the complete table, Unknown has the largest frequency, \(50\), so it is the overall mode. 2. Among known responses only, Option A has the largest frequency, \(40\), so it is the modal known preference. 3. The overall mode highlights the prevalence of missing or unknown classification, while the known-response mode summarizes the observed preferences.

Answer

Overall modal category: Unknown. Modal known-response category: Option A. Report both when data completeness and the leading known preference are both relevant.
54865712
A two-category table initially has A \(30\) and B \(20\). Later, \(50\) additional observations are added to B while A remains unchanged. Find A’s frequency and relative frequency before and after the addition. Explain how a category’s frequency can stay constant while its relative frequency changes.

Hints

- Write the total number of observations at each time. - Keep A’s numerator unchanged while updating the denominator. - Distinguish an absolute count from a share of the whole.

Solution

1. Initially, the total is \(50\), so A has frequency \(30\) and relative frequency \(\frac{30}{50}=0.60\). 2. After the addition, B has frequency \(70\) and the total is \(100\). 3. A still has frequency \(30\), but its new relative frequency is \(\frac{30}{100}=0.30\). 4. Relative frequency depends on both the category count and the total count.

Answer

Before: A frequency \(30\), relative frequency \(0.60\). After: A frequency \(30\), relative frequency \(0.30\). The frequency stays fixed, but the larger total reduces A’s share.
54865912
An ordinal response table is listed out of order. <table><thead><tr><th>Response</th><th>Frequency</th></tr></thead><tbody><tr><td>Agree</td><td>\(34\)</td></tr><tr><td>Strongly disagree</td><td>\(10\)</td></tr><tr><td>Strongly agree</td><td>\(22\)</td></tr><tr><td>Neutral</td><td>\(18\)</td></tr><tr><td>Disagree</td><td>\(16\)</td></tr></tbody></table> Reorder the categories from least to greatest agreement, add cumulative frequencies, and identify the median category.

Hints

- Put the categories in their meaningful order before accumulating counts. - Keep a running total down the reordered table. - Locate the middle observations using the final total.

Solution

1. The ordered categories and frequencies are Strongly disagree \(10\), Disagree \(16\), Neutral \(18\), Agree \(34\), and Strongly agree \(22\). 2. The cumulative frequencies are \(10\), \(26\), \(44\), \(78\), and \(100\). 3. The \(50\)th and \(51\)st observations fall in Agree, so Agree is the median category.

Answer

<table><thead><tr><th>Response</th><th>Frequency</th><th>Cumulative frequency</th></tr></thead><tbody><tr><td>Strongly disagree</td><td>\(10\)</td><td>\(10\)</td></tr><tr><td>Disagree</td><td>\(16\)</td><td>\(26\)</td></tr><tr><td>Neutral</td><td>\(18\)</td><td>\(44\)</td></tr><tr><td>Agree</td><td>\(34\)</td><td>\(78\)</td></tr><tr><td>Strongly agree</td><td>\(22\)</td><td>\(100\)</td></tr></tbody></table> The median category is Agree.
54866012
A categorical distribution has exact relative frequencies \(\frac{2}{9}\), \(\frac{1}{3}\), and \(\frac{4}{9}\). Find the smallest possible sample size greater than \(50\), and give the corresponding category frequencies.

Hints

- Express all proportions using a common denominator. - Determine which sample sizes make every category count an integer. - Apply the lower bound only after identifying the required multiples.

Solution

1. All three relative frequencies have denominator \(9\) when written with a common denominator: \(\frac{2}{9}\), \(\frac{3}{9}\), and \(\frac{4}{9}\). 2. The sample size must be a multiple of \(9\). The smallest multiple of \(9\) greater than \(50\) is \(54\). 3. The frequencies are \(\frac{2}{9}\cdot54=12\), \(\frac{3}{9}\cdot54=18\), and \(\frac{4}{9}\cdot54=24\).

Answer

The smallest possible sample size is \(54\), with category frequencies \(12\), \(18\), and \(24\).
54866112
A table reports six category percentages, rounded to the nearest whole percent, as \(17\%\), \(17\%\), \(17\%\), \(17\%\), \(17\%\), and \(16\%\). The displayed total is \(101\%\). Can the table still represent a valid complete distribution? Give one possible set of exact percentages.

Hints

- Treat each displayed value as a rounded interval rather than an exact percentage. - Try small adjustments below the values that were rounded upward. - Verify both the exact total and the resulting rounded entries.

Solution

1. Each displayed percentage may differ from its exact value because of rounding. 2. One possible set of exact percentages is \(16.8\%\) for each of the first five categories and \(16.0\%\) for the sixth. 3. These exact percentages sum to \(5\cdot16.8\%+16.0\%=100.0\%\) and round to the displayed values.

Answer

Yes. For example, exact percentages \(16.8\%\), \(16.8\%\), \(16.8\%\), \(16.8\%\), \(16.8\%\), and \(16.0\%\) sum to \(100\%\) and round to the reported percentages.
54866312
A city records the primary way commuters travel to work in two different years. <table><thead><tr><th>Travel mode</th><th>Year 1</th><th>Year 2</th></tr></thead><tbody><tr><td>Car</td><td>\(60\)</td><td>\(84\)</td></tr><tr><td>Bus</td><td>\(45\)</td><td>\(84\)</td></tr><tr><td>Bicycle</td><td>\(30\)</td><td>\(48\)</td></tr><tr><td>Walk</td><td>\(15\)</td><td>\(24\)</td></tr></tbody></table> For bus commuters, find the change in frequency and the change in relative frequency from Year 1 to Year 2. Express the relative-frequency change in percentage points.

Hints

- Find the total number of commuters separately for each year. - Compare both the category counts and the category shares. - A change in percentages is stated in percentage points.

Solution

1. The Year 1 total is \(60+45+30+15=150\), so the bus relative frequency is \(\frac{45}{150}=0.30\). 2. The Year 2 total is \(84+84+48+24=240\), so the bus relative frequency is \(\frac{84}{240}=0.35\). 3. The frequency changed by \(84-45=39\) commuters, and the relative frequency changed by \(0.35-0.30=0.05\), or \(5\) percentage points.

Answer

The bus frequency increased by \(39\) commuters, and its relative frequency increased by \(5\) percentage points, from \(30\%\) to \(35\%\).
54866412
A survey currently has these response counts: Category A, \(48\); Category B, \(39\); Category C, \(33\). Additional respondents will all choose Category B. What is the smallest number of additional respondents needed to make Category B the unique modal category? Then give the new relative-frequency distribution, rounded to three decimal places.

Hints

- A unique mode must have a frequency strictly greater than every other frequency. - After finding the added count, update the total before calculating shares. - Check that the rounded relative frequencies sum to approximately \(1\).

Solution

1. Category B must exceed Category A's count of \(48\). The smallest increase satisfies \(39+x>48\), so \(x=10\). 2. The new counts are \(48\), \(49\), and \(33\), with total \(48+49+33=130\). 3. The relative frequencies are \(\frac{48}{130}\approx0.369\), \(\frac{49}{130}\approx0.377\), and \(\frac{33}{130}\approx0.254\).

Answer

The smallest number is \(10\). The new relative frequencies are Category A: \(0.369\), Category B: \(0.377\), and Category C: \(0.254\).
54866512
A preference table lists \(44\) responses for Option A, \(31\) for Option B, and \(25\) for Option C. The table accidentally omits the category “No preference,” which represents exactly \(20\%\) of all responses. Find the total sample size, the frequency for “No preference,” and the complete relative-frequency distribution.

Hints

- Determine what percentage of the full sample is represented by the categories already listed. - Use the listed total as that fraction of the unknown sample size. - Divide each completed category count by the full sample size.

Solution

1. The listed frequencies total \(44+31+25=100\). 2. Because “No preference” is \(20\%\) of all responses, the listed categories represent \(80\%\). Thus \(0.80n=100\), giving \(n=125\). 3. The omitted frequency is \(0.20\cdot125=25\). 4. The relative frequencies are \(\frac{44}{125}=0.352\), \(\frac{31}{125}=0.248\), \(\frac{25}{125}=0.200\), and \(\frac{25}{125}=0.200\).

Answer

The sample size is \(125\), and “No preference” has frequency \(25\). The relative frequencies are Option A: \(0.352\), Option B: \(0.248\), Option C: \(0.200\), and No preference: \(0.200\).
54864512
A report says that \(18\%\) of a sample of \(250\) households selected one category, with the percentage rounded to the nearest whole percent. Find every possible integer category frequency.

Hints

- Translate the rounded percentage into an interval of exact percentages. - Apply both ends of that interval to the sample size. - List only whole-number counts that satisfy the rounding condition.

Solution

1. A percentage that rounds to \(18\%\) must be at least \(17.5\%\) and less than \(18.5\%\). 2. For \(250\) households, the possible counts satisfy \(43.75\le c<46.25\). 3. The integer possibilities are \(44\), \(45\), and \(46\), giving exact percentages \(17.6\%\), \(18.0\%\), and \(18.4\%\).

Answer

The possible category frequencies are \(44\), \(45\), and \(46\).
54865212
A frequency table has four categories and a total of \(120\) observations. Category B has twice the frequency of Category A. Category C has frequency \(30\), and Category D has relative frequency \(0.20\). Find every category frequency and relative frequency.

Hints

- Convert the given relative frequency to a count first. - Remove the known category counts from the total. - Use the stated multiplicative relationship to split the remaining observations.

Solution

1. Category D has frequency \(0.20\cdot120=24\). 2. Categories A and B together have \(120-30-24=66\) observations. 3. Since B has twice A’s frequency, the \(66\) observations form three equal parts: A \(=22\) and B \(=44\). 4. The relative frequencies are A \(\frac{22}{120}\approx0.183\), B \(\frac{44}{120}\approx0.367\), C \(\frac{30}{120}=0.250\), and D \(0.200\).

Answer

<table><thead><tr><th>Category</th><th>Frequency</th><th>Relative frequency</th></tr></thead><tbody><tr><td>A</td><td>\(22\)</td><td>\(0.183\)</td></tr><tr><td>B</td><td>\(44\)</td><td>\(0.367\)</td></tr><tr><td>C</td><td>\(30\)</td><td>\(0.250\)</td></tr><tr><td>D</td><td>\(24\)</td><td>\(0.200\)</td></tr></tbody></table>

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