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Joint, marginal, and conditional relative frequency

Learning goals

  • Computes joint relative frequencies using the grand total.
  • Computes marginal relative frequencies from row or column totals.
  • Computes conditional relative frequencies using the correct conditioning total.
  • Distinguishes joint, marginal, and conditional percentages.
  • Uses conditional distributions to compare groups.

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Joint, marginal, and conditional relative frequency 1
A headline says, “\(30\%\) of all survey respondents were satisfied customers who used chat.” Which relative frequency is this: joint, marginal, or conditional? What denominator does the statement require?

Hints

- Identify whether one category or a category pair is being described. - Notice the phrase “of all survey respondents.” - Conditional statements restrict the denominator to a subgroup.

Solution

1. The statement describes the intersection of two categories: satisfied and used chat. 2. It is a joint relative frequency. 3. Its denominator is the grand total of all survey respondents.

Answer

It is a joint relative frequency, using all survey respondents as the denominator.
Joint, marginal, and conditional relative frequency 2
The table summarizes whether students in a school club also have a part-time job. <table><tr><th></th><th>Part-time job</th><th>No part-time job</th><th>Total</th></tr><tr><td>Club member</td><td>\(18\)</td><td>\(42\)</td><td>\(60\)</td></tr><tr><td>Not a club member</td><td>\(30\)</td><td>\(30\)</td><td>\(60\)</td></tr><tr><td>Total</td><td>\(48\)</td><td>\(72\)</td><td>\(120\)</td></tr></table> What denominator should be used for the conditional relative frequency of having a part-time job among club members, and what is that conditional relative frequency?

Hints

- The words “among club members” identify the group that forms the denominator. - Use the total for that restricted group rather than the grand total. - The numerator is the count that satisfies both descriptions.

Solution

1. The condition is being a club member, so restrict the denominator to the club-member row total, \(60\). 2. Of those \(60\) students, \(18\) have a part-time job. 3. The conditional relative frequency is \(\frac{18}{60}=0.30\).

Answer

The denominator is \(60\), and the conditional relative frequency is \(0.30\), or \(30\%\).
Joint, marginal, and conditional relative frequency 3
A park cleanup had \(30\) volunteers from two neighborhoods. Each chose one cleanup activity. <table> <tr><th></th><th>Collected litter</th><th>Planted trees</th></tr> <tr><td>Riverside</td><td>\(9\)</td><td>\(6\)</td></tr> <tr><td>Hillcrest</td><td>\(12\)</td><td>\(3\)</td></tr> </table> What fraction of **all volunteers** came from Hillcrest and collected litter? Give a fraction in simplest form.

Hints

- One cell represents people who meet both descriptions. - Which count represents everybody at the cleanup?

Solution

1. The Hillcrest and collected-litter cell contains \(12\) volunteers. 2. The denominator is all \(30\) volunteers. The joint relative frequency is \(\frac{12}{30}=\frac{2}{5}\).

Answer

\(\frac{12}{30}=\frac{2}{5}\).
Joint, marginal, and conditional relative frequency 4
A school survey records each student's one elective choice (art or music) and whether they play a school sport. <table> <tr><th></th><th>Plays a sport</th><th>Does not play a sport</th></tr> <tr><td>Art</td><td>\(6\)</td><td>\(12\)</td></tr> <tr><td>Music</td><td>\(9\)</td><td>\(3\)</td></tr> </table> Among students taking art, what fraction play a school sport?

Hints

- Would adding more music students change the chance for someone who takes art? - How could you check that your fraction describes only the students eligible for this choice?

Solution

1. There are \(6+12=18\) art students. 2. Of those, \(6\) play a sport, so the conditional relative frequency is \(\frac{6}{18}=\frac{1}{3}\).

Answer

\(\frac{1}{3}\).
Joint, marginal, and conditional relative frequency 5
A survey includes \(50\) students. Of these, \(20\) ride the bus, and \(8\) of the bus riders arrive early. A student calculates \(\frac{8}{20}\). Is this a joint, marginal, or conditional relative frequency? Explain the denominator.

Hints

- Would including students who do not ride the bus change the meaning of this fraction? - Does the fraction describe all surveyed students or a smaller group?

Solution

1. The denominator \(20\) counts only bus riders, not all surveyed students. 2. The fraction \(\frac{8}{20}\) describes early arrival within the bus-rider group, so it is conditional.

Answer

Conditional relative frequency. The denominator is all \(20\) bus riders.
Joint, marginal, and conditional relative frequency 6
The table summarizes science fair projects. <table> <tr><th></th><th>used a physical prototype</th><th>did not use a physical prototype</th><th>Total</th></tr> <tr><td>team project</td><td>\(20\)</td><td>\(15\)</td><td>\(35\)</td></tr> <tr><td>individual project</td><td>\(12\)</td><td>\(18\)</td><td>\(30\)</td></tr> <tr><td>Total</td><td>\(32\)</td><td>\(33\)</td><td>\(65\)</td></tr> </table> Find the joint relative frequency of “team project and used a physical prototype,” the marginal relative frequency of “team project,” and the conditional relative frequency of “used a physical prototype” among those who are “team project.” Give decimals to three places.

Hints

- A joint relative frequency uses the grand total as its denominator. - A marginal relative frequency uses the corresponding row or column total count over the grand total. - A conditional relative frequency uses only the stated subgroup as its denominator.

Solution

1. Joint relative frequency: \(20/65\approx 0.308\). 2. Marginal relative frequency for “team project”: \(35/65\approx 0.538\). 3. Conditional relative frequency: \(20/35\approx 0.571\).

Answer

Joint: \(0.308\). Marginal: \(0.538\). Conditional: \(0.571\).
Joint, marginal, and conditional relative frequency 7
A joint relative-frequency table has columns \(C_1\) and \(C_2\). The column margins are \(P(C_1)=0.35\) and \(P(C_2)=0.65\). Within column \(C_1\), \(60\%\) of observations are in row \(R_1\); within column \(C_2\), \(30\%\) are in row \(R_1\). Construct the complete joint relative-frequency table, including row and column margins.

Hints

- A within-column percentage must be multiplied by that column’s marginal share. - Fill the complementary row within each column before calculating row margins. - Check that all four joint entries add to \(1\).

Solution

1. In column \(C_1\), \(P(R_1\cap C_1)=0.35\cdot0.60=0.21\), so \(P(R_2\cap C_1)=0.35-0.21=0.14\). 2. In column \(C_2\), \(P(R_1\cap C_2)=0.65\cdot0.30=0.195\), so \(P(R_2\cap C_2)=0.65-0.195=0.455\). 3. Row margins are \(0.21+0.195=0.405\) and \(0.14+0.455=0.595\).

Answer

<table> <tr><th></th><th>\(C_1\)</th><th>\(C_2\)</th><th>Total</th></tr> <tr><td>\(R_1\)</td><td>\(0.210\)</td><td>\(0.195\)</td><td>\(0.405\)</td></tr> <tr><td>\(R_2\)</td><td>\(0.140\)</td><td>\(0.455\)</td><td>\(0.595\)</td></tr> <tr><td>Total</td><td>\(0.350\)</td><td>\(0.650\)</td><td>\(1.000\)</td></tr> </table>
Joint, marginal, and conditional relative frequency 8
Two high-school clubs survey their members about using the school study room. Each student belongs to exactly one of the two clubs. Some members leave the question unanswered. <table> <tr><th></th><th>Yes</th><th>No</th><th>Unknown</th></tr> <tr><td>Art club</td><td>\(30\)</td><td>\(10\)</td><td>\(10\)</td></tr> <tr><td>Chess club</td><td>\(20\)</td><td>\(20\)</td><td>\(10\)</td></tr> </table> The report will calculate joint relative frequencies using only students who answered yes or no. Make that two-way table, including row and column totals. Explain why you must use a new denominator.

Hints

- Which students are included in the revised table? - Find the total number of usable responses first. - Should the four joint relative frequencies sum to \(1\)?

Solution

1. Removing the unknown responses leaves \(30+10+20+20=80\) answered surveys. 2. Divide each of the four retained counts by \(80\), giving \(0.375,0.125,0.25,0.25\). 3. Row totals are \(0.50\) and \(0.50\); column totals are \(0.625\) and \(0.375\). The original \(100\) includes \(20\) unanswered surveys.

Answer

<table> <tr><th></th><th>Yes</th><th>No</th><th>Total</th></tr> <tr><td>Art club</td><td>\(0.375\)</td><td>\(0.125\)</td><td>\(0.50\)</td></tr> <tr><td>Chess club</td><td>\(0.250\)</td><td>\(0.250\)</td><td>\(0.50\)</td></tr> <tr><td>Total</td><td>\(0.625\)</td><td>\(0.375\)</td><td>\(1\)</td></tr> </table> The new denominator is \(80\), because \(20\) students did not answer yes or no.
Joint, marginal, and conditional relative frequency 9
Two row groups have the same conditional distribution: <table> <tr><th></th><th>choice X</th><th>choice Y</th><th>choice Z</th></tr> <tr><td>Group 1</td><td>\(0.20\)</td><td>\(0.50\)</td><td>\(0.30\)</td></tr> <tr><td>Group 2</td><td>\(0.20\)</td><td>\(0.50\)</td><td>\(0.30\)</td></tr> </table> The entries are row conditional relative frequencies. What descriptive conclusion can be made about the association between group and choice? Why are row totals unnecessary for this comparison?

Hints

- Compare corresponding percentages across the two rows. - Association appears as a change in the conditional distribution. - Normalization removes the effect of unequal group sizes.

Solution

1. Each choice has the same conditional relative frequency in both groups. 2. Therefore the displayed conditional distributions show no descriptive association between group and choice. 3. Row totals are unnecessary because the comparison is already normalized within each group.

Answer

The two conditional distributions are identical, so the table shows no descriptive association between group and choice. Row totals are not needed because each row is already expressed relative to its own group.
Joint, marginal, and conditional relative frequency 10
Convert the following count table to column conditional relative frequencies. <table> <tr><th></th><th>completed survey</th><th>did not complete</th></tr> <tr><td>email invitation</td><td>\(72\)</td><td>\(18\)</td></tr> <tr><td>text invitation</td><td>\(48\)</td><td>\(42\)</td></tr> </table> Each column must sum to \(1\).

Hints

- Find each column total before dividing. - Use a different denominator for each column. - Check each column, not each row, for a sum of \(1\).

Solution

1. The completed-survey column total is \(120\), giving \(72/120=0.60\) and \(48/120=0.40\). 2. The did-not-complete column total is \(60\), giving \(18/60=0.30\) and \(42/60=0.70\). 3. Each column sums to \(1\).

Answer

<table> <tr><th></th><th>completed survey</th><th>did not complete</th></tr> <tr><td>email invitation</td><td>\(0.60\)</td><td>\(0.30\)</td></tr> <tr><td>text invitation</td><td>\(0.40\)</td><td>\(0.70\)</td></tr> </table>
Joint, marginal, and conditional relative frequency 11
A school surveyed students about how they got to school and whether they arrived late last Friday. <table> <tr><th></th><th>Late</th><th>On time</th><th>Total</th></tr> <tr><td>Bus</td><td>\(18\)</td><td>\(42\)</td><td>\(60\)</td></tr> <tr><td>Walk</td><td>\(16\)</td><td>\(64\)</td><td>\(80\)</td></tr> <tr><td>Total</td><td>\(34\)</td><td>\(106\)</td><td>\(140\)</td></tr> </table> a) Find the percent late within each travel group. b) Which group had a higher late rate? Someone says walkers were more likely to be late because there were nearly as many late walkers as late bus riders. Explain the mistake.

Hints

- How many bus riders and walkers were surveyed? - Find the share late within each group before comparing. - Why might similar counts give different percentages?

Solution

1. Among bus riders, \(18/60=30\%\) were late. Among walkers, \(16/80=20\%\) were late. 2. The bus group had the higher rate, even though the two late counts are close. 3. A group rate divides by that group's total, not by all \(140\) students or by the count in another group.

Answer

a) Bus: \(30\%\); walk: \(20\%\). b) Bus riders. The groups contain different numbers of students, so comparing late counts alone is misleading.
Joint, marginal, and conditional relative frequency 12
A sample has \(400\) observations and the joint relative-frequency table: <table> <tr><th></th><th>yes</th><th>no</th></tr> <tr><td>Group A</td><td>\(0.18\)</td><td>\(0.27\)</td></tr> <tr><td>Group B</td><td>\(0.32\)</td><td>\(0.23\)</td></tr> </table> Convert every joint relative frequency to a count and add all margins.

Hints

- Every joint frequency uses the same grand-total denominator. - Multiply rather than divide when recovering counts. - Verify that the four counts sum to the sample size.

Solution

1. Multiply each joint relative frequency by \(400\): \(72\), \(108\), \(128\), and \(92\). 2. Row totals are \(180\) and \(220\). 3. Column totals are \(200\) and \(200\), with grand total \(400\).

Answer

<table> <tr><th></th><th>yes</th><th>no</th><th>Total</th></tr> <tr><td>Group A</td><td>\(72\)</td><td>\(108\)</td><td>\(180\)</td></tr> <tr><td>Group B</td><td>\(128\)</td><td>\(92\)</td><td>\(220\)</td></tr> <tr><td>Total</td><td>\(200\)</td><td>\(200\)</td><td>\(400\)</td></tr> </table>
Joint, marginal, and conditional relative frequency 13
A table displays these percentages: <table> <tr><th></th><th>uses app</th><th>does not use app</th></tr> <tr><td>under 30</td><td>\(40\%\)</td><td>\(60\%\)</td></tr> <tr><td>30 or older</td><td>\(25\%\)</td><td>\(75\%\)</td></tr> </table> Can these entries be interpreted as joint relative frequencies? What interpretation is supported by the row sums?

Hints

- Check what total the full table has. - Then check each row and each column separately. - The location of the \(100\%\) sum identifies the denominator structure.

Solution

1. The four entries sum to \(200\%\), so they cannot all be joint relative frequencies for one sample. 2. Each row sums to \(100\%\). 3. The supported interpretation is row conditional relative frequencies: app use within each age group.

Answer

No. They are row conditional relative frequencies, because each row—not the whole table—sums to \(100\%\).
Joint, marginal, and conditional relative frequency 14
Convert this count table to a complete joint relative-frequency table. <table> <tr><th></th><th>morning</th><th>evening</th></tr> <tr><td>weekday</td><td>\(45\)</td><td>\(30\)</td></tr> <tr><td>weekend</td><td>\(15\)</td><td>\(10\)</td></tr> </table> Include margins and express every value as a decimal.

Hints

- Find the grand total before converting any cell. - Joint relative frequencies all use that one denominator. - Margins can be found by adding relative frequencies or converting count totals.

Solution

1. The grand total is \(100\). 2. Divide every count by \(100\), giving joint cells \(0.45,0.30,0.15,0.10\). 3. Row margins are \(0.75,0.25\); column margins are \(0.60,0.40\).

Answer

<table> <tr><th></th><th>morning</th><th>evening</th><th>Total</th></tr> <tr><td>weekday</td><td>\(0.45\)</td><td>\(0.30\)</td><td>\(0.75\)</td></tr> <tr><td>weekend</td><td>\(0.15\)</td><td>\(0.10\)</td><td>\(0.25\)</td></tr> <tr><td>Total</td><td>\(0.60\)</td><td>\(0.40\)</td><td>\(1.00\)</td></tr> </table>
Joint, marginal, and conditional relative frequency 15
A certification program compares pass rates for two preparation formats in two years. <table> <tr><th></th><th>Self-paced</th><th>Instructor-led</th></tr> <tr><td>Year 1</td><td>\(36/60\)</td><td>\(28/40\)</td></tr> <tr><td>Year 2</td><td>\(63/90\)</td><td>\(36/60\)</td></tr> </table> For each year, compute the self-paced pass rate minus the instructor-led pass rate. Then describe how the conditional-rate gap changed from Year 1 to Year 2.

Hints

- Form each rate within its own year-and-format group. - Keep the subtraction order the same in both years. - Compare both the magnitude and the sign of the two gaps.

Solution

1. In Year 1, the rates are \(0.60\) and \(0.70\), so the gap is \(0.60-0.70=-0.10\). 2. In Year 2, the rates are \(0.70\) and \(0.60\), so the gap is \(0.70-0.60=0.10\). 3. The gap increased by \(0.20\), or \(20\) percentage points, and its direction reversed.

Answer

Year 1 gap: \(-10\) percentage points. Year 2 gap: \(+10\) percentage points. The self-paced-minus-instructor-led gap increased by \(20\) percentage points and reversed direction.
Joint, marginal, and conditional relative frequency 16
A customer report says, “\(80\%\) of the customers who renewed their subscriptions signed up online.” A manager restates this as “\(80\%\) of online customers renewed.” Is that justified? Explain which group each percentage would be based on.

Hints

- Which group follows the word “of” in the original statement? - Which group is the whole in the manager's version? - Can one count give two different percentages when the denominators differ?

Solution

1. The report uses all customers who renewed as its reference group. 2. The manager's statement uses all online customers as the reference group. 3. The groups can be different sizes, so the first percentage does not determine the second.

Answer

No. The \(80\%\) is a share of customers who renewed. To find the percentage of online customers who renewed, we would need the number of online customers as the denominator.
Joint, marginal, and conditional relative frequency 17
A row-conditional relative-frequency table compares two groups across three outcomes. <table> <tr><th></th><th>Outcome X</th><th>Outcome Y</th><th>Outcome Z</th></tr> <tr><td>Group 1</td><td>\(0.20\)</td><td>\(0.50\)</td><td>\(0.30\)</td></tr> <tr><td>Group 2</td><td>\(0.35\)</td><td>\(0.40\)</td><td>\(0.25\)</td></tr> </table> For each outcome, find the absolute difference between the two conditional frequencies. Which outcome contributes the largest visible difference between the group profiles? Explain why unequal group sizes would not change these three within-row comparisons.

Hints

- Compare corresponding entries rather than raw group counts. - Use absolute differences so direction does not affect the size of a gap. - Recall what denominator is used in a row-conditional table.

Solution

1. The absolute differences are \(|0.20-0.35|=0.15\), \(|0.50-0.40|=0.10\), and \(|0.30-0.25|=0.05\). 2. Outcome X has the largest conditional-frequency difference. 3. Each row is already normalized within its own group, so changing the number of observations in a group would change joint frequencies but not these within-row rates.

Answer

The absolute differences are \(0.15,0.10,\) and \(0.05\). Outcome X differs most. Unequal row sizes do not alter row-conditional comparisons.
Joint, marginal, and conditional relative frequency 18
Panel a) shows morning-shift employees and panel b) shows evening-shift employees. In each panel, R means remote and O means on-site. a) Find the conditional relative frequency of remote work within each shift. b) Which shift has the higher remote-work conditional relative frequency, and by how many percentage points?
Figure for problem 556191

Hints

- Treat each panel as one conditioning group. - Within a panel, the denominator is the sum of its R and O counts. - Compare the two resulting proportions with a percentage-point difference.

Solution

1. In panel a), the morning shift has \(30\) remote and \(50\) on-site employees, for \(80\) total. Its remote conditional relative frequency is \(\frac{30}{80}=0.375\). 2. In panel b), the evening shift has \(40\) remote and \(60\) on-site employees, for \(100\) total. Its remote conditional relative frequency is \(\frac{40}{100}=0.40\). 3. The evening-shift rate exceeds the morning-shift rate by \(0.40-0.375=0.025\), or \(2.5\) percentage points.

Answer

a) Morning: \(0.375\); evening: \(0.40\). b) The evening shift is higher by \(2.5\) percentage points.
Joint, marginal, and conditional relative frequency 19
Find the marginal relative frequency of bus riders among all students. State which bars you counted.
Figure for problem 563286

Hints

- A marginal total ignores one of the two category labels; which one is irrelevant here? - Why should you not use only one bus bar? - Check whether your denominator covers students who did not ride the bus.

Solution

1. Both bus bars belong in the bus margin: \(12+4=16\). 2. The six bars total \(12+4+6+8+7+3=40\). 3. The marginal bus share is \(16/40=0.40\).

Answer

Both bus bars contribute \(12+4=16\) riders. All six bars total \(40\) students, so \(\frac{16}{40}=0.40\), or \(40\%\).
Joint, marginal, and conditional relative frequency 20
Of \(20\) hikers in the Pine Trail group, \(8\) bike to the trailhead. Of \(40\) hikers in the River Trail group, \(8\) bike there. Which hiking group has the higher relative frequency of bike riders? Explain why the equal bike counts do not settle the comparison.

Hints

- Could equally many cyclists make up different shares of two groups? - Imagine adding more noncyclists to one group. What happens to its bike-riding rate? - Are you comparing raw counts or parts of two different wholes?

Solution

1. The Pine Trail rate is \(\frac{8}{20}=40\%\). 2. The River Trail rate is \(\frac{8}{40}=20\%\). 3. Pine Trail has the higher rate because the same count is a larger share of its smaller group.

Answer

Pine Trail: \(40\%\), compared with River Trail: \(20\%\). The groups have different sizes.
Joint, marginal, and conditional relative frequency 21
A school recorded each student's trip to school and whether they stayed for an after-school club. <table> <tr><th></th><th>Stayed</th><th>Did not stay</th><th>Total</th></tr> <tr><td>Bus</td><td>\(\square\)</td><td>\(\square\)</td><td>\(80\)</td></tr> <tr><td>Walk</td><td>\(42\)</td><td>\(78\)</td><td>\(120\)</td></tr> <tr><td>Total</td><td>\(\square\)</td><td>\(\square\)</td><td>\(200\)</td></tr> </table> Among bus riders, \(35\%\) stayed for a club. Complete the two missing bus cells and the column totals. Why is \(200\) not the right number to multiply by \(35\%\)?

Hints

- Which group does the percentage describe? - How is the other bus cell related to the bus row total? - Can the two column totals confirm your completed row?

Solution

1. The conditional rate refers to bus riders, so \(0.35\cdot80=28\) stayed. 2. The remaining bus riders are \(80-28=52\). 3. Column totals are \(28+42=70\) and \(52+78=130\). The grand total \(200\) is not the bus group's size.

Answer

Bus stayed: \(28\); bus did not stay: \(52\); column totals: \(70\) stayed and \(130\) did not stay. The rate is based on \(80\) bus riders, not all \(200\) students.
Joint, marginal, and conditional relative frequency 22
The table shows \(60\) visitors to a museum workshop. <table> <tr><th></th><th>First visit</th><th>Returning</th></tr> <tr><td>Teen</td><td>\(15\)</td><td>\(5\)</td></tr> <tr><td>Adult</td><td>\(10\)</td><td>\(30\)</td></tr> </table> a) Among teens, what fraction are first-time visitors? b) Among first-time visitors, what fraction are teens? c) Why are the two fractions different?

Hints

- Highlight the group named after “among” in each question. - The two questions use the same cell but not the same total. - One denominator is a row sum; the other is a column sum.

Solution

1. The teen total is \(15+5=20\); the first-time total is \(15+10=25\). 2. Among teens, \(\frac{15}{20}=\frac{3}{4}\) are first-time visitors. 3. Among first-time visitors, \(\frac{15}{25}=\frac{3}{5}\) are teens. 4. The same interior count is divided by different group totals.

Answer

a) \(\frac{3}{4}\). b) \(\frac{3}{5}\). c) The reference group changes from all teens to all first-time visitors.
Joint, marginal, and conditional relative frequency 23
Among morning-session students, what fraction passed the quiz? What fraction of afternoon-session students passed? Which session had the higher pass rate?
Figure for problem 563292

Hints

- Does each session have the same number of students? - Which bars tell you how many students were in one session altogether? - Could the larger pass count belong to the group with the lower pass rate?

Solution

1. In the morning, \(30\) passed and \(10\) did not, so \(\frac{30}{40}=\frac{3}{4}\) passed. 2. In the afternoon, \(20\) passed and \(10\) did not, so \(\frac{20}{30}=\frac{2}{3}\) passed. 3. Since \(\frac{3}{4}>\frac{2}{3}\), the morning group had the higher pass rate.

Answer

Morning: \(\frac{3}{4}\) (\(75\%\)); afternoon: \(\frac{2}{3}\) (about \(66.7\%\)). Morning was higher.
Joint, marginal, and conditional relative frequency 24
Two campuses used the same two categorical variables. Their joint relative-frequency tables are shown below. Campus A, \(n=80\): <table> <tr><th></th><th>Completed</th><th>Did not complete</th></tr> <tr><td>Online</td><td>\(0.30\)</td><td>\(0.20\)</td></tr> <tr><td>In person</td><td>\(0.10\)</td><td>\(0.40\)</td></tr> </table> Campus B, \(n=220\): <table> <tr><th></th><th>Completed</th><th>Did not complete</th></tr> <tr><td>Online</td><td>\(0.10\)</td><td>\(0.30\)</td></tr> <tr><td>In person</td><td>\(0.25\)</td><td>\(0.35\)</td></tr> </table> Construct the pooled joint relative-frequency table for all \(300\) participants. Explain why averaging corresponding cells of the two displayed tables would be incorrect.

Hints

- Recover each campus’s cell counts before combining the data. - A pooled relative frequency uses the combined grand total. - Ask what weighting an ordinary average of the two table entries would impose.

Solution

1. Convert Campus A to counts: \(\begin{pmatrix}24&16\\8&32\end{pmatrix}\). 2. Convert Campus B to counts: \(\begin{pmatrix}22&66\\55&77\end{pmatrix}\). 3. Add corresponding counts to obtain \(\begin{pmatrix}46&82\\63&109\end{pmatrix}\) out of \(300\). 4. Divide each count by \(300\): the pooled joint frequencies are approximately \(\begin{pmatrix}0.153&0.273\\0.210&0.363\end{pmatrix}\). 5. A cellwise average would give equal weight to campuses of sizes \(80\) and \(220\), so it would not represent the combined population.

Answer

<table> <tr><th></th><th>Completed</th><th>Did not complete</th></tr> <tr><td>Online</td><td>\(0.153\)</td><td>\(0.273\)</td></tr> <tr><td>In person</td><td>\(0.210\)</td><td>\(0.363\)</td></tr> </table> The entries are rounded. Pool counts first because the campuses have unequal sample sizes.
Joint, marginal, and conditional relative frequency 25
A recreation center recorded \(137\) separate visits. Each visit was classified by day and whether the pool was used. <table> <tr><th></th><th>used the pool</th><th>did not use the pool</th><th>Total</th></tr> <tr><td>weekday visit</td><td>\(36\)</td><td>\(27\)</td><td>\(63\)</td></tr> <tr><td>weekend visit</td><td>\(32\)</td><td>\(42\)</td><td>\(74\)</td></tr> <tr><td>Total</td><td>\(68\)</td><td>\(69\)</td><td>\(137\)</td></tr> </table> Anouk uses \(42/69\) as the joint relative frequency for “weekend visit and did not use the pool.” Explain her error. Find the correct joint relative frequency to three decimal places. Also find the percentage of no-pool visits that happened on weekends, to one decimal place.

Hints

- Which denominator represents all visits? - What group of visits does the column total \(69\) describe? - Check that the two requested rates refer to different reference groups.

Solution

1. A joint relative frequency uses all \(137\) visits, so it is \(42/137\approx0.307\). 2. The \(69\) in Anouk's denominator counts only visits when the pool was not used. 3. Within that group, the weekend share is \(42/69\approx0.609\), or about \(60.9\%\).

Answer

Joint relative frequency: \(42/137\approx0.307\). Anouk used a subgroup total instead of the grand total. Among visits without pool use, the weekend share is \(42/69\approx0.609\), or about \(60.9\%\).
Joint, marginal, and conditional relative frequency 26
Two school surveys record whether students bring lunch from home and whether they take the bus. West High: <table> <tr><th></th><th>Bus</th><th>Not bus</th><th>Total</th></tr> <tr><td>Lunch from home</td><td>\(0.25\)</td><td>\(0.25\)</td><td>\(0.50\)</td></tr> <tr><td>Other lunch</td><td>\(0.25\)</td><td>\(0.25\)</td><td>\(0.50\)</td></tr> <tr><td>Total</td><td>\(0.50\)</td><td>\(0.50\)</td><td>\(1.00\)</td></tr> </table> East High: <table> <tr><th></th><th>Bus</th><th>Not bus</th><th>Total</th></tr> <tr><td>Lunch from home</td><td>\(0.45\)</td><td>\(0.05\)</td><td>\(0.50\)</td></tr> <tr><td>Other lunch</td><td>\(0.05\)</td><td>\(0.45\)</td><td>\(0.50\)</td></tr> <tr><td>Total</td><td>\(0.50\)</td><td>\(0.50\)</td><td>\(1.00\)</td></tr> </table> Check that the two schools have the same row and column totals. Among students who bring lunch from home, what percentage take the bus at each school? What do the matching totals fail to show?

Hints

- Compare the outside totals before looking inside each table. - Which row is the reference group for the requested percentage? - Can the same row totals hide different cell values?

Solution

1. Each row total and each column total is \(0.50\) at both schools. 2. At West High, the bus share among students who bring lunch is \(0.25/0.50=0.50\), or \(50\%\). 3. At East High, that share is \(0.45/0.50=0.90\), or \(90\%\). 4. Matching row and column totals do not show how the categories are paired within the table.

Answer

Both schools have row and column totals of \(0.50\) each. Among students who bring lunch, \(50\%\) at West High and \(90\%\) at East High take the bus. The totals do not show this difference.
Joint, marginal, and conditional relative frequency 27
A count table initially has rows \(A=(20,30)\) and \(B=(30,20)\). Every record in row A is duplicated once, while row B is unchanged. Compare the row-conditional distribution for A, the row marginal for A, and the first-column marginal before and after duplication.

Hints

- Separate within-row denominators from the grand-total denominator. - Duplicating every record in one row preserves that row’s internal proportions. - Recompute margins using the new grand total.

Solution

1. Initially, A’s conditional distribution is \((20/50,30/50)=(0.40,0.60)\), its row margin is \(0.50\), and the first-column margin is \(50/100=0.50\). 2. After duplication, A becomes \((40,60)\), B remains \((30,20)\), and the new total is \(150\). 3. A’s conditional distribution remains \((0.40,0.60)\). 4. A’s row margin becomes \(100/150=2/3\), and the first-column margin becomes \(70/150=7/15\approx0.467\).

Answer

A’s row-conditional distribution remains \((0.40,0.60)\). Its row margin changes from \(0.50\) to \(\frac23\), and the first-column margin changes from \(0.50\) to \(\frac{7}{15}\approx0.467\).
Joint, marginal, and conditional relative frequency 28
A survey of \(200\) residents records location and support for a transit proposal. Some responses are unresolved. <table> <tr><th></th><th>Support</th><th>Oppose</th><th>Unresolved</th><th>Total</th></tr> <tr><td>Urban</td><td>\(72\)</td><td>\(36\)</td><td>\(12\)</td><td>\(120\)</td></tr> <tr><td>Rural</td><td>\(28\)</td><td>\(44\)</td><td>\(8\)</td><td>\(80\)</td></tr> </table> Assume every unresolved response will eventually be classified as either support or oppose. a) Find the possible interval for the overall support relative frequency. b) Find the possible interval for the conditional support relative frequency among urban residents. c) Explain why reporting either unresolved group as all support or all oppose gives a bound rather than an estimate.

Hints

- Keep unresolved observations in the denominator because they remain part of the survey. - To obtain an extreme rate, place every unresolved case on the same side. - Distinguish a range of logically possible values from a prediction.

Solution

1. There are \(100\) confirmed supporters and \(20\) unresolved responses. The final support count can range from \(100\) to \(120\), so the overall relative frequency lies in \([0.50,0.60]\). 2. Among urban residents, the support count can range from \(72\) to \(84\) out of \(120\). The conditional interval is \([0.60,0.70]\). 3. Assigning every unresolved case to one endpoint gives the smallest or largest possible value. It does not describe a typical or most likely allocation.

Answer

a) Overall support relative frequency: \([0.50,0.60]\). b) Urban conditional support relative frequency: \([0.60,0.70]\). c) The endpoint assignments are worst-case bounds, not estimates of how unresolved responses will actually split.
Joint, marginal, and conditional relative frequency 29
A training program has two departments. Department A has \(20\) employees and an \(80\%\) completion rate. Department B has \(180\) employees and a \(40\%\) completion rate. a) Find the unweighted mean of the two department completion rates. b) Find the completion relative frequency among all \(200\) employees. c) Explain what population each result describes and why neither calculation is automatically “the correct average.”

Hints

- Decide whether departments or employees are the units receiving equal weight. - Convert each department rate to a count before pooling employees. - Interpret a numerical average by naming its reference population.

Solution

1. The unweighted department mean is \((0.80+0.40)/2=0.60\). 2. Completed counts are \(0.80\cdot20=16\) and \(0.40\cdot180=72\). The employee-level rate is \((16+72)/200=0.44\). 3. The value \(0.60\) treats the two departments as equally weighted units. The value \(0.44\) treats each employee as equally weighted. The appropriate result depends on the stated reference population.

Answer

a) Unweighted department mean: \(0.60\). b) Overall employee completion relative frequency: \(0.44\). c) The first describes an equally weighted “typical department”; the second describes a randomly selected employee.
Joint, marginal, and conditional relative frequency 30
A report gives \(P(A)=0.40\), \(P(B)=0.50\), and \(P(A\cap B)=0.18\). It also states that \(P(B\mid A)=0.60\). a) Check whether all four values can describe the same two-way relative-frequency table. b) If the first three values are correct, find the correct value of \(P(B\mid A)\). c) Construct the corresponding \(2\times2\) joint relative-frequency table.

Hints

- A conditional relative frequency, its conditioning margin, and the corresponding joint cell must agree. - Use the given margins to determine the other cells after checking the disputed value. - Verify every row sum, column sum, and the grand total.

Solution

1. The first three values imply \(P(B\mid A)=P(A\cap B)/P(A)=0.18/0.40=0.45\), not \(0.60\). The four claims are inconsistent. 2. The remaining cells are \(P(A\cap B^c)=0.40-0.18=0.22\), \(P(A^c\cap B)=0.50-0.18=0.32\), and \(P(A^c\cap B^c)=1-(0.18+0.22+0.32)=0.28\). 3. These cells have the stated margins and total \(1\).

Answer

a) No; the reported conditional value is inconsistent. b) \(P(B\mid A)=0.45\). c) <table> <tr><th></th><th>\(B\)</th><th>\(B^c\)</th><th>Total</th></tr> <tr><td>\(A\)</td><td>\(0.18\)</td><td>\(0.22\)</td><td>\(0.40\)</td></tr> <tr><td>\(A^c\)</td><td>\(0.32\)</td><td>\(0.28\)</td><td>\(0.60\)</td></tr> <tr><td>Total</td><td>\(0.50\)</td><td>\(0.50\)</td><td>\(1.00\)</td></tr> </table>
Joint, marginal, and conditional relative frequency 31
A joint relative-frequency table shows four cells rounded to the nearest hundredth: \(0.24\), \(0.26\), \(0.25\), and \(0.24\). Add the displayed values. Does their total mean some observations are missing? Give one set of unrounded cell values that adds to \(1\) and rounds to these values.

Hints

- What total must a complete joint relative-frequency table have before rounding? - Can more than one number round to the same hundredth? - Try adjusting the four values slightly while keeping their total fixed.

Solution

1. All cells in a complete joint relative-frequency table must total \(1\) before rounding. 2. One possible unrounded set is \(0.242\), \(0.262\), \(0.252\), and \(0.244\). 3. These values sum to \(1.000\), and each rounds to the matching displayed value. 4. Therefore rounding alone can account for the displayed sum of \(0.99\); there is no evidence that observations are missing.

Answer

The displayed total is \(0.99\). No observations need be missing: for example, \(0.242\), \(0.262\), \(0.252\), and \(0.244\) add to \(1.000\) and round to the displayed cells.
Joint, marginal, and conditional relative frequency 32
Two departments report internship placements. Department A: \(36\) placements among \(60\) applicants. Department B: \(45\) placements among \(90\) applicants. Which department has the larger joint relative frequency in the combined sample? Which has the larger conditional placement rate within its own applicants? Explain why the rankings need not match.

Hints

- Joint frequencies use the combined grand total. - Conditional rates use each department’s own applicant total. - Unequal group sizes can reverse the ranking.

Solution

1. In the combined sample of \(150\), joint placement frequencies are \(\frac{36}{150}=0.24\) for A and \(\frac{45}{150}=0.30\) for B. 2. Conditional placement rates are \(\frac{36}{60}=0.60\) for A and \(\frac{45}{90}=0.50\) for B. 3. B has the larger joint share because it has more applicants; A has the larger within-department rate.

Answer

Department B has the larger joint relative frequency, \(0.30\) versus \(0.24\). Department A has the larger conditional placement rate, \(0.60\) versus \(0.50\).
Joint, marginal, and conditional relative frequency 33
A cell in a relative-frequency table has joint relative frequency \(j>0\). The row containing that cell represents a fraction \(b\) of the whole, where \(0<b\le1\). The within-row relative frequency for that cell is \(j/b\). Can \(j/b\) be smaller than \(j\)? When are the two values equal?

Hints

- Which denominator describes the whole table, and which describes just the row? - Can a row's share of the table be greater than \(1\)? - Think about when dividing by a fraction leaves a positive number unchanged.

Solution

1. The row accounts for at most the whole table, so \(0<b\le1\). 2. Dividing the positive \(j\) by \(b\le1\) cannot make it smaller, so \(j/b\ge j\). 3. The two values are equal when \(b=1\). If \(b<1\), the within-row frequency is strictly larger.

Answer

No. \(j/b\ge j\), with equality only when \(b=1\). If the row is less than the entire table, the within-row value is larger.
Joint, marginal, and conditional relative frequency 34
Row marginal relative frequencies are \(0.30\) for Group A and \(0.70\) for Group B. Their row conditional distributions for outcomes \(X\) and \(Y\) are \((0.80,0.20)\) and \((0.40,0.60)\). Find the column conditional distribution of group membership among observations with outcome \(X\).

Hints

- Convert row-conditional values to joint frequencies first. - Add the relevant joint cells for the column margin. - Renormalize within the column.

Solution

1. Joint \(X\) frequencies are \(0.30\cdot0.80=0.24\) for A and \(0.70\cdot0.40=0.28\) for B. 2. The \(X\) marginal is \(0.24+0.28=0.52\). 3. Within outcome \(X\), the group distribution is \(\frac{0.24}{0.52}=\frac{6}{13}\) for A and \(\frac{0.28}{0.52}=\frac{7}{13}\) for B.

Answer

Among observations with outcome \(X\), Group A has relative frequency \(\frac{6}{13}\) and Group B has relative frequency \(\frac{7}{13}\).
Joint, marginal, and conditional relative frequency 35
A company has two shifts. The day shift has an on-time rate of \(0.92\), and the night shift has an on-time rate of \(0.80\). Overall, the on-time rate is \(0.875\). Find the marginal relative frequency of employees on the day shift.

Hints

- Express the overall rate as a weighted combination of the two shift rates. - Let one group’s marginal share be the unknown. - The two group shares must add to \(1\).

Solution

1. Let \(x\) be the day-shift share. Then \(0.92x+0.80(1-x)=0.875\). 2. Solving gives \(0.12x=0.075\), so \(x=0.625\). 3. The night-shift share is \(0.375\).

Answer

The day-shift marginal relative frequency is \(0.625\), or \(62.5\%\).
Joint, marginal, and conditional relative frequency 36
A clinic reports vaccination rates of \(0.74\) for Group A and \(0.61\) for Group B. Group A has \(200\) people and Group B has \(300\). Find the percentage-point difference in rates and the difference in vaccinated counts.

Hints

- A rate difference and a count difference answer different questions. - Convert each rate using its own group size. - Do not infer the larger count from the larger percentage.

Solution

1. The rate difference is \(0.74-0.61=0.13\), or \(13\) percentage points. 2. Vaccinated counts are \(0.74\cdot200=148\) and \(0.61\cdot300=183\). 3. Group B has \(35\) more vaccinated people despite its lower rate.

Answer

The rate difference is \(13\) percentage points in favor of Group A. Group B has \(35\) more vaccinated people, \(183\) versus \(148\).
Joint, marginal, and conditional relative frequency 37
Two school music groups asked each member whether they prefer afternoon or morning rehearsals. The table shows shares of everyone surveyed. Each student belongs to just one of the two groups. <table> <tr><th></th><th>Afternoon</th><th>Morning</th></tr> <tr><td>Jazz band</td><td>\(0.30\)</td><td>\(0.20\)</td></tr> <tr><td>Choir</td><td>\(0.10\)</td><td>\(0.40\)</td></tr> </table> a) Among jazz band students, what percentage prefer afternoon rehearsals? What percentage of choir students prefer afternoon rehearsals? b) Among **all students who prefer afternoons**, what percentage are in jazz band? c) Explain why the percentage in part b) is different from the jazz band percentage in part a).

Hints

- Find each music group's total before calculating percentages within it. - For part b), which category names the group being considered? - Does the same cell have to represent the same percentage when its denominator changes?

Solution

1. The jazz band row sums to \(0.50\), so \(0.30/0.50=60\%\) prefer afternoons. 2. The choir row sums to \(0.50\), so \(0.10/0.50=20\%\) prefer afternoons. 3. The afternoon column sums to \(0.40\). Among those students, \(0.30/0.40=75\%\) are in jazz band. 4. The two jazz band percentages use different reference groups: all jazz band students versus all afternoon-preferring students.

Answer

a) Jazz band: \(60\%\); choir: \(20\%\). b) \(75\%\). c) Part a) uses all jazz band students, while part b) uses all students who prefer afternoons.
Joint, marginal, and conditional relative frequency 38
A dashboard displays the value \(0.27\), rounded to the nearest hundredth using the usual school rule that a halfway value rounds up. The full sample contains \(200\) observations, and the relevant row contains \(120\) observations. a) If \(0.27\) is a joint relative frequency, what integer cell counts are possible? b) If \(0.27\) is instead the conditional relative frequency within that row, what integer cell counts are possible? c) Explain why the two interpretations give different possible counts even though the displayed decimal is the same.

Hints

- Identify the denominator for each interpretation before doing any rounding work. - Translate the displayed decimal into the interval of exact values that would round to it. - A table cell contains a whole-number count, so keep only integers compatible with each denominator.

Solution

1. A value that rounds to \(0.27\) lies in \([0.265,0.275)\). 2. For a joint relative frequency, the denominator is the full sample of \(200\). Thus the cell count \(k\) must satisfy \(53\le k<55\), giving \(k=53\) or \(k=54\). 3. For a row-conditional relative frequency, the denominator is \(120\). Thus \(31.8\le k<33\), so the only integer possibility is \(k=32\). 4. Joint and conditional relative frequencies use different denominators, so the same rounded decimal need not represent the same count.

Answer

a) \(53\) or \(54\) b) \(32\) c) A joint relative frequency uses all \(200\) observations as its denominator, while the row-conditional relative frequency uses only the \(120\) observations in that row.
Joint, marginal, and conditional relative frequency 39
Two rows of a conditional relative-frequency table use the same three displayed percentages, each rounded to the nearest whole percent. <table> <tr><th>Group</th><th>Outcome 1</th><th>Outcome 2</th><th>Outcome 3</th><th>Row total</th></tr> <tr><td>A</td><td>\(17\%\)</td><td>\(42\%\)</td><td>\(42\%\)</td><td>\(12\)</td></tr> <tr><td>B</td><td>\(17\%\)</td><td>\(42\%\)</td><td>\(42\%\)</td><td>\(24\)</td></tr> </table> a) Explain why each displayed row may total \(101\%\) without making the table invalid. b) Recover the three integer cell counts in each row. c) There are \(36\) observations altogether. For Outcome 2, compute the joint relative frequency for Group A and for Group B. Explain why the nearly identical conditional percentages do not imply equal joint relative frequencies.

Hints

- Conditional percentages in a row use that row total as their denominator. - Convert each rounded percentage back to a count that is compatible with the stated row total. - For a joint relative frequency, switch to the grand total as the denominator.

Solution

1. Rounding each conditional percentage separately can make the displayed row total differ slightly from \(100\%\). 2. In Group A, the counts are \(2,5,5\), since \(2/12\approx16.67\%\) and \(5/12\approx41.67\%\). 3. In Group B, the counts are \(4,10,10\), since \(4/24\approx16.67\%\) and \(10/24\approx41.67\%\). 4. For Outcome 2, the joint relative frequencies are \(5/36\approx0.139\) for Group A and \(10/36\approx0.278\) for Group B. 5. Conditional percentages use different row denominators, while joint relative frequencies use the common grand total of \(36\).

Answer

a) The \(101\%\) total is a rounding effect. b) Group A: \((2,5,5)\). Group B: \((4,10,10)\). c) Group A and Outcome 2: \(\frac{5}{36}\approx0.139\). Group B and Outcome 2: \(\frac{10}{36}\approx0.278\). The conditional percentages use different row totals, so they do not determine equal joint shares of the full sample.
Joint, marginal, and conditional relative frequency 40
The table shows exact joint relative frequencies for customers at a bookstore. <table> <tr><th></th><th>Uses store app</th><th>Does not use app</th></tr> <tr><td>New customer</td><td>\(12.5\%\)</td><td>\(37.5\%\)</td></tr> <tr><td>Returning customer</td><td>\(25\%\)</td><td>\(25\%\)</td></tr> </table> a) Find all row and column marginal relative frequencies. b) Find the percentage of **new customers** who use the app and the percentage of **app users** who are returning customers. c) A clerk says, “Since the table says \(25\%\) for returning app users, only \(25\%\) of returning customers use the app.” Is that correct? Explain.

Hints

- Find the marginal totals before working within a smaller group. - In each part, identify whether the reference group is a row, a column, or the whole table. - Does an interior percentage necessarily tell you the percentage within one row?

Solution

1. Row margins are \(50\%\) new and \(50\%\) returning; column margins are \(37.5\%\) use the app and \(62.5\%\) do not. 2. App use among new customers is \(12.5/50=25\%\). 3. Returning customers among app users are \(25/37.5=\frac{2}{3}\). 4. The clerk incorrectly treats a joint share as a within-row share. App use among returning customers is \(25/50=50\%\).

Answer

a) New: \(50\%\); returning: \(50\%\); app use: \(37.5\%\); no app: \(62.5\%\). b) New customers who use the app: \(25\%\). App users who are returning customers: \(\frac{2}{3}\), or about \(66.7\%\). c) No. The \(25\%\) is a joint share of all customers; \(50\%\) of returning customers use the app.
Joint, marginal, and conditional relative frequency 41
Two groups have conditional success rates \(0.75\) and \(0.35\). The share of the population in the first group is not known exactly, but it lies between \(0.20\) and \(0.50\). Find the complete range of possible marginal success rates. State where the smallest and largest values occur.

Hints

- Express the overall rate as a weighted average using one unknown group share. - Use the fact that the two group shares add to \(1\). - Determine whether the weighted average increases or decreases as the unknown share changes.

Solution

1. Let \(w\) be the first group’s population share. The marginal rate is \(0.75w+0.35(1-w)=0.35+0.40w\). 2. This expression increases as \(w\) increases. 3. At \(w=0.20\), the marginal rate is \(0.35+0.40(0.20)=0.43\). 4. At \(w=0.50\), it is \(0.35+0.40(0.50)=0.55\).

Answer

The possible marginal success rates form the interval \([0.43,0.55]\). The minimum occurs when the first group’s share is \(0.20\), and the maximum occurs when it is \(0.50\).
Joint, marginal, and conditional relative frequency 42
A table has three rows with counts: \( (18,12),\quad(30,20),\quad(24,36). \) The first two rows will be combined. Find the conditional distribution of the two columns within the combined row, and compare it with the third row’s conditional distribution.

Hints

- Add counts before computing the combined conditional percentages. - Use the new combined row total as its denominator. - Normalize the comparison row separately.

Solution

1. The combined first-two rows are \((18+30,12+20)=(48,32)\), with total \(80\). 2. Its conditional distribution is \((0.60,0.40)\). 3. The third row totals \(60\), with distribution \((0.40,0.60)\).

Answer

The combined row has conditional distribution \((0.60,0.40)\). The third row has \((0.40,0.60)\).
Joint, marginal, and conditional relative frequency 43
Two full tables describe different populations. Population A: <table> <tr><th></th><th>Outcome 1</th><th>Outcome 2</th></tr> <tr><td>Eligible</td><td>\(30\)</td><td>\(20\)</td></tr> <tr><td>Not eligible</td><td>\(10\)</td><td>\(40\)</td></tr> </table> Population B: <table> <tr><th></th><th>Outcome 1</th><th>Outcome 2</th></tr> <tr><td>Eligible</td><td>\(30\)</td><td>\(20\)</td></tr> <tr><td>Not eligible</td><td>\(40\)</td><td>\(10\)</td></tr> </table> A report filters both populations to eligible observations and then converts each eligible row to conditional relative frequencies. a) Find the filtered conditional distribution for each population. b) Find the full-population relative frequency of Outcome 1 in each population. c) Explain what information the filtering hides.

Hints

- After filtering, use only the retained row as the reference population. - For the full-population rate, restore the excluded row and use the grand total. - Compare what remains visible before and after renormalization.

Solution

1. In both populations, the eligible row is \((30,20)\), so the filtered distribution is \((30/50,20/50)=(0.60,0.40)\). 2. Population A has \((30+10)/100=0.40\) in Outcome 1. Population B has \((30+40)/100=0.70\). 3. Filtering discards the not-eligible row. Renormalization then makes the retained rows identical even though the full tables differ substantially.

Answer

a) Both filtered distributions are \((0.60,0.40)\). b) Full-population Outcome 1 rates are \(0.40\) for Population A and \(0.70\) for Population B. c) Filtering hides all differences in the excluded row and can make different full populations look identical.
Joint, marginal, and conditional relative frequency 44
A report initially shows these success counts: <table> <tr><th></th><th>Success</th><th>Failure</th><th>Total</th></tr> <tr><td>Group A</td><td>\(20\)</td><td>\(10\)</td><td>\(30\)</td></tr> <tr><td>Group B</td><td>\(19\)</td><td>\(10\)</td><td>\(29\)</td></tr> </table> One successful observation was assigned to Group A but actually belongs to Group B. a) Compare the two conditional success rates before correction. b) Correct the table and compare the rates again. c) Explain how one reclassification reverses the ranking.

Hints

- A reclassification transfers one observation rather than deleting it. - Recompute both the numerator and denominator for each affected group. - Compare the rates before and after the correction, not just the success counts.

Solution

1. Before correction, Group A’s rate is \(20/30\approx0.667\), while Group B’s is \(19/29\approx0.655\), so Group A is higher. 2. Move one success from Group A to Group B. The corrected rows are \((19,10)\) and \((20,10)\). 3. Corrected rates are \(19/29\approx0.655\) for Group A and \(20/30\approx0.667\) for Group B, so Group B is higher. 4. The original rates were close, and the correction changes both the numerator and denominator of each group in opposite directions.

Answer

a) Before correction: Group A \(\approx0.667\), Group B approximately \(0.655\); Group A is higher. b) After correction: Group A approximately \(0.655\), Group B \(\approx0.667\); Group B is higher. c) Moving one successful observation changes both groups and reverses the ranking.
Joint, marginal, and conditional relative frequency 45
Two populations have the same row-conditional distributions: Group A \((0.75,0.25)\) and Group B \((0.30,0.70)\). Construct one joint relative-frequency table when the row margins are \((0.50,0.50)\), and another when the row margins are \((0.20,0.80)\). Compare the column margins and explain what the conditional profiles do not determine.

Hints

- Multiply each row’s margin by its conditional profile. - Add down columns after constructing each joint table. - Compare what changes when only the row weights change.

Solution

1. With row margins \((0.50,0.50)\), the joint rows are \((0.375,0.125)\) and \((0.150,0.350)\); column margins are \((0.525,0.475)\). 2. With row margins \((0.20,0.80)\), the joint rows are \((0.150,0.050)\) and \((0.240,0.560)\); column margins are \((0.390,0.610)\). 3. The same within-row profiles can produce different joint and marginal distributions because row weights differ.

Answer

With row margins \((0.50,0.50)\): <table> <tr><th></th><th>Outcome 1</th><th>Outcome 2</th><th>Total</th></tr> <tr><td>Group A</td><td>\(0.375\)</td><td>\(0.125\)</td><td>\(0.500\)</td></tr> <tr><td>Group B</td><td>\(0.150\)</td><td>\(0.350\)</td><td>\(0.500\)</td></tr> <tr><td>Total</td><td>\(0.525\)</td><td>\(0.475\)</td><td>\(1.000\)</td></tr> </table> With row margins \((0.20,0.80)\): <table> <tr><th></th><th>Outcome 1</th><th>Outcome 2</th><th>Total</th></tr> <tr><td>Group A</td><td>\(0.150\)</td><td>\(0.050\)</td><td>\(0.200\)</td></tr> <tr><td>Group B</td><td>\(0.240\)</td><td>\(0.560\)</td><td>\(0.800\)</td></tr> <tr><td>Total</td><td>\(0.390\)</td><td>\(0.610\)</td><td>\(1.000\)</td></tr> </table> Conditional profiles alone do not determine joint or column-marginal frequencies.
Joint, marginal, and conditional relative frequency 46
A sample consists of \(25\%\) Group A with success rate \(0.60\), \(50\%\) Group B with success rate \(0.40\), and \(25\%\) Group C with success rate \(0.20\). Find each group’s contribution to the marginal success frequency and identify which group contributes most.

Hints

- A group’s contribution combines its size and within-group rate. - Compute joint success frequencies for all groups. - Do not rank contributions by conditional rate alone.

Solution

1. Contributions are \(0.25\cdot0.60=0.15\), \(0.50\cdot0.40=0.20\), and \(0.25\cdot0.20=0.05\). 2. The marginal success frequency is \(0.40\). 3. Group B contributes most, even though Group A has the highest conditional rate.

Answer

The contributions are \(0.15,0.20,\) and \(0.05\); Group B contributes the most.
Joint, marginal, and conditional relative frequency 47
Two community centers offer basic and advanced bike-repair courses. In each center, \(80\%\) of basic-course students pass the final skills test and \(50\%\) of advanced-course students pass. At Maple Center, \(30\%\) of students take the basic course. At River Center, \(60\%\) take the basic course. Everyone takes exactly one course. What percentage passes at each center? Why are the overall pass rates different?

Hints

- How many students out of 100 at each center would take each course? - Think about the students in the two courses separately before combining their results. - Do the centers have the same proportion of students taking the easier test?

Solution

1. At Maple Center, the pass rate is \(0.30\cdot0.80+0.70\cdot0.50=0.59\). 2. At River Center, it is \(0.60\cdot0.80+0.40\cdot0.50=0.68\). 3. River Center has more students in the basic course, which has the higher pass rate.

Answer

Maple Center: \(59\%\). River Center: \(68\%\). River Center has more students in the basic course, which has the higher pass rate.
Joint, marginal, and conditional relative frequency 48
A report gives this percentage table: <table> <tr><th></th><th>yes</th><th>no</th></tr> <tr><td>Group A</td><td>\(20\%\)</td><td>\(30\%\)</td></tr> <tr><td>Group B</td><td>\(40\%\)</td><td>\(60\%\)</td></tr> </table> The analyst says the first row uses percentages of the grand total, while the second row uses percentages within Group B. Explain why the table is not a coherent relative-frequency table and what must be added to make it interpretable.

Hints

- Determine the reference group for each displayed percentage. - A table’s margins are meaningful only when the entries share a compatible denominator system. - Mixed normalizations require explicit labels rather than one unlabeled table.

Solution

1. The first row’s denominator is the full sample, while the second row’s denominator is only Group B. 2. Values with different denominator systems cannot be compared or summed as one table. 3. The table must use one consistent normalization, or each entry must be explicitly labeled with its denominator and corresponding margins.

Answer

The table mixes joint and row-conditional percentages, so it is not coherent as one relative-frequency table. Use a single denominator convention or label every denominator explicitly.
Joint, marginal, and conditional relative frequency 49
At a city art center, \(20\%\) of students take advanced art and \(80\%\) take regular art. The center asks \(1000\) randomly chosen students from each type of class whether they bring their own supplies. In advanced art, \(126\) say yes; in regular art, \(31\) say yes. Assume each class's survey rate holds for all students in that class type. Estimate the percentage of all the center's students who bring supplies. Why isn't \(\frac{157}{2000}\) the right overall estimate?

Hints

- Did the researchers ask the two kinds of students in the same proportion they occur at the center? - Imagine the center had \(100\) students. How many would take each kind of class? - Would a simple average of the class rates represent that mix?

Solution

1. The within-group shares are \(126/1000=0.126\) and \(31/1000=0.031\). 2. Combine them using the shares of students in each class type: \(0.20\cdot0.126+0.80\cdot0.031=0.0500\). 3. The unadjusted survey share \(\frac{157}{2000}=0.0785\) treats the two classes as equally common, even though only one-fifth of students are in advanced art.

Answer

Estimated share: \(0.0500\), or \(5\%\). The raw survey share \(0.0785\) gives equal weight to the two types, even though only \(20\%\) take advanced art.
Joint, marginal, and conditional relative frequency 50
Two school district help desks handle hardware and software requests. Here are their results. <table> <tr><th>Desk</th><th>Request type</th><th>Requests</th><th>Resolved</th></tr> <tr><td>North</td><td>Hardware</td><td>\(200\)</td><td>\(40\)</td></tr> <tr><td>North</td><td>Software</td><td>\(800\)</td><td>\(720\)</td></tr> <tr><td>South</td><td>Hardware</td><td>\(800\)</td><td>\(200\)</td></tr> <tr><td>South</td><td>Software</td><td>\(200\)</td><td>\(190\)</td></tr> </table> a) For each desk, what percentage of hardware requests were resolved? What percentage of software requests were resolved? b) Find the overall percentage resolved at each desk. c) Which desk has the higher rate for each type? Why does that desk have a lower overall rate?

Hints

- Why would dividing by all requests give a different result from dividing by one request type? - Look for a way to combine the resolved counts without averaging the two percentages. - Could the mix of easy and hard requests affect the overall comparison?

Solution

1. Hardware rates are \(40/200=20\%\) at North and \(200/800=25\%\) at South. 2. Software rates are \(720/800=90\%\) at North and \(190/200=95\%\) at South. 3. North resolved \(760\) of \(1000\) requests, or \(76\%\). South resolved \(390\) of \(1000\), or \(39\%\). 4. South does better within both types, but most of its requests concern hardware, the type less often resolved at both desks. North handles mostly software requests.

Answer

a) Hardware: North \(20\%\), South \(25\%\). Software: North \(90\%\), South \(95\%\). b) North resolved \(76\%\), and South resolved \(39\%\). c) South has higher rates for both types. Its overall rate is lower because it handles mostly hardware requests.
Joint, marginal, and conditional relative frequency 51
A \(2\times2\) joint relative-frequency table has row margin \(P(A)=0.65\) and column margin \(P(B)=0.55\). Let \(x=P(A\cap B)\). a) Express every cell of the table in terms of \(x\). b) Determine the complete interval of values of \(x\) that makes every cell nonnegative. c) Find the value of \(x\) for which the row and column variables are independent, and verify that it lies in the feasible interval.

Hints

- Use the row and column margins to write the three remaining cells from one unknown intersection. - Every joint relative frequency must be at least zero. - Independence selects one particular interior table from the feasible family.

Solution

1. The four cells are \(x\), \(0.65-x\), \(0.55-x\), and \(1-0.65-0.55+x=x-0.20\). 2. Nonnegativity requires \(x\ge0\), \(x\le0.65\), \(x\le0.55\), and \(x\ge0.20\). Therefore \(0.20\le x\le0.55\). 3. Independence requires \(x=P(A)P(B)=0.65\cdot0.55=0.3575\), which is inside the feasible interval.

Answer

a) <table> <tr><th></th><th>\(B\)</th><th>\(B^c\)</th><th>Total</th></tr> <tr><td>\(A\)</td><td>\(x\)</td><td>\(0.65-x\)</td><td>\(0.65\)</td></tr> <tr><td>\(A^c\)</td><td>\(0.55-x\)</td><td>\(x-0.20\)</td><td>\(0.35\)</td></tr> <tr><td>Total</td><td>\(0.55\)</td><td>\(0.45\)</td><td>\(1\)</td></tr> </table> b) \(x\in[0.20,0.55]\). c) Independence occurs at \(x=0.3575\).
Joint, marginal, and conditional relative frequency 52
A tutoring center compares Program A and Program B for students with strong or limited prior preparation. The results are: <table> <tr><th></th><th>Program A</th><th>Program B</th></tr> <tr><td>Strong preparation</td><td>\(9/10\) passed</td><td>\(72/90\) passed</td></tr> <tr><td>Limited preparation</td><td>\(3/10\) passed</td><td>\(2/10\) passed</td></tr> </table> a) Compare the conditional pass rates for the two programs within each preparation group. b) Compare the overall pass rates after pooling the preparation groups. c) Explain why the direction of the comparison reverses and why the pooled comparison alone is misleading.

Hints

- Compute each pass rate using the total for its own preparation-group-and-program cell. - Pool counts within each program only after comparing the two preparation groups separately. - Compare the preparation-group composition of the two programs to explain the reversal.

Solution

1. Among students with strong preparation, Program A has pass rate \(\frac{9}{10}=0.90\), while Program B has \(\frac{72}{90}=0.80\). Program A is higher. 2. Among students with limited preparation, Program A has \(\frac{3}{10}=0.30\), while Program B has \(\frac{2}{10}=0.20\). Program A is again higher. 3. Overall, Program A has \(\frac{12}{20}=0.60\), while Program B has \(\frac{74}{100}=0.74\). The pooled comparison favors Program B. 4. Program B has a much larger share of students with strong preparation, the group with higher pass rates under either program. Unequal group composition creates Simpson’s paradox: the pooled association reverses the within-group associations.

Answer

a) Strong preparation: Program A \(0.90\), Program B \(0.80\). Limited preparation: Program A \(0.30\), Program B \(0.20\). Program A is higher in both groups. b) Overall: Program A \(0.60\), Program B \(0.74\). c) Program B includes a much larger proportion of strongly prepared students. The unequal weights create Simpson’s paradox, so the pooled rates hide the within-group pattern.

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