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Display and summarize data

Learning goals

  • Chooses an appropriate display for the variable type and distribution.
  • Computes relevant measures of center and spread.
  • Describes shape, center, spread, and unusual features.
  • Uses graphical and numerical summaries together.
  • Compares groups using consistent statistics when relevant.

Click problems to add them to your worksheet.

Display and summarize data 1
A wildlife center recorded how many minutes rescued birds spent at a feeding station. The staff wants a display that shows every exact value and makes repeated values easy to see. Choose the most appropriate display from a dot plot, histogram, or box plot. Explain why the other two displays are less suitable for this purpose.

Hints

- Which display keeps each measurement separate? - Think about how repeated measurements would look in the three displays. - Which displays combine several measurements into a smaller summary?

Solution

1. A dot plot puts one mark at each observation, so individual feeding times remain visible and repeated values make stacks. 2. A histogram combines measurements into intervals, hiding individual times. 3. A box plot summarizes the distribution using quartiles and other key values instead of showing every observation.

Answer

A dot plot is most appropriate because it shows exact feeding times and repeated values. A histogram groups values into intervals; a box plot summarizes the data.
Display and summarize data 2
A report lists the five-number summary for a set of trail elevations as \(420, 510, 495, 560, 610\) feet. Explain why this cannot be a valid five-number summary. Identify the specific ordering condition that fails, and state one possible correction if the values \(495\) and \(510\) were accidentally reversed.

Hints

- Recall the order of the five values in the summary. - Compare each listed value with the one immediately before and after it. - A possible correction should restore nondecreasing order.

Solution

1. A five-number summary must be ordered as minimum, \(Q_1\), median, \(Q_3\), maximum, so the values cannot decrease from one position to the next. 2. The listed \(Q_1=510\) is greater than the listed median \(495\), which violates the required order. 3. If those two entries were reversed, the corrected summary would be \(420, 495, 510, 560, 610\), which is in nondecreasing order.

Answer

The summary is invalid because \(Q_1\) cannot exceed the median. If \(495\) and \(510\) were reversed, a possible corrected summary would be \(420, 495, 510, 560, 610\) feet.
Display and summarize data 3
The numbers of books read by seven students during a month were \(1, 2, 2, 3, 4, 5, 11\). Find the mean, median, mode, and range.

Hints

- Add all seven values before dividing by the number of students. - Use the middle position of the ordered list for the median. - Check repeated values and the two endpoints for the remaining summaries.

Solution

1. The sum is \(1+2+2+3+4+5+11=28\), so the mean is \(28\div 7=4\). 2. The ordered list has seven values, so the median is the fourth value, \(3\). 3. The value \(2\) occurs most often, so the mode is \(2\). 4. The range is \(11-1=10\).

Answer

Mean: \(4\) Median: \(3\) Mode: \(2\) Range: \(10\)
Display and summarize data 4
For the data set \(12, 18, 21, 29\), find the minimum, maximum, and range.

Hints

- Locate the smallest observation. - Locate the largest observation. - Subtract the minimum from the maximum to find the spread.

Solution

1. The minimum is \(12\). 2. The maximum is \(29\). 3. The range is \(29-12=17\).

Answer

Minimum: \(12\) Maximum: \(29\) Range: \(17\)
Display and summarize data 5
A dashboard reports that a set of response times has minimum \(5\) seconds, maximum \(12\) seconds, and mean \(14\) seconds. Determine whether the report can be correct. Justify your conclusion using a property of the mean.

Hints

- Locate the reported mean relative to the two endpoints. - Think of the mean as a balance point of the observed values. - A balance point cannot lie beyond every observation on the same side.

Solution

1. Every observation lies between \(5\) and \(12\) seconds. 2. The mean of a data set must lie between its minimum and maximum. 3. Since \(14>12\), the reported mean is impossible, so at least one summary value is incorrect.

Answer

The report cannot be correct. A mean of \(14\) seconds cannot occur when every value is between \(5\) and \(12\) seconds.
Display and summarize data 6
A data set is \(1,2,2,3,3,3,4,5,5\). Chiamaka made the dot plot shown. Does the dot plot represent the data correctly? If not, identify the values whose dot counts must be corrected and state the correct counts.
Figure for problem 551384

Hints

- Tally how many times each value appears in the raw data. - Compare each tally with the height of the corresponding stack of dots. - A matching total number of dots is not enough; each value must have the correct frequency.

Solution

1. In the data set, the value \(3\) occurs \(3\) times, but the dot plot shows only \(2\) dots at \(3\). 2. In the data set, the value \(5\) occurs \(2\) times, but the dot plot shows \(3\) dots at \(5\). 3. Therefore, the dot plot is not correct. It needs \(3\) dots at \(3\) and \(2\) dots at \(5\).

Answer

No. The value \(3\) should have \(3\) dots, and the value \(5\) should have \(2\) dots.
Display and summarize data 7
A school records students' favorite lunch choice and the number of minutes they wait in line. Which display is better for each variable: a bar chart of categories or a histogram? Explain each choice.

Hints

- Ask whether each answer is a category or a number. - What would the horizontal axis represent for lunch choices? - Which display shows how numerical measurements fall into intervals?

Solution

1. Favorite lunch choice is a category, so a bar chart can show how many students chose each option. 2. Waiting time is a numerical measurement, so a histogram can group times into intervals and show their distribution.

Answer

Lunch choices: bar chart, because the values are categories. Waiting times: histogram, because they are measured numbers that can be grouped into intervals.
Display and summarize data 8
A game store sold this many packs of cards on \(10\) afternoons: \(3,5,5,3,4,5,6,3,4,5\) If you make a dot plot, how many dots go above \(3\), \(4\), \(5\), and \(6\)?

Hints

- Which different numbers appear in the data? - A repeated number makes another dot directly above that value. - Check that your dot counts add up to the number of afternoons.

Solution

1. Count each distinct value in the list: \(3\) appears three times, \(4\) appears twice, \(5\) appears four times, and \(6\) appears once. 2. Put that many dots in the stack for each value. The stacks total \(3+2+4+1=10\) observations.

Answer

Above \(3\): \(3\) dots; \(4\): \(2\) dots; \(5\): \(4\) dots; \(6\): \(1\) dot.
Display and summarize data 9
A box plot summarizes daily pages read by a class. a) Read the first and third quartiles. b) About what fraction of observations, by ordered position, lies between those quartiles?
Figure for problem 560447

Hints

- The edges of the box mark two different quartiles. - Think of the ordered observations divided into four equal parts. - How many of those quarters lie inside the box?

Solution

1. The left side of the box is \(Q_1=15\) pages and the right side is \(Q_3=25\) pages. 2. The interval from the first to third quartile represents the middle \(50\%\) of observations by rank.

Answer

a) \(Q_1=15\) pages and \(Q_3=25\) pages b) About \(\frac{1}{2}\), or \(50\%\), of the ordered observations
Display and summarize data 10
A library recorded how long visitors stayed. <table><tr><th>Time (minutes)</th><th>Total visitors below this time</th></tr><tr><td>Less than \(5\)</td><td>\(3\)</td></tr><tr><td>Less than \(10\)</td><td>\(7\)</td></tr><tr><td>Less than \(15\)</td><td>\(11\)</td></tr><tr><td>Less than \(20\)</td><td>\(14\)</td></tr></table> a) How many visitors stayed less than \(10\) minutes? b) Nina adds \(3+7\) and says \(10\) visitors stayed less than \(10\) minutes. Explain her mistake.

Hints

- Does each row count only the latest time interval, or everyone below its time limit? - Could one visitor be included in more than one row? - When is it valid to add two counts without counting anyone twice?

Solution

1. The row for less than \(10\) minutes counts \(7\) visitors. 2. All \(3\) visitors who stayed less than \(5\) minutes are already included in those \(7\). Adding the two counts includes those \(3\) visitors twice.

Answer

a) \(7\) visitors. b) The \(3\) visitors who stayed less than \(5\) minutes are already part of the \(7\), so \(3+7\) counts them twice.
Display and summarize data 11
A box plot has third quartile \(Q_3=14\) and interquartile range \(4\). The upper outlier fence is \(Q_3+1.5\cdot\text{IQR}\). Find the fence. Would a value of \(20\) be an outlier under this rule? Explain.

Hints

- Is a fence an observed score, or a cutoff used to check scores? - Compare the cutoff with the value you are testing. - Think about whether being on a boundary is the same as passing it.

Solution

1. The upper fence is \(14+1.5\cdot4=20\). 2. A value equal to the fence is not beyond it. Thus \(20\) is not an outlier under the upper-fence rule.

Answer

Upper fence: \(20\). A value of \(20\) is not an outlier because it is on the fence, not beyond it.
Display and summarize data 12
In a modified box plot, open circles mark outliers. The whiskers stop at the smallest and largest values that are not outliers. a) What is the largest score that is not an outlier? b) What is the largest score overall?
Figure for problem 560493

Hints

- Compare the two rightmost positions on the scale. - Does the right whisker reach the farthest plotted mark? - Check the tick marks under both positions before choosing your answers.

Solution

1. The right whisker ends at \(12\), so the largest non-outlier score is \(12\). 2. The separate open circle is at \(18\), so the largest score overall is \(18\). 3. A modified box plot shows outliers separately rather than extending a whisker to them.

Answer

a) \(12\) points b) \(18\) points
Display and summarize data 13
A climbing gym timed visitors completing a beginner route. Which time interval contains the most visitors, and how many visitors are in it?
Figure for problem 560496

Hints

- Find the bar that reaches highest on the frequency axis. - Read both the left and right boundaries of that bar. - Read its height from the vertical scale.

Solution

1. The tallest bar covers times from \(5\) to less than \(10\) minutes. 2. Its frequency is \(6\), so that interval contains the most visitors.

Answer

\(5\) to less than \(10\) minutes; \(6\) visitors.
Display and summarize data 14
Two teams compete in a quiz. Team A has \(n\) members with a mean score of \(a\) points. Team B has \(m\) members with a mean score of \(b\) points. a) Write an expression for the total number of points earned by both teams. b) Write an expression for the mean score of all participants. c) Simplify the expression from part b) when the teams have the same number of members, so \(n=m\). Identify the resulting mean.

Hints

- A mean tells you how many points one team member earned on average. How could you use that to describe the whole team's points? - Should a member of one team count more than a member of the other in the combined mean? - What changes when the teams have the same number of members?

Solution

1. Team A earned \(na\) points and Team B earned \(mb\) points, for a combined total of \(na+mb\). 2. There are \(n+m\) participants, so the combined mean is \(\frac{na+mb}{n+m}\). 3. If \(m=n\), this becomes \(\frac{na+nb}{2n}=\frac{a+b}{2}\), the arithmetic mean of the two team means.

Answer

a) \(na+mb\) points b) \(\frac{na+mb}{n+m}\) points c) \(\frac{a+b}{2}\), the arithmetic mean of the two team means
Display and summarize data 15
Two marching-band sections were evaluated for timing accuracy. Their timing errors, measured in milliseconds from the target beat, were summarized as follows. <table><tr><th>Section</th><th>Mean error</th><th>Population standard deviation</th></tr><tr><td>Brass</td><td>\(6\,\text{ms}\)</td><td>\(11\,\text{ms}\)</td></tr><tr><td>Woodwinds</td><td>\(6\,\text{ms}\)</td><td>\(4\,\text{ms}\)</td></tr></table> a) Which section was more consistent? Explain. b) Does the smaller standard deviation prove that every woodwind timing error was closer to \(6\,\text{ms}\) than every brass timing error? Explain.

Hints

- First compare the measures of center, then compare the measures of spread. - Think about what a smaller distance from the mean means for a group as a whole. - A summary statistic does not list the location of every observation.

Solution

1. The two sections have the same mean error, so their centers are equal. 2. The woodwinds have the smaller standard deviation, so their timing errors were generally closer to their mean. The woodwinds were more consistent. 3. Standard deviation summarizes overall spread. It does not compare every individual value across the two groups, so it cannot prove that every woodwind error was closer to the mean than every brass error.

Answer

a) The woodwinds were more consistent because their population standard deviation was smaller. b) No. A smaller standard deviation describes the group’s overall spread; individual values from the two sections could still overlap.
Display and summarize data 16
A city tested two bus routes during the morning commute. The travel times, in minutes, have these five-number summaries. <table><tr><th>Route</th><th>Minimum</th><th>\(Q_1\)</th><th>Median</th><th>\(Q_3\)</th><th>Maximum</th></tr><tr><td>East</td><td>\(12\)</td><td>\(15\)</td><td>\(18\)</td><td>\(22\)</td><td>\(31\)</td></tr><tr><td>West</td><td>\(10\)</td><td>\(17\)</td><td>\(19\)</td><td>\(21\)</td><td>\(24\)</td></tr></table> a) Find the range and interquartile range for each route. b) Which route has the lower typical travel time? c) Which route is more consistent in its middle half? Support your answer with a statistic.

Hints

- Use the endpoints of a five-number summary to measure total spread. - Use the two quartiles to measure the spread of the middle half. - A typical value can be compared using the medians shown in the table.

Solution

1. Route East has range \(31-12=19\) minutes and IQR \(22-15=7\) minutes. 2. Route West has range \(24-10=14\) minutes and IQR \(21-17=4\) minutes. 3. Route East has the lower median, \(18\) minutes compared with \(19\) minutes, so it has the lower typical travel time. 4. Route West has the smaller IQR, so the middle half of its travel times is more consistent.

Answer

a) Route East: range \(19\) minutes, IQR \(7\) minutes. Route West: range \(14\) minutes, IQR \(4\) minutes. b) Route East has the lower typical travel time because its median is \(18\) minutes. c) Route West is more consistent in the middle half because its IQR is \(4\) minutes, compared with \(7\) minutes for Route East.
Display and summarize data 17
A machine produced five test strips with thickness deviations, in micrometers, of \(3, 3, 5, 7, 7\). These are all the strips in the batch. a) Find the mean deviation. b) Find the population variance. c) Find the population standard deviation. Round to the nearest hundredth.

Hints

- Begin by averaging the five deviations. - For the variance, square each distance from the mean and average those squared distances. - Take the square root of the variance only after the variance is found.

Solution

1. The mean is \(\frac{3+3+5+7+7}{5}=5\) micrometers. 2. The squared deviations from the mean are \(4, 4, 0, 4, 4\), with sum \(16\). 3. The population variance is \(\frac{16}{5}=3.2\) square micrometers. 4. The population standard deviation is \(\sqrt{3.2}\approx 1.79\) micrometers.

Answer

a) \(5\) micrometers b) \(3.2\) square micrometers c) Approximately \(1.79\) micrometers
Display and summarize data 18
A box plot summarizes how many lines \(80\) students spoke in a theater workshop. a) About how many observations make up the middle half of the ordered data? b) What is the median number of lines? Explain what the median divides in the ordered list. c) Can the box plot tell exactly how many students spoke more than \(20\) lines? Explain.
Figure for problem 544003

Hints

- Think about what fraction of ordered observations the box represents. - What does the middle line mark in an ordered data set? - Does a quartile or whisker show every value between its endpoints?

Solution

1. The box marks the middle half of observations by rank, or about \(50\%\) of \(80\): approximately \(40\) observations. 2. The median is \(13\) lines. It separates the lower \(40\) ordered positions from the upper \(40\); with repeated values, it does not imply exactly \(40\) students spoke at most \(13\) lines. 3. The third quartile is \(18\) lines, but the plot does not show how values above \(18\) are distributed. The right whisker ends at \(27\), not necessarily at the largest data value, so the count above \(20\) cannot be found exactly.

Answer

a) About \(40\) observations b) \(13\) lines; the median separates the lower and upper halves of the ordered data by position. c) No. The box plot does not give the individual counts above \(20\).
Display and summarize data 19
The two histograms show the same set of kayak trip lengths. Plot a) uses \(1\)-mile intervals, and plot b) uses \(5\)-mile intervals. Explain how both histograms can be accurate. Which display is more useful for identifying separate clusters, and what tradeoff comes with that choice?
Figure for problem 544004

Hints

- Compare which values are combined into a single bar in each plot. - Look for a gap that is visible with narrow intervals but absorbed by wider intervals. - Decide whether the goal favors local detail or a simpler overall shape.

Solution

1. A histogram groups observations according to its interval boundaries. Changing the interval width changes which values are combined in each bar. 2. In plot b), the wider intervals combine values that appear as separated groups in plot a). 3. Plot a) is more useful for identifying the separate clusters because its \(1\)-mile intervals preserve more local detail. Its tradeoff is that it has more bars and can look more irregular.

Answer

Both displays are accurate because they group the same observations into different interval widths. Plot a), with \(1\)-mile intervals, shows the separate clusters more clearly, but it produces a more detailed and potentially noisier display.
Display and summarize data 20
Two engineering classes tracked how often students used the makerspace in one month. <table><tr><th>Visits</th><th>Class A</th><th>Class B</th></tr><tr><td>\(0\)–\(2\)</td><td>\(6\)</td><td>\(8\)</td></tr><tr><td>\(3\)–\(5\)</td><td>\(12\)</td><td>\(20\)</td></tr><tr><td>\(6\)–\(8\)</td><td>\(6\)</td><td>\(12\)</td></tr></table> a) Find the relative frequency for each interval in each class. b) Compare the two distributions. Why are the table counts alone misleading?

Hints

- Divide each count by the number of students in that class. - Compare the same interval in each class after changing counts to percentages. - Why might a larger class have larger counts even with similar patterns?

Solution

1. For Class A, the relative frequencies are \(6/24=25\%\), \(12/24=50\%\), and \(6/24=25\%\). 2. For Class B, they are \(8/40=20\%\), \(20/40=50\%\), and \(12/40=30\%\). 3. Both classes have \(50\%\) in the middle interval. Class B has a higher percentage in the highest interval; Class A has a higher percentage in the lowest. 4. The classes have different totals, so their raw counts do not fairly compare the distributions.

Answer

a) Class A: \(25\%\), \(50\%\), \(25\%\). Class B: \(20\%\), \(50\%\), \(30\%\). b) The middle percentages match, but Class B has more students proportionally in the highest interval. Raw counts are misleading because the classes differ in size.
Display and summarize data 21
A calibration report says that the mean error for a set of digital scales was \(2.4\,\text{g}\) and the population standard deviation was \(0\,\text{g}\). What must be true about every recorded error? Explain why no other data set can have those two summaries.

Hints

- Think about what standard deviation measures relative to the mean. - Ask what must happen to every deviation for their average square to be zero. - Use the given mean to determine the common value.

Solution

1. A population standard deviation of \(0\) means every observation has zero distance from the mean. 2. Since the mean is \(2.4\,\text{g}\), every recorded error must equal \(2.4\,\text{g}\). 3. If even one value differed from \(2.4\,\text{g}\), its squared deviation would be positive, making the standard deviation greater than \(0\).

Answer

Every recorded error was exactly \(2.4\,\text{g}\). Any different value would create positive spread and a nonzero standard deviation.
Display and summarize data 22
The box plots compare turnaround times, in hours, for two laboratories. For these plots, the whiskers mark the actual minimum and maximum values, and there are no values beyond the whiskers. a) Compare the medians and interquartile ranges. b) Compare the ranges. c) Explain how the box plots can have identical boxes but noticeably different whiskers.
Figure for problem 544010

Hints

- Read the three values that determine each box before looking at the whiskers. - Use the stated minimum and maximum whisker endpoints to calculate total spread. - Match each visual part of a box plot to the statistic it represents.

Solution

1. Both median lines are at \(10\) hours. Both boxes extend from \(8\) to \(12\), so each interquartile range is \(12-8=4\) hours. 2. Lab X has range \(16-4=12\) hours. Lab Y has range \(28-1=27\) hours. 3. The matching boxes show the same center and middle-half spread. The different minimum and maximum values create different whisker lengths and show that Lab Y has much greater overall spread.

Answer

a) Both labs have median \(10\) hours and IQR \(4\) hours. b) Lab X has range \(12\) hours; Lab Y has range \(27\) hours. c) The boxes match because \(Q_1\), the median, and \(Q_3\) match. The whiskers differ because the minimum and maximum values differ.
Display and summarize data 23
Rafael ordered eight package masses as \(2, 4, 5, 7, 8, 10, 12, 20\) kilograms. Rafael reported \(Q_1=5\) and \(Q_3=12\) by choosing one middle value from each half. Using the convention that the quartile of an even-sized half is the mean of its two middle values, correct \(Q_1\), \(Q_3\), and the interquartile range.

Hints

- Split the ordered list into two equal halves. - Each half contains an even number of observations. - Apply the stated quartile convention before subtracting the quartiles.

Solution

1. The lower half is \(2, 4, 5, 7\). Its two middle values are \(4\) and \(5\), so \(Q_1=\frac{4+5}{2}=4.5\). 2. The upper half is \(8, 10, 12, 20\). Its two middle values are \(10\) and \(12\), so \(Q_3=\frac{10+12}{2}=11\). 3. The interquartile range is \(11-4.5=6.5\) kilograms.

Answer

\(Q_1=4.5\,\text{kg}\), \(Q_3=11\,\text{kg}\), and \(\text{IQR}=6.5\,\text{kg}\).
Display and summarize data 24
A community theater recorded the number of days needed to build each of nine stage sets: \(3, 4, 5, 7, 8, 10, 12, 13, 30\) For quartiles, exclude the overall median from both halves. a) Find the five-number summary. b) Find the range and interquartile range. c) Which measure of spread is less affected by the unusually large value \(30\): the range or the interquartile range? Explain.

Hints

- Keep the list in order and locate the middle observation first. - Apply the stated rule when forming the lower and upper halves. - Compare which spread measure uses the maximum value.

Solution

1. The ordered data have median \(8\). The lower half is \(3, 4, 5, 7\), so \(Q_1=\frac{4+5}{2}=4.5\). The upper half is \(10, 12, 13, 30\), so \(Q_3=\frac{12+13}{2}=12.5\). 2. The five-number summary is \(3, 4.5, 8, 12.5, 30\). 3. The range is \(30-3=27\) days. The interquartile range is \(12.5-4.5=8\) days. 4. The interquartile range is less affected because it uses the middle half of the data and does not depend on the maximum.

Answer

a) \(3, 4.5, 8, 12.5, 30\) b) Range: \(27\) days. IQR: \(8\) days. c) The interquartile range is less affected because it does not use the extreme maximum.
Display and summarize data 25
The dot plot shows the number of new podcast episodes released each week by Cedar Grove Student Radio during a \(13\)-week term. Find the mean, median, and mode. Then describe the distribution's symmetry.
Figure for problem 544014

Hints

- Count the dots above each possible number of episodes. - Use the frequency counts to compute the total and locate the middle observation. - For symmetry, compare frequencies at values equally far from the center you determine.

Solution

1. The dot plot shows frequencies \(1, 3, 5, 3, 1\) for \(0, 1, 2, 3, 4\) episodes. 2. The total number of episodes is \(0\cdot 1+1\cdot 3+2\cdot 5+3\cdot 3+4\cdot 1=26\). The mean is \(26\div 13=2\). 3. With \(13\) weeks, the median is the seventh ordered value. The five weeks with \(2\) episodes cover the middle position, so the median is \(2\). 4. The most frequent value is \(2\), so the mode is \(2\). 5. Frequencies at equal distances from \(2\) match, so the distribution is symmetric about \(2\).

Answer

Mean: \(2\) episodes Median: \(2\) episodes Mode: \(2\) episodes The distribution is symmetric about \(2\).
Display and summarize data 26
An environmental scientist wants to show both the overall shape of river-depth measurements and a compact comparison with measurements from another river. Explain why using both a histogram and a box plot can be more useful than either display alone. Give one kind of information each display shows better.

Hints

- Think about what a histogram shows at each interval. - Which features are easier to compare in two box plots? - Does a box plot show each peak or gap in a distribution?

Solution

1. A histogram shows frequencies across intervals, making peaks, gaps, clusters, or skewness visible. 2. A box plot summarizes the median, quartiles and overall spread, allowing compact side-by-side comparisons. 3. Together the displays show both detailed shape and key summary values.

Answer

Histograms show clusters, gaps and other details of shape. Box plots let us compare medians and interquartile ranges quickly. Together they reveal more than either one alone.
Display and summarize data 27
The histograms show two data sets. a) Describe the direction of skewness in each plot. b) For each set, predict whether the mean is greater than or less than the median. c) Explain why the relationship between the mean and median alone would not prove the exact shape of a distribution.
Figure for problem 544018

Hints

- Locate the longer, thinner tail in each histogram. - Think about which center is pulled more strongly toward extreme values. - Separate a useful pattern from a conclusion that must always hold.

Solution

1. Plot a) has most observations at lower values and a tail extending to higher values, so Set A is right-skewed. 2. High values in the right tail pull the mean above the median. For Set A, the mean is greater than the median. 3. Plot b) has most observations at higher values and a tail extending to lower values, so Set B is left-skewed. 4. Low values in the left tail pull the mean below the median. For Set B, the mean is less than the median. 5. Different distributions can share the same mean and median, so those two summaries alone cannot establish every feature of shape, such as gaps, clusters, or multiple peaks.

Answer

a) Set A is right-skewed; Set B is left-skewed. b) Set A: mean greater than median. Set B: mean less than median. c) The comparison is a clue, not proof, because two summary statistics do not determine the full distribution.
Display and summarize data 28
Six instrument readings have mean \(50\). Their deviations from the mean are \(-7, -3, -1, 1, 3, 7\). a) Reconstruct the six readings. b) Find the median. c) Describe the symmetry of the distribution and explain why the mean and median match.

Hints

- A deviation tells how far and in which direction a value lies from the mean. - Pair negative and positive deviations of equal size. - Use the two middle reconstructed values for the median.

Solution

1. Add each deviation to \(50\). The readings are \(43, 47, 49, 51, 53, 57\). 2. The median is the mean of the third and fourth values: \(\frac{49+51}{2}=50\). 3. Each value below \(50\) is paired with a value the same distance above \(50\). The distribution is symmetric about \(50\), so both the balance point and the middle location are \(50\).

Answer

a) \(43, 47, 49, 51, 53, 57\) b) Median: \(50\) c) The data are symmetric about \(50\), so the mean and median are both \(50\).
Display and summarize data 29
A set of water-flow measurements has mean \(26\), median \(24\), population standard deviation \(5\), and interquartile range \(7\). Each value \(x\) is recentered by defining \(z=x-26\). Find the mean, median, population standard deviation, and interquartile range of the \(z\)-values. Explain the purpose of recentering.

Hints

- Treat the transformation as the same shift applied to every observation. - Decide which summaries represent position and which represent distance. - Interpret what a transformed value of \(0\) means.

Solution

1. Subtracting \(26\) from every value subtracts \(26\) from measures of center. The new mean is \(26-26=0\), and the new median is \(24-26=-2\). 2. Subtracting a constant does not change distances between observations, so the population standard deviation remains \(5\) and the interquartile range remains \(7\). 3. Recentering places the original mean at \(0\), so positive and negative \(z\)-values show positions above and below that mean.

Answer

Mean: \(0\) Median: \(-2\) Population standard deviation: \(5\) Interquartile range: \(7\) Recentering makes \(0\) represent the original mean without changing spread.
Display and summarize data 30
A delivery service grouped the distances of \(40\) package deliveries into equal-width intervals. <table><tr><th>Distance (miles)</th><th>Deliveries</th></tr><tr><td>\(0\) to less than \(5\)</td><td>\(7\)</td></tr><tr><td>\(5\) to less than \(10\)</td><td>\(11\)</td></tr><tr><td>\(10\) to less than \(15\)</td><td>\(9\)</td></tr><tr><td>\(15\) to less than \(20\)</td><td>\(8\)</td></tr><tr><td>\(20\) to less than \(25\)</td><td>\(\square\)</td></tr></table> a) Find the missing frequency. b) Find the relative frequency of that interval as a percentage. c) Explain how the total number of deliveries can be used to check the table.

Hints

- Do the distance intervals cover all the deliveries, without overlap? - How can the overall total help you check whether one interval is missing a count? - What does the percentage for one interval compare with the whole group?

Solution

1. The four known counts sum to \(7+11+9+8=35\). 2. The missing count is \(40-35=5\). 3. Its relative frequency is \(\frac{5}{40}=0.125=12.5\%\). 4. The intervals cover every delivery exactly once, so the frequency column must total \(40\).

Answer

a) \(5\) deliveries b) \(12.5\%\) c) The five interval counts must add up to \(40\).
Display and summarize data 31
The box plot summarizes a data set with unequal quartile-segment lengths. For this display, the whiskers mark the actual minimum and maximum values, and there are no values beyond the whiskers. Does the longest quartile segment contain the most observations? State what the unequal segment lengths indicate.
Figure for problem 544035

Hints

- Identify the four sections created by the stated minimum, quartiles, median, and maximum. - Recall what proportion of ordered observations lies in each quartile section. - Interpret horizontal length as distance on the measurement scale.

Solution

1. The four quartile sections from the minimum through \(Q_1\), the median, \(Q_3\), and the maximum each represent approximately \(25\%\) of the ordered observations, regardless of their visual lengths. 2. The longest segment therefore does not contain a larger proportion of the data. 3. A longer segment means that its observations are spread across a wider range of values; a shorter segment means that its observations are more concentrated.

Answer

The statement is false. Each quartile segment represents approximately \(25\%\) of the data. A longer segment indicates greater spread, not a larger count.
Display and summarize data 32
The box plot has a labeled numerical scale but does not show the original observations. For this display, the whiskers mark the actual minimum and maximum values, and there are no values beyond the whiskers. For each statistic, state whether it can be determined exactly from the box plot: minimum, mean, median, interquartile range, range, and population standard deviation. Give each value that can be determined, and explain why the others cannot.
Figure for problem 544038

Hints

- Read the stated minimum, quartiles, median, and maximum from the scale. - Form any requested differences from quartiles or endpoints. - Identify which statistics require all observations rather than only the five-number summary.

Solution

1. The box plot directly marks the minimum \(2\), first quartile \(5\), median \(7\), third quartile \(10\), and maximum \(14\). 2. The interquartile range is \(10-5=5\), and the range is \(14-2=12\). 3. The mean and population standard deviation depend on all individual observations and their frequencies. The box plot does not preserve that information, so neither can be determined exactly.

Answer

Can be determined exactly: minimum \(2\), median \(7\), interquartile range \(5\), and range \(12\). Cannot be determined exactly: mean and population standard deviation, because the individual observations and frequencies are not shown.
Display and summarize data 33
Two cross-country teams compare finishing ranks, where a lower rank is better. Team A: median rank \(12\), interquartile range \(6\) Team B: median rank \(15\), interquartile range \(3\) Compare the teams’ typical performance and consistency. Explain why “larger” does not automatically mean “better” for the median in this context.

Hints

- Interpret the direction of the measurement before comparing centers. - Use the median for typical performance and the interquartile range for middle-half consistency. - Keep “better result” separate from “larger number.”

Solution

1. Team A has the lower median rank, \(12\) compared with \(15\), so Team A had the better typical finish. 2. Team B has the smaller interquartile range, \(3\) compared with \(6\), so Team B’s middle half was more consistent. 3. Rank values describe position, and smaller numbers represent stronger finishes. The context determines how to interpret the direction of the median.

Answer

Team A had the better typical finish because its median rank was lower. Team B was more consistent because its IQR was smaller.
Display and summarize data 34
Data Set B is created by adding \(100\) to every value in Data Set A. Compare the shapes, means, medians, ranges, interquartile ranges, and population standard deviations of the two distributions. Also describe how their histograms would be related if the histogram for Set B used interval boundaries exactly \(100\) greater than the corresponding boundaries for Set A.

Hints

- Imagine translating every point on a number line without stretching the scale. - Separate statistics based on location from statistics based on distance. - For the histogram, compare corresponding bins whose boundaries are translated by the same amount as the data.

Solution

1. Adding \(100\) shifts every observation the same distance, so the distribution's shape is unchanged. 2. The mean and median of Set B are each \(100\) greater than the corresponding statistic for Set A. 3. Differences between values do not change, so the range, interquartile range, and population standard deviation are unchanged. 4. Translating every interval boundary by \(100\) preserves which observations fall in corresponding bins, so Set B's histogram has the same bar pattern shifted \(100\) units to the right.

Answer

The two distributions have the same shape and the same range, IQR, and population standard deviation. Set B's mean and median are each \(100\) larger. With corresponding interval boundaries shifted by \(100\), its histogram is Set A's histogram shifted \(100\) units to the right.
Display and summarize data 35
A school recorded arrival status by commute type. <table><tr><th>Commute type</th><th>On time</th><th>Late</th><th>Total</th></tr><tr><td>School bus</td><td>\(54\)</td><td>\(6\)</td><td>\(60\)</td></tr><tr><td>Other commute</td><td>\(72\)</td><td>\(18\)</td><td>\(90\)</td></tr><tr><td>Total</td><td>\(126\)</td><td>\(24\)</td><td>\(150\)</td></tr></table> a) Find the conditional relative-frequency distribution of arrival status for each commute group. b) Compare the on-time rates. c) Find the overall on-time rate and explain why it does not replace the two conditional rates.

Hints

- Divide each row by its own row total. - Compare the same outcome across the two commute groups. - Use the grand total only for the overall rate, not for a conditional rate.

Solution

1. For school-bus riders, the conditional distribution is \(\frac{54}{60}=90\%\) on time and \(\frac{6}{60}=10\%\) late. 2. For other commuters, it is \(\frac{72}{90}=80\%\) on time and \(\frac{18}{90}=20\%\) late. 3. The on-time rate is \(10\) percentage points higher for school-bus riders. 4. Overall, \(\frac{126}{150}=84\%\) of students arrived on time. This pooled percentage combines groups of different sizes and does not show the within-group difference.

Answer

a) School bus: \(90\%\) on time, \(10\%\) late. Other commute: \(80\%\) on time, \(20\%\) late. b) The school-bus on-time rate is \(10\) percentage points higher. c) Overall on-time rate: \(84\%\). The overall rate combines the groups and hides their conditional difference.
Display and summarize data 36
A school surveyed textbook format preference by grade level. <table><tr><th>Grade</th><th>Digital</th><th>Print</th><th>Total</th></tr><tr><td>9</td><td>\(36\)</td><td>\(24\)</td><td>\(60\)</td></tr><tr><td>10</td><td>\(30\)</td><td>\(30\)</td><td>\(60\)</td></tr><tr><td>11</td><td>\(18\)</td><td>\(42\)</td><td>\(60\)</td></tr><tr><td>Total</td><td>\(84\)</td><td>\(96\)</td><td>\(180\)</td></tr></table> a) Find the conditional relative-frequency distribution of textbook preference within each grade. b) Describe the association between grade level and textbook preference. c) Which grade shows the largest preference gap, and what is that gap in percentage points?

Hints

- Divide each grade row by its row total. - Compare the digital percentages in grade order. - Find a preference gap by subtracting the two percentages within one grade.

Solution

1. Grade 9: \(\frac{36}{60}=60\%\) digital and \(40\%\) print. 2. Grade 10: \(50\%\) digital and \(50\%\) print. 3. Grade 11: \(\frac{18}{60}=30\%\) digital and \(70\%\) print. 4. Digital preference decreases as grade level increases in this survey, while print preference increases, so the variables are associated. 5. Grade 11 has the largest gap: \(70\%-30\%=40\) percentage points in favor of print.

Answer

a) Grade 9: \(60\%\) digital, \(40\%\) print. Grade 10: \(50\%\) digital, \(50\%\) print. Grade 11: \(30\%\) digital, \(70\%\) print. b) Higher grade levels show lower digital preference and higher print preference in this survey. c) Grade 11, with a \(40\)-percentage-point gap in favor of print.
Display and summarize data 37
The histogram shows \(16\) pottery-firing times. Identify the interval containing the median. Can the exact median be found from the histogram? Explain.
Figure for problem 544183

Hints

- Determine which two ordered positions define the median. - Add bar frequencies from left to right until both positions are included. - Distinguish knowing an interval from knowing the exact observations inside it.

Solution

1. With \(16\) observations, the median depends on the eighth and ninth ordered values. 2. The first interval contains \(4\) observations. The first two intervals together contain \(11\), so both middle positions lie in \([10,20)\). 3. The histogram gives only interval counts, not the individual values within \([10,20)\), so the exact median cannot be determined.

Answer

The median lies in the interval \([10,20)\) hours. Its exact value cannot be found because the individual firing times inside that interval are unknown.
Display and summarize data 38
The histogram shows the number of minutes spent tuning instruments before rehearsal. a) Identify the modal interval. b) What percentage of rehearsals required less than \(15\) minutes of tuning? c) Can the exact mode be determined? Explain.
Figure for problem 544184

Hints

- Read the interval represented by the tallest bar. - Add the frequencies of all bars completely to the left of \(15\). - Distinguish the most frequent interval from the most frequent exact value.

Solution

1. The tallest bar has frequency \(8\), so the modal interval is \([5,10)\). 2. The total frequency is \(3+8+6+3=20\). The number below \(15\) minutes is \(3+8+6=17\), so the percentage is \(\frac{17}{20}=85\%\). 3. The histogram shows only interval counts, not the individual values within the modal interval. Therefore, the exact most frequent value cannot be determined.

Answer

a) \([5,10)\) minutes b) \(85\%\) c) No. The interval with the greatest frequency is known, but the exact values inside it are not.
Display and summarize data 39
The sorted data set is \(2,4,8,8,12,16,20,22\). Use the median of the lower four values for \(Q_1\) and the median of the upper four values for \(Q_3\). Kārlis's box plot is intended to use its whiskers for the actual minimum and maximum, with no separate outliers. Kārlis made the box plot shown. Exactly one of its five displayed summary values is incorrect. Identify that value and give the correct five-number summary.
Figure for problem 551385

Hints

- Check the minimum, median, and maximum against the ordered list before checking the quartiles. - Apply the stated lower-half and upper-half rule separately. - Compare each computed summary value with the corresponding location on the displayed box plot.

Solution

1. The minimum is \(2\) and the maximum is \(22\). 2. The median is \(\frac{8+12}{2}=10\). 3. The lower half is \(2,4,8,8\), so \(Q_1=\frac{4+8}{2}=6\). 4. The upper half is \(12,16,20,22\), so \(Q_3=\frac{16+20}{2}=18\). 5. The displayed value \(Q_3=20\) is incorrect. The correct five-number summary is \((2, 6, 10, 18, 22)\).

Answer

The incorrect displayed value is \(Q_3\). It should be \(18\), not \(20\). Correct five-number summary: \((2, 6, 10, 18, 22)\).
Display and summarize data 40
The following are the numbers of minutes \(17\) students spent on a practice set: \(3, 5, 6, 7, 8, 9, 12, 14, 15, 16, 18, 22, 25, 28, 33, 37, 45\). The candidate histogram below was constructed using the intervals \([0,10)\), \([10,20)\), \([20,30)\), \([30,40)\), and \([40,50)\). One observation was placed in the wrong interval, so exactly two bar heights are wrong. a) Identify the two incorrect bars and give their corrected heights. b) Describe the overall shape of the corrected histogram.
Figure for problem 554952

Hints

- Count how many raw observations belong in each stated interval. - Read the current heights from the candidate histogram and compare them with your interval counts. - Judge the shape only after correcting the two bars.

Solution

1. The correct interval frequencies are \(6\) for \([0,10)\), \(5\) for \([10,20)\), \(3\) for \([20,30)\), \(2\) for \([30,40)\), and \(1\) for \([40,50)\). 2. The candidate shows height \(4\) for \([10,20)\), so that bar must increase to \(5\). 3. The candidate shows height \(4\) for \([20,30)\), so that bar must decrease to \(3\). 4. The corrected frequencies decrease toward larger values, producing a longer right tail, so the distribution is right-skewed.

Answer

a) \([10,20)\): change height \(4\) to \(5\). \([20,30)\): change height \(4\) to \(3\). b) Right-skewed.
Display and summarize data 41
A cycling club recorded these ten commute times, in minutes: \(4,6,7,8,10,12,13,15,18,20\) Use the median of the lower five values for \(Q_1\) and the median of the upper five values for \(Q_3\). Construct a horizontal box plot on paper. To make the construction checkable, report the value where you place each graphical feature: a) left whisker endpoint b) left edge of the box c) median line d) right edge of the box e) right whisker endpoint Then state the interval covered by the box itself.

Hints

- Find the five-number summary using the stated quartile convention. - In a horizontal box plot, distinguish the whisker endpoints from the two edges of the box. - The median is drawn as a line inside the box, not as a whisker endpoint.

Solution

1. The minimum is \(4\) and the maximum is \(20\). 2. The median is \(\frac{10+12}{2}=11\). 3. The lower five values are \(4,6,7,8,10\), so \(Q_1=7\). 4. The upper five values are \(12,13,15,18,20\), so \(Q_3=15\). 5. A horizontal box plot therefore places the left whisker endpoint at \(4\), the left box edge at \(7\), the median line at \(11\), the right box edge at \(15\), and the right whisker endpoint at \(20\). The box spans \([7,15]\).

Answer

a) \(4\) b) \(7\) c) \(11\) d) \(15\) e) \(20\) The box itself spans \([7,15]\).
Display and summarize data 42
The dot plot below shows the numbers of pages Marisol read during her first eight study sessions. During four more sessions, she read \(2, 4, 4, 8\) pages. a) Add one dot for each new session. After doing so, what are the final stack heights above \(2\), \(4\), and \(8\)? b) After the additions, state the mode and identify any integers from \(2\) through \(9\) that have no dots.
Figure for problem 555060

Hints

- Read the current height of each affected stack from the displayed dot plot before adding anything. - Repeated new observations add multiple dots to the same vertical stack. - Determine the mode and gaps from the completed plot, not from the four new observations alone.

Solution

1. In the displayed plot, the starting stack heights are \(1\) above \(2\), \(2\) above \(4\), and \(1\) above \(8\). 2. The four new observations add one dot at \(2\), two dots at \(4\), and one dot at \(8\). 3. The final stack heights are therefore \(2\) above \(2\), \(4\) above \(4\), and \(2\) above \(8\). 4. The completed plot has its tallest stack at \(4\), so the mode is \(4\). The values \(6\) and \(7\) have no dots.

Answer

a) Final stack heights: above \(2\): \(2\); above \(4\): \(4\); above \(8\): \(2\) b) Mode: \(4\). No dots: \(6\) and \(7\).
Display and summarize data 43
The table shows how many reading-club members borrowed fewer than each number of books. <table><tr><th>Books borrowed</th><th>Cumulative count</th></tr><tr><td>Fewer than \(4\)</td><td>\(2\)</td></tr><tr><td>Fewer than \(8\)</td><td>\(9\)</td></tr><tr><td>Fewer than \(12\)</td><td>\(15\)</td></tr><tr><td>Fewer than \(16\)</td><td>\(18\)</td></tr></table> Find the number of members who borrowed \(4\)–\(7\) books and the number who borrowed \(8\)–\(11\) books.

Hints

- Each cumulative total includes every earlier group. - To isolate a single interval, compare totals at its upper and lower boundaries. - Check that the two counts are nonnegative and smaller than the relevant cumulative totals.

Solution

1. The group from \(4\) to \(7\) is the difference between the totals below \(8\) and below \(4\): \(9-2=7\). 2. The group from \(8\) to \(11\) is the difference between the totals below \(12\) and below \(8\): \(15-9=6\).

Answer

\(4\)–\(7\) books: \(7\) members; \(8\)–\(11\) books: \(6\) members.
Display and summarize data 44
A large set of water-temperature readings has \(Q_1=8^\circ\text{C}\) and \(Q_3=12^\circ\text{C}\). Use the \(1.5\cdot\text{IQR}\) rule to decide which of these values would be marked as outliers: \(2^\circ\text{C}\), \(18^\circ\text{C}\), and \(21^\circ\text{C}\). Values exactly on a fence are not outliers.

Hints

- First calculate the spread between the quartiles. - Find both the lower and upper fences. - Compare each proposed reading with the strict outside-the-fence condition.

Solution

1. The IQR is \(12-8=4^\circ\text{C}\). 2. The fences are \(8-1.5\cdot4=2^\circ\text{C}\) and \(12+1.5\cdot4=18^\circ\text{C}\). 3. Both \(2^\circ\text{C}\) and \(18^\circ\text{C}\) lie on the fences, not beyond them. Only \(21^\circ\text{C}\) is flagged.

Answer

Only \(21^\circ\text{C}\) is an outlier under the rule; the other two values are exactly on the fences.
Display and summarize data 45
The bars show the number of bicycles rented on two days. a) Which day had more rentals? b) Why would comparing the colored areas of the bars give a misleading impression? How should the bar widths be drawn for a fair comparison?
Figure for problem 560452

Hints

- Read the heights of the bars using the vertical axis. - Notice that one bar covers a much wider horizontal span. - In a bar chart, what graphical feature should encode the count?

Solution

1. Tuesday's bar reaches \(12\) rentals, whereas Monday's reaches \(10\), so Tuesday had more. 2. The Monday bar is much wider even though its height is smaller. Its larger area can make it look as though Monday has more rentals. 3. Bars that compare counts should have equal widths; use their heights to compare totals.

Answer

a) Tuesday, with \(12\) rentals compared with \(10\) on Monday b) Unequal bar widths can make the larger area look like the larger count. Both bars should have the same width.
Display and summarize data 46
A delivery driver recorded the length in miles of \(12\) local trips: \(2,5,8,4,10,13,14,18,9,1,7,12\) Group the trips into these intervals and give the frequency of each: \(0\) to less than \(5\), \(5\) to less than \(10\), \(10\) to less than \(15\), and \(15\) to less than \(20\) miles.

Hints

- Check which interval contains each listed distance. - Treat the left endpoint as included and the right endpoint as excluded. - Add the completed interval counts to check that every trip was counted exactly once.

Solution

1. The first interval contains \(1,2,4\), for a count of \(3\). 2. The next two intervals contain \(5,7,8,9\) and \(10,12,13,14\), for \(4\) trips each. 3. The final interval contains \(18\), for \(1\) trip. The frequencies total \(3+4+4+1=12\).

Answer

\(0\) to less than \(5\) miles: \(3\); \(5\) to less than \(10\) miles: \(4\); \(10\) to less than \(15\) miles: \(4\); \(15\) to less than \(20\) miles: \(1\).
Display and summarize data 47
A café tracked how long customers waited for their orders. The ogive shows the cumulative number of waits below each time limit. a) How many customers waited less than \(10\) minutes? b) How many customers waited at least \(10\) minutes but less than \(15\) minutes? Explain how you found this number from the graph.
Figure for problem 560492

Hints

- An ogive shows how many observations fall below a given interval boundary. - Read the heights of the curve at the two time limits. - To isolate one interval, think about which earlier observations are already included in the later total.

Solution

1. The ogive reaches a cumulative count of \(8\) at \(10\) minutes, so \(8\) customers waited less than \(10\) minutes. 2. At \(15\) minutes the cumulative count is \(14\). Subtract the earlier total: \(14-8=6\). 3. The difference gives the customers in the interval from \(10\) to less than \(15\) minutes because cumulative totals include earlier intervals.

Answer

a) \(8\) customers b) \(6\) customers; subtract the cumulative count at \(10\) minutes from the count at \(15\) minutes.
Display and summarize data 48
A class recorded quiz scores. The first quartile is \(Q_1=5\) and the third quartile is \(Q_3=11\). The lowest score is \(3\), and the three highest scores are \(11\), \(12\), and \(25\). A modified box plot uses the fences \(Q_1-1.5\cdot\text{IQR}\) and \(Q_3+1.5\cdot\text{IQR}\). Values outside the fences are plotted as separate outliers. Each whisker ends at the most extreme score that is not an outlier. a) Find the lower and upper fences. b) Where do the two whiskers end? Which score is plotted as an outlier?

Hints

- Start by finding the difference between the two quartiles. - Compare the lowest and highest scores with the two fences. - A whisker ends at a score in the data, not at the numerical fence itself.

Solution

1. The interquartile range is \(11-5=6\). 2. The lower fence is \(5-1.5\cdot6=-4\) and the upper fence is \(11+1.5\cdot6=20\). 3. The lowest score, \(3\), is inside the fences. The largest score inside the fences is \(12\), so the whiskers end at \(3\) and \(12\). 4. The score \(25\) is above the upper fence, so it is plotted separately as an outlier.

Answer

a) Lower fence: \(-4\); upper fence: \(20\). b) Whisker endpoints: \(3\) and \(12\); outlier: \(25\).
Display and summarize data 49
A survey asked \(200\) students how many minutes their trip to school takes. The histogram shows the relative frequencies in \(10\)-minute intervals. a) State the relative frequency for each of the five intervals. b) How many students have a travel time from \(10\) to \(20\) minutes? c) Estimate the mean travel time. Use each interval midpoint, such as \(5\) minutes for the interval from \(0\) to \(10\) minutes.
Figure for problem 535131

Hints

- What does a height of 35 on this percentage axis mean? - How could you use the number surveyed to change a percentage into a number of students? - When all you know is a time interval, what single time could stand for that interval in an estimate?

Solution

1. The bar heights are \(20\%\), \(35\%\), \(25\%\), \(15\%\), and \(5\%\) for the consecutive \(10\)-minute intervals. 2. The second interval represents \(0.35\cdot 200=70\) students. 3. The midpoint estimate is \(5\cdot 0.20+15\cdot 0.35+25\cdot 0.25+35\cdot 0.15+45\cdot 0.05=20\) minutes.

Answer

a) \(0\)–\(10\): \(20\%\); \(10\)–\(20\): \(35\%\); \(20\)–\(30\): \(25\%\); \(30\)–\(40\): \(15\%\); \(40\)–\(50\): \(5\%\) b) \(70\) students c) Approximately \(20\) minutes
Display and summarize data 50
Two prototype delivery drones were tested for flight time, in minutes, before returning to base. Drone A: \(18, 19, 20, 20, 21, 22, 60\) Drone B: \(14, 17, 19, 20, 21, 23, 26\) For quartiles, when a data set has an odd number of observations, exclude the overall median from both halves. a) For comparing a typical flight time and consistency, decide whether mean and standard deviation or median and interquartile range are more appropriate. b) Find the median and interquartile range for each drone. c) Compare the two distributions in context.

Hints

- Look for any value that is far from the rest before choosing summary statistics. - Use the stated quartile convention after locating the overall median. - Compare both center and spread rather than using only one statistic.

Solution

1. Drone A has an extreme value of \(60\), so resistant statistics are more appropriate. 2. For Drone A, the median is \(20\). The lower half is \(18, 19, 20\), so \(Q_1=19\). The upper half is \(21, 22, 60\), so \(Q_3=22\). Thus, \(\text{IQR}=22-19=3\). 3. For Drone B, the median is \(20\). The lower half is \(14, 17, 19\), so \(Q_1=17\). The upper half is \(21, 23, 26\), so \(Q_3=23\). Thus, \(\text{IQR}=23-17=6\). 4. Both drones have the same median flight time, but Drone A has the smaller IQR. Its middle half is more tightly clustered, although the \(60\)-minute value is unusually high.

Answer

a) Median and interquartile range are more appropriate because Drone A has an extreme value. b) Drone A: median \(20\) minutes, IQR \(3\) minutes. Drone B: median \(20\) minutes, IQR \(6\) minutes. c) The typical flight times are the same, but Drone A is more consistent in the middle half of its data. Drone A also has an unusually long \(60\)-minute result.
Display and summarize data 51
A ceramics class counted small glaze flaws on a batch of tiles. <table><tr><th>Flaws on a tile</th><th>Number of tiles</th></tr><tr><td>\(0\)</td><td>\(4\)</td></tr><tr><td>\(1\)</td><td>\(x\)</td></tr><tr><td>\(2\)</td><td>\(5\)</td></tr><tr><td>\(3\)</td><td>\(1\)</td></tr></table> The mean number of flaws was \(1.2\). a) Find \(x\). b) Find the median and mode of the completed data set.

Hints

- Express the total number of tiles and the total number of flaws in terms of the unknown frequency. - Use the definition of mean to connect those two totals. - After finding the frequency, use positions in the ordered data rather than writing every value separately.

Solution

1. The total number of flaws is \(x+2\cdot 5+3\cdot 1=x+13\). The total number of tiles is \(4+x+5+1=x+10\). 2. Use the stated mean: \(\frac{x+13}{x+10}=1.2\). Then \(x+13=1.2x+12\), so \(1=0.2x\) and \(x=5\). 3. There are \(15\) tiles. The median is the eighth ordered value. Positions \(5\) through \(9\) are tiles with \(1\) flaw, so the median is \(1\). 4. Both \(1\) flaw and \(2\) flaws occur \(5\) times, more often than any other value. The data set has two modes: \(1\) and \(2\).

Answer

a) \(x=5\) b) Median: \(1\) flaw. Modes: \(1\) flaw and \(2\) flaws.
Display and summarize data 52
Two dot plots show the number of bonus tokens earned in eight rounds of two arcade games. Plot a) shows Game A, and plot b) shows Game B. a) Show that the two data sets have the same mean, median, and range. b) Describe one important difference in their shapes. c) Would a box plot show that difference as clearly as the dot plots? Explain.
Figure for problem 544000

Hints

- Read the value represented by every dot before calculating the summaries. - Compare where the dots stack and where gaps occur. - Consider which details disappear when data are reduced to a five-number summary.

Solution

1. Reading the dot plots gives Game A: \(2, 2, 2, 5, 5, 8, 8, 8\) and Game B: \(2, 3, 4, 5, 5, 6, 7, 8\). 2. Each data set has sum \(40\), so each mean is \(40\div 8=5\). 3. In each ordered set, the fourth and fifth values are both \(5\), so each median is \(5\). Each range is \(8-2=6\). 4. Game A has clusters at the low and high ends, with fewer observations in the middle. Game B is spread more evenly across the interval. 5. A box plot would not show the difference as clearly because it does not display the frequency at each exact value.

Answer

a) Both data sets have mean \(5\), median \(5\), and range \(6\). b) Game A is concentrated near \(2\) and \(8\), while Game B is distributed more evenly. c) No. The dot plots show the stacks and gaps at exact values, while a box plot would summarize the data.
Display and summarize data 53
At a museum, one guide led \(18\) visitors through a short exhibit with a mean viewing time of \(7.5\) minutes. A second guide led \(12\) visitors through the same exhibit with a mean viewing time of \(10.5\) minutes. Find the mean viewing time for all visitors. Explain why simply averaging \(7.5\) and \(10.5\) would not be correct.

Hints

- Would the two groups have equal influence on the overall average? Why? - What information is missing if you know only the two averages? - How might each group's number of visitors help you recover that information?

Solution

1. The first group contributed \(18\cdot7.5=135\) total visitor-minutes. 2. The second group contributed \(12\cdot10.5=126\) total visitor-minutes. 3. The total is \(261\) visitor-minutes over \(18+12=30\) visitors. 4. The combined mean is \(261\div30=8.7\) minutes. 5. Simply averaging the two means would give equal weight to different-sized groups.

Answer

The combined mean is \(8.7\) minutes. Averaging the two group means directly is wrong because the groups have different sizes.
Display and summarize data 54
A rowing coach recorded recovery times, in minutes, as \(12, 14, 15, 16, 18, 105\). The final entry was a typing error and should have been \(15\). a) Find the mean and median before the correction. b) Find the mean and median after the correction. c) Which measure of center was more resistant to the incorrect extreme value? Explain using the changes.

Hints

- Calculate each center once with the incorrect entry and once with the corrected entry. - Compare the size of the two changes rather than only the final values. - A resistant statistic changes relatively little when one extreme observation changes greatly.

Solution

1. Before correction, the sum is \(180\), so the mean is \(180\div 6=30\). The median is \(\frac{15+16}{2}=15.5\). 2. After replacing \(105\) with \(15\), the ordered data are \(12, 14, 15, 15, 16, 18\). The sum is \(90\), so the mean is \(90\div 6=15\). The median is \(\frac{15+15}{2}=15\). 3. The mean changes by \(15\) minutes, while the median changes by only \(0.5\) minute. The median is more resistant to the extreme value.

Answer

a) Before correction: mean \(30\) minutes, median \(15.5\) minutes. b) After correction: mean \(15\) minutes, median \(15\) minutes. c) The median was more resistant because it changed by only \(0.5\) minute, while the mean changed by \(15\) minutes.
Display and summarize data 55
A greenhouse recorded daily temperatures in degrees Celsius. The data had mean \(18^\circ\text{C}\), median \(17^\circ\text{C}\), population standard deviation \(4^\circ\text{C}\), and interquartile range \(6^\circ\text{C}\). The temperatures are converted to degrees Fahrenheit using \(F=1.8C+32\). Find the four summary statistics in degrees Fahrenheit.

Hints

- Apply the conversion to locations in the distribution. - A difference between two temperatures is affected by the scale factor but not by the added constant. - Separate the measures of center from the measures of spread.

Solution

1. Measures of center follow the full conversion rule. The mean becomes \(1.8\cdot 18+32=64.4^\circ\text{F}\), and the median becomes \(1.8\cdot 17+32=62.6^\circ\text{F}\). 2. Measures of spread are multiplied by \(1.8\) but are not increased by \(32\). The population standard deviation becomes \(1.8\cdot 4=7.2^\circ\text{F}\), and the interquartile range becomes \(1.8\cdot 6=10.8^\circ\text{F}\).

Answer

Mean: \(64.4^\circ\text{F}\) Median: \(62.6^\circ\text{F}\) Population standard deviation: \(7.2^\circ\text{F}\) Interquartile range: \(10.8^\circ\text{F}\)
Display and summarize data 56
A quality score is calculated from the number of defects \(x\) by the rule \(y=100-2x\). For one production week, the defect counts had mean \(12\), median \(11\), population standard deviation \(3\), and interquartile range \(4\). Find the mean, median, population standard deviation, and interquartile range of the quality scores.

Hints

- Apply the scoring rule to measures that locate the center. - A negative scale factor reverses order but does not make spread negative. - Use the magnitude of the scale factor for distance-based summaries.

Solution

1. The mean follows the linear transformation: \(100-2\cdot 12=76\). 2. Because the rule reverses the order but remains linear, the median transforms to \(100-2\cdot 11=78\). 3. Measures of spread are multiplied by the absolute value of the scale factor. The population standard deviation is \(2\cdot 3=6\), and the interquartile range is \(2\cdot 4=8\).

Answer

Mean: \(76\) Median: \(78\) Population standard deviation: \(6\) Interquartile range: \(8\)
Display and summarize data 57
Four calibration offsets are \(6, 8, 10, 12\) units. A fifth offset equal to the original mean is added to the data set. a) Find the original mean and population standard deviation. b) Find the new mean and population standard deviation. c) Explain why the center stays fixed while the spread decreases.

Hints

- Find the balance point of the original four values first. - Compare the squared-deviation total before and after adding the new value. - The denominator in the population variance changes even when the numerator does not.

Solution

1. The original mean is \((6+8+10+12)\div 4=9\). 2. The squared deviations from \(9\) are \(9, 1, 1, 9\), with sum \(20\). The original population standard deviation is \(\sqrt{20\div 4}=\sqrt{5}\approx 2.24\). 3. Adding the original mean gives \(6, 8, 9, 10, 12\). The new mean remains \(9\). 4. The added value has deviation \(0\), so the squared-deviation sum remains \(20\). The new population standard deviation is \(\sqrt{20\div 5}=2\). 5. The added observation is at the center, so it does not shift the mean and reduces the average squared distance from the mean.

Answer

a) Original mean: \(9\). Original population standard deviation: \(\sqrt{5}\approx 2.24\). b) New mean: \(9\). New population standard deviation: \(2\). c) The new value is exactly at the mean, so it adds no deviation while increasing the population size.
Display and summarize data 58
A school counted the students riding bus routes 1, 2, and 3. The chart shows the results. Someone says route 3 carries about three times as many students as route 1. Is that true? Use the counts and explain why the bars can give that impression.
Figure for problem 544019

Hints

- Read the bar values from the numbered y-axis rather than comparing just the visible lengths. - Compare the full heights as a ratio. - What part of each bar would be shown if the axis started at zero?

Solution

1. The counts are \(42\), \(44\) and \(46\). Route 3 has only \(46-42=4\) more riders than route 1. 2. Their ratio is \(\frac{46}{42}\approx1.10\), not \(3\). 3. Because the vertical axis starts at \(40\), only the lengths above \(40\) are shown; these visible lengths are \(2\) and \(6\), with ratio \(3\), making the full counts look farther apart.

Answer

No. Route 3 has \(46\) riders and route 1 has \(42\), about \(1.10\) times as many. The axis starts at \(40\), exaggerating the small difference.
Display and summarize data 59
Five scores in an arcade contest are \(62, 68, 71, 74, 80\). The first score was written down incorrectly. After it is corrected, the mean is \(75\). a) Find the corrected value. b) Find the new median. c) Explain why the median changes from \(71\) to \(74\).

Hints

- How does a mean relate to the sum of all five scores? - Which four scores stay the same after the correction? - Where does the corrected score belong in the ordered list?

Solution

1. A mean of \(75\) for five scores requires a total of \(5\cdot75=375\). 2. The unchanged scores total \(68+71+74+80=293\), so the corrected score is \(375-293=82\). 3. In order, the new scores are \(68,71,74,80,82\), with median \(74\). 4. The smallest score was replaced by the largest, shifting the four unchanged scores one position to the left in the ordered list. This moves \(74\) into the middle position.

Answer

a) \(82\) b) \(74\) c) Replacing the lowest score with \(82\) moves \(74\) from fourth to third in the ordered list.
Display and summarize data 60
Two collections have the following summaries. Collection A: mean \(10\), population standard deviation \(4\) Collection B: mean \(100\), population standard deviation \(4\) a) What can be concluded about their absolute spread around their means? b) Can the summaries alone justify the claim that the two collections are equally variable relative to their typical sizes? Explain.

Hints

- Interpret standard deviation in the original measurement units. - Compare the size of the spread with the size of each center. - Distinguish an absolute difference from a difference relative to scale.

Solution

1. Equal population standard deviations mean that observations in both collections have the same typical absolute distance from their respective means, measured in the original units. 2. The means are on very different scales. A spread of \(4\) is large compared with \(10\) but small compared with \(100\). 3. Therefore, the collections have equal absolute standard deviations, but the given summaries do not support calling their relative variability equal.

Answer

a) Both collections have the same absolute population standard deviation of \(4\) units. b) No. The same absolute spread represents different proportions of means \(10\) and \(100\), so equal relative variability does not follow.
Display and summarize data 61
The cumulative-frequency graph summarizes the number of support requests handled by \(14\) student volunteers. a) Find the ordinary frequency for each value \(1, 2, 3, 4, 5\). b) Using the median-of-halves convention, find \(Q_1\), the median, \(Q_3\), and the interquartile range.
Figure for problem 544023

Hints

- Read the cumulative total at each integer value. - Recover each ordinary frequency by subtracting the previous cumulative total. - Use the resulting frequencies to locate positions in the ordered data.

Solution

1. Read cumulative frequencies \(2, 5, 9, 12, 14\) at \(x=1,2,3,4,5\). 2. Subtract consecutive cumulative frequencies. The ordinary frequencies for \(1,2,3,4,5\) are \(2,3,4,3,2\), respectively. 3. The ordered list has \(14\) values. Positions \(7\) and \(8\) are both \(3\), so the median is \(3\). 4. The lower half has \(7\) values, and its fourth value is \(2\), so \(Q_1=2\). The upper half has \(7\) values, and its fourth value is \(4\), so \(Q_3=4\). 5. The interquartile range is \(4-2=2\).

Answer

a) Frequencies: \(1\mapsto 2\), \(2\mapsto 3\), \(3\mapsto 4\), \(4\mapsto 3\), \(5\mapsto 2\). b) \(Q_1=2\), median \(=3\), \(Q_3=4\), and \(\text{IQR}=2\).
Display and summarize data 62
Six lantern tests had a mean burn time of \(12\) hours. Several additional tests, each lasting exactly \(18\) hours, were added to the data set. The new mean became \(14\) hours. How many \(18\)-hour tests were added?

Hints

- Convert the original mean into an original total. - Represent both the new total and the new number of tests using one unknown. - Use the new mean to form an equation.

Solution

1. The original total burn time is \(6\cdot 12=72\) hours. 2. Let \(k\) be the number of added tests. The new mean gives \(\frac{72+18k}{6+k}=14\). 3. Solve: \(72+18k=84+14k\), so \(4k=12\) and \(k=3\).

Answer

\(3\) additional tests were added.
Display and summarize data 63
The numbers of minutes needed to reset eight escape-room puzzles were \(4, 5, 6, 7, 8, 9, 10, 24\). Use the \(1.5\cdot\text{IQR}\) rule to identify any outlier. Then state the endpoints of the whiskers in a modified box plot.

Hints

- Find the quartiles before looking for unusual values. - Build a lower and upper boundary from the interquartile range. - Modified whiskers stop at the most extreme values that are not outliers.

Solution

1. The lower half is \(4,5,6,7\), so \(Q_1=\frac{5+6}{2}=5.5\). The upper half is \(8,9,10,24\), so \(Q_3=\frac{9+10}{2}=9.5\). 2. The interquartile range is \(9.5-5.5=4\). 3. The fences are \(5.5-1.5\cdot 4=-0.5\) and \(9.5+1.5\cdot 4=15.5\). 4. The value \(24\) exceeds the upper fence, so it is an outlier. The smallest and largest non-outliers are \(4\) and \(10\), so the whiskers end there.

Answer

The outlier is \(24\) minutes. The modified-box-plot whiskers end at \(4\) minutes and \(10\) minutes.
Display and summarize data 64
Six trail markers were placed at distances \(2, 4, 5, 7, 9, 11\) miles from a visitor center. A report is revised by removing the nearest and farthest markers. Compare the mean, median, and range before and after the two endpoints are removed.

Hints

- Calculate the three summaries for the complete set first. - Remove only the two endpoint values, then recalculate. - Compare how location statistics and endpoint-based spread respond differently.

Solution

1. For all six distances, the mean is \(\frac{38}{6}=\frac{19}{3}\approx 6.33\) miles, the median is \(\frac{5+7}{2}=6\) miles, and the range is \(11-2=9\) miles. 2. After removing \(2\) and \(11\), the data are \(4,5,7,9\). The mean is \(\frac{25}{4}=6.25\) miles, the median is \(\frac{5+7}{2}=6\) miles, and the range is \(9-4=5\) miles. 3. The median is unchanged, the mean decreases slightly, and the range decreases substantially.

Answer

Before removal: mean \(\frac{19}{3}\approx 6.33\) miles, median \(6\) miles, range \(9\) miles. After removal: mean \(6.25\) miles, median \(6\) miles, range \(5\) miles.
Display and summarize data 65
A robotics team counted the number of sensor resets during \(12\) test runs. <table><tr><th>Resets</th><th>Frequency</th></tr><tr><td>\(0\)</td><td>\(2\)</td></tr><tr><td>\(1\)</td><td>\(a\)</td></tr><tr><td>\(2\)</td><td>\(b\)</td></tr><tr><td>\(3\)</td><td>\(1\)</td></tr></table> The mean number of resets was \(1.5\). Find \(a\) and \(b\).

Hints

- Use the total number of runs to form one equation from the frequencies. - Convert the mean into the total number of resets. - Solve the two conditions together.

Solution

1. The frequencies total \(12\), so \(2+a+b+1=12\), which gives \(a+b=9\). 2. A mean of \(1.5\) for \(12\) runs requires \(18\) total resets. Thus, \(1\cdot a+2\cdot b+3\cdot 1=18\), so \(a+2b=15\). 3. Subtract \(a+b=9\) from \(a+2b=15\) to get \(b=6\). Then \(a=9-6=3\).

Answer

\(a=3\) and \(b=6\).
Display and summarize data 66
One rehearsal group has practice counts \(1,1,1\), and a second group has practice counts \(9,9,9,9,9\). Find the two group medians, their average, and the actual median of the combined data. Determine whether averaging the two group medians gives the combined median, and explain why or why not.

Hints

- Find each group’s middle independently first. - Then combine all observations and reorder them. - Unequal group sizes affect which observations occupy the combined middle positions.

Solution

1. The first group median is \(1\), and the second group median is \(9\). 2. Their average is \(\frac{1+9}{2}=5\). 3. The combined ordered data are \(1,1,1,9,9,9,9,9\). The fourth and fifth values are both \(9\), so the combined median is \(9\). 4. Because the group sizes differ, averaging the two medians does not locate the middle of the combined data.

Answer

The group medians are \(1\) and \(9\), and their average is \(5\). The actual combined median is \(9\), so averaging the group medians does not give the combined median when the group sizes differ.
Display and summarize data 67
Ten water samples were reported to have a mean mineral concentration of \(31.2\,\text{mg/L}\). Later, one entry of \(84\,\text{mg/L}\) was found to be a typing error; the correct value was \(24\,\text{mg/L}\). Find the corrected mean without reconstructing the other nine measurements.

Hints

- Recover the reported total from the reported mean and sample size. - Remove the incorrect contribution and insert the corrected one. - The number of samples does not change.

Solution

1. The reported total concentration is \(10\cdot 31.2=312\,\text{mg/L}\) when the ten numerical entries are added. 2. Replace the incorrect entry in the total: \(312-84+24=252\). 3. Divide by \(10\): the corrected mean is \(252\div 10=25.2\,\text{mg/L}\).

Answer

The corrected mean is \(25.2\,\text{mg/L}\).
Display and summarize data 68
Five measurements have deviations from their mean of \(-5, -2, 1, 4,\) and \(d\). a) Find \(d\). b) Find the population standard deviation of the measurements. Round to the nearest hundredth.

Hints

- Express each deviation as an observation minus the mean. - Use the definition of the mean to simplify the sum of all deviations. - After finding the missing deviation, average the squared deviations and take a square root.

Solution

1. If the measurements are \(x_1,\ldots,x_5\) with mean \(\bar{x}\), then the sum of their deviations is \((x_1-\bar{x})+\cdots+(x_5-\bar{x})=(x_1+\cdots+x_5)-5\bar{x}=0\). 2. Therefore, \(-5-2+1+4+d=0\), so \(-2+d=0\) and \(d=2\). 3. The squared deviations total \(25+4+1+16+4=50\). 4. The population variance is \(50\div 5=10\), so the population standard deviation is \(\sqrt{10}\approx 3.16\).

Answer

a) \(d=2\) b) The population standard deviation is approximately \(3.16\).
Display and summarize data 69
A five-value data set has mean \(6\), median \(6\), range \(8\), and interquartile range \(8\). Use the median-of-halves convention. Which option matches all four summaries? A: \(2,4,6,8,10\) B: \(1,5,6,7,11\) C: \(2,2,6,10,10\)

Hints

- Check the easiest summaries first to eliminate options efficiently. - Apply the stated quartile convention consistently. - An option must satisfy every condition, not only the mean and median.

Solution

1. Option A has mean \(6\), median \(6\), and range \(8\). Its quartiles are \(3\) and \(9\), so its IQR is \(6\). 2. Option B has mean \(6\) and median \(6\), but its range is \(11-1=10\), so it does not match. 3. Option C has mean \(\frac{30}{5}=6\), median \(6\), and range \(10-2=8\). Its quartiles are \(2\) and \(10\), so its IQR is \(8\). 4. Only Option C matches every summary.

Answer

Option C: \(2,2,6,10,10\).
Display and summarize data 70
Consider these two ordered data sets. Use the convention that, for an odd number of observations, the overall median is excluded when finding \(Q_1\) and \(Q_3\). Set A: \(0,2,2,4,6,6,8\) Set B: \(0,2,2,4,4,6,8\) a) Show that both sets have the same five-number summary. b) Find each mean. c) Explain what this example shows about the information lost in a box plot.

Hints

- Apply the stated quartile convention to the lower and upper halves after identifying the overall median. - Use all seven values when finding each mean. - Compare which information appears in a five-number summary with which information the mean uses.

Solution

1. In each set, the median is \(4\). Excluding that median, the lower half has median \(2\) and the upper half has median \(6\). Both sets therefore have minimum \(0\), \(Q_1=2\), median \(4\), \(Q_3=6\), and maximum \(8\). 2. Set A has sum \(28\), so its mean is \(28\div 7=4\). 3. Set B has sum \(26\), so its mean is \(26\div 7=\frac{26}{7}\approx 3.71\). 4. The same five-number summary can come from different interior values and different means. A box plot does not preserve every observation.

Answer

a) Both five-number summaries are \(0,2,4,6,8\). b) Set A mean: \(4\). Set B mean: \(\frac{26}{7}\approx 3.71\). c) A box plot can be identical for data sets with different individual values and different means.
Display and summarize data 71
The ordered values \(2,5,7,9\) are joined by a fifth value \(x\), where \(x\ge 9\). Find \(x\) so that the mean of the five values equals their median.

Hints

- Use the condition on \(x\) to determine the median before solving. - Convert “mean equals median” into an equation. - Check that the solution remains in the stated position in the ordered list.

Solution

1. Because \(x\ge 9\), the completed order is \(2,5,7,9,x\), so the median is \(7\). 2. Set the mean equal to \(7\): \(\frac{2+5+7+9+x}{5}=7\). 3. Then \(23+x=35\), so \(x=12\). 4. The value \(12\) satisfies the condition \(x\ge 9\), and the completed data have both mean and median \(7\).

Answer

\(x=12\)
Display and summarize data 72
An ordered data set has at least five values. Its minimum is increased by \(5\), its maximum is decreased by \(5\), and all other values stay unchanged. The revised endpoints still remain the smallest and largest values. Determine what happens to the mean, median, and range.

Hints

- Track the net change in the total used for the mean. - Ask whether either endpoint determines the middle position. - Consider how moving both endpoints inward changes their difference.

Solution

1. The total changes by \(+5-5=0\), so the mean is unchanged. 2. Only the two endpoints change. Because the data set has at least five values and the order is preserved, neither endpoint occupies a middle position, so the median is unchanged. 3. The minimum moves up by \(5\) and the maximum moves down by \(5\), so the distance between them decreases by \(10\). The range decreases by \(10\).

Answer

The mean and median remain unchanged. The range decreases by \(10\).
Display and summarize data 73
A survey compared participation in music lessons with honor-roll status. <table><tr><th>Music lessons</th><th>Honor roll</th><th>Not on honor roll</th><th>Total</th></tr><tr><td>Yes</td><td>\(24\)</td><td>\(16\)</td><td>\(40\)</td></tr><tr><td>No</td><td>\(36\)</td><td>\(44\)</td><td>\(80\)</td></tr><tr><td>Total</td><td>\(60\)</td><td>\(60\)</td><td>\(120\)</td></tr></table> a) Among students taking music lessons, what percentage are on the honor roll? b) Among honor-roll students, what percentage take music lessons? c) Explain why the two percentages are different, and compare honor-roll rates for students with and without music lessons.

Hints

- Name the group described after the word “among”; that group supplies the denominator. - Do not reverse the conditioning group when forming a percentage. - For the final comparison, calculate the same outcome rate within both lesson groups.

Solution

1. Among students taking music lessons, the honor-roll percentage is \(\frac{24}{40}=60\%\). 2. Among honor-roll students, the music-lesson percentage is \(\frac{24}{60}=40\%\). 3. The percentages use different conditioning groups and therefore different denominators. 4. The honor-roll rate without music lessons is \(\frac{36}{80}=45\%\). The rate is \(15\) percentage points higher among students taking music lessons, showing an association in this survey.

Answer

a) \(60\%\) b) \(40\%\) c) The denominators differ: part a conditions on music lessons, while part b conditions on honor-roll status. Honor-roll rates are \(60\%\) with music lessons and \(45\%\) without, a \(15\)-percentage-point difference.
Display and summarize data 74
Riverside High recorded orientation completion for \(90\) students choosing between in-person and online orientation. Of the \(90\) students, \(54\) chose in-person orientation. A total of \(45\) students completed orientation, and \(36\) students both chose in-person orientation and completed it. a) Construct a two-way frequency table with rows for in-person and online orientation and columns for completed and not completed. Include row and column totals. b) Find the completion rate within each orientation group to the nearest tenth of a percent. c) Describe the association shown by the conditional rates.

Hints

- Start with the one intersection count that is given directly, then use row and column totals to fill neighboring cells. - Check that every row total, column total, and the grand total agree before calculating percentages. - For conditional completion rates, divide by the total within each orientation group rather than by \(90\).

Solution

1. The in-person/completed cell is \(36\). Since \(54\) students chose in-person orientation, the in-person/not-completed cell is \(54-36=18\). 2. The online total is \(90-54=36\). Since \(45\) students completed orientation overall, the online/completed cell is \(45-36=9\). 3. The online/not-completed cell is \(36-9=27\). The not-completed column total is \(18+27=45\). 4. The in-person completion rate is \(\frac{36}{54}\approx 66.7\%\). The online completion rate is \(\frac{9}{36}=25.0\%\). 5. Completion is associated with orientation format in this group: the completion rate is about \(41.7\) percentage points higher for students who chose in-person orientation.

Answer

a) <table><tr><th>Orientation</th><th>Completed</th><th>Not completed</th><th>Total</th></tr><tr><td>In-person</td><td>\(36\)</td><td>\(18\)</td><td>\(54\)</td></tr><tr><td>Online</td><td>\(9\)</td><td>\(27\)</td><td>\(36\)</td></tr><tr><td>Total</td><td>\(45\)</td><td>\(45\)</td><td>\(90\)</td></tr></table> b) In-person: \(66.7\%\); online: \(25.0\%\) c) Orientation format and completion are associated in this group; the in-person completion rate is about \(41.7\) percentage points higher.
Display and summarize data 75
Ten students recorded whole-number hours spent on an art project. The grouped data are shown. <table><tr><th>Hours</th><th>Frequency</th></tr><tr><td>\(0\)–\(4\)</td><td>\(3\)</td></tr><tr><td>\(5\)–\(9\)</td><td>\(4\)</td></tr><tr><td>\(10\)–\(14\)</td><td>\(3\)</td></tr></table> a) Find the smallest possible mean and the largest possible mean consistent with the table. b) Find the midpoint estimate of the mean. c) Explain why the exact mean cannot be determined.

Hints

- For a lower bound, place every observation at the smallest value allowed in its interval. - For an upper bound, use the largest allowed value in each interval. - A midpoint estimate replaces every observation in an interval with the interval’s midpoint.

Solution

1. The smallest total occurs when every value is at the lower endpoint of its interval: \(3\cdot 0+4\cdot 5+3\cdot 10=50\). The smallest possible mean is \(50\div 10=5\) hours. 2. The largest total occurs when every value is at the upper endpoint: \(3\cdot 4+4\cdot 9+3\cdot 14=90\). The largest possible mean is \(90\div 10=9\) hours. 3. Using interval midpoints \(2, 7, 12\), the estimated total is \(3\cdot 2+4\cdot 7+3\cdot 12=70\), so the midpoint estimate is \(70\div 10=7\) hours. 4. The table does not show the individual values within each interval, so different data sets can produce different exact means.

Answer

a) Smallest possible mean: \(5\) hours. Largest possible mean: \(9\) hours. b) Midpoint estimate: \(7\) hours. c) The exact values inside the intervals are unknown, so the exact mean is not determined.
Display and summarize data 76
The ordered data set \(4,6,8,10,12,12,14,16,20\) has one new value \(x\) added. Find all values of \(x\) for which the median of the resulting ten-value data set is still \(12\).

Hints

- Identify which ordered positions determine the median after one value is added. - Consider separately where the new value falls relative to the two existing \(12\)s. - The middle pair must average to exactly \(12\).

Solution

1. With ten values, the median is the mean of the fifth and sixth ordered values. 2. If \(x\ge 12\), the two original \(12\)s remain in positions five and six or earlier adjacent middle positions, so the middle pair is \(12,12\) and the median is \(12\). 3. If \(x<12\), the new value is inserted before the two \(12\)s. The fifth value is then less than \(12\), while the sixth value is \(12\), so their mean is less than \(12\). 4. Therefore, the required values are \(x\ge 12\).

Answer

\(x\ge 12\)

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