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Display and summarize data

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5439999
A wildlife center recorded the number of minutes each of \(18\) 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

- Focus on the phrase “every exact value.” - Think about whether each display shows individual observations or combines them. - Consider how repeated values would appear in each display.

Solution

1. A dot plot places one mark for each observation, so all \(18\) exact values remain visible and repeated values form stacks. 2. A histogram groups values into intervals, so exact observations may be hidden. 3. A box plot summarizes the distribution with five key values, so it does not show each observation or all repetitions.

Answer

A dot plot is most appropriate. It preserves every exact feeding time and makes repeated values visible. A histogram groups values, and a box plot gives only a summary.
5440079
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.
5440169
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: \(10\)
5440289
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: \(17\)
5440299
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.
5513849
A data set is \(1,2,2,3,3,3,4,5,5\). A student 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.
5139249
The function \(T(x)=x^2+5\) is given. a) Find \(T(1)\), \(T(2)\), \(T(3)\), \(T(4)\), and \(T(5)\). b) Which of these inputs gives the greatest function value?

Hints

- Substitute each listed input for \(x\) in the function rule. - Keep the input and its output paired as you work. - For part b), compare the outputs after you have evaluated all five inputs.

Solution

1. Substitute each input into \(T(x)=x^2+5\): \(T(1)=6\), \(T(2)=9\), \(T(3)=14\), \(T(4)=21\), and \(T(5)=30\). 2. Compare the five outputs. The greatest is \(30\), which occurs at input \(x=5\).

Answer

a) \(T(1)=6\), \(T(2)=9\), \(T(3)=14\), \(T(4)=21\), and \(T(5)=30\) b) \(x=5\)
5240679
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

- Multiply each team's mean by its number of members to find its total points. - Divide the combined point total by the combined number of participants. - In part c), substitute \(m=n\), factor, and simplify.

Solution

a) Team A earned \(na\) total points, and Team B earned \(mb\) total points. Together, they earned \(na+mb\) points. b) There are \(n+m\) participants, so the combined mean is \(\frac{na+mb}{n+m}\). c) If \(n=m\), then \(\frac{na+nb}{n+n}=\frac{n(a+b)}{2n}=\frac{a+b}{2}\). This is 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
5439969
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.
5439979
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.
5440019
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
5440039
A box plot summarizes the number of lines spoken by each of \(80\) students in a theater workshop. a) Approximately how many students spoke between the first and third quartiles? b) Approximately how many students spoke at most the median number of lines? c) Can the box plot determine exactly how many students spoke more than \(20\) lines? Explain.
Figure for problem 544003

Hints

- Identify the portion of a box plot represented by the box itself. - Use the median line to reason about the lower half of the ordered data. - Check whether the display shows individual observations between a quartile and a whisker endpoint.

Solution

1. The box extends from the first quartile, \(9\), to the third quartile, \(18\). It contains the middle \(50\%\) of the data, so approximately \(0.50\cdot 80=40\) students are in that interval. 2. The median is \(13\) and divides the ordered data into two halves. Approximately \(0.50\cdot 80=40\) students spoke at most \(13\) lines. 3. The value \(20\) lies between the third quartile, \(18\), and the maximum, \(27\). A box plot does not show how observations within that quartile are distributed, so the exact count above \(20\) cannot be determined.

Answer

a) Approximately \(40\) students b) Approximately \(40\) students c) No. The box plot places \(20\) in the upper quarter, but it does not show the individual values within that quarter.
5440049
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.
5440089
Two engineering classes tracked how many times each student used the makerspace in one month. The grouped counts are shown. <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> Class A has \(24\) students, and Class B has \(40\) students. a) Convert each row to a relative frequency for its class. b) Compare the two distributions. Why would comparing only the bar counts be misleading?

Hints

- Divide each interval count by the total number of students in that class. - Compare percentages in corresponding intervals. - Ask how unequal group sizes affect raw bar heights.

Solution

1. For Class A, the relative frequencies are \(\frac{6}{24}=25\%\), \(\frac{12}{24}=50\%\), and \(\frac{6}{24}=25\%\). 2. For Class B, the relative frequencies are \(\frac{8}{40}=20\%\), \(\frac{20}{40}=50\%\), and \(\frac{12}{40}=30\%\). 3. Both classes have \(50\%\) of students in the middle interval. Class B has a larger percentage in the highest interval, while Class A has a larger percentage in the lowest interval. 4. Raw counts are affected by the unequal class sizes, so relative frequencies provide the fair comparison.

Answer

a) Class A: \(25\%\), \(50\%\), \(25\%\). Class B: \(20\%\), \(50\%\), \(30\%\). b) The middle percentages are equal, but Class B has relatively more students in the highest interval. Counts alone are misleading because Class B has more students.
5440099
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.
5440109
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.
5440129
A student ordered eight package masses as \(2, 4, 5, 7, 8, 10, 12, 20\) kilograms. The student 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 \(\text{IQR}=6.5\,\text{kg}\).
5440139
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.
5440149
The dot plot shows the number of new podcast episodes released each week by a student radio station 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 those frequencies to find the total and the middle position. - Compare stacks at values equally far from \(2\).

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 \(2\).
5440159
An environmental scientist wants to report both the overall shape of \(240\) river-depth measurements and a compact comparison with measurements from a second river. Explain why using both a histogram and a box plot can be more informative than using either display alone. State one feature each display can reveal better than the other.

Hints

- List what visual information is retained by each display. - Think about detailed shape versus compact comparison. - Identify a feature that depends on frequencies within intervals and another that depends on quartiles.

Solution

1. A histogram shows how observations are distributed across intervals, so it can reveal features such as clusters, gaps, skewness, or multiple peaks. 2. A box plot gives a compact display of the median, quartiles, overall spread, and possible outliers, making side-by-side comparisons efficient. 3. Using both displays combines detailed shape information with concise numerical-position information.

Answer

A histogram is better for seeing detailed shape, including clusters or multiple peaks. A box plot is better for compactly comparing medians, interquartile ranges, and possible outliers. Together they provide information that either display alone may hide.
5440189
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.
5440259
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 \(50\).
5440309
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.
5440329
A histogram of \(40\) package-delivery distances has five equal-width intervals. Four bar frequencies are \(7, 11, 9,\) and \(8\). The frequency label on the fifth bar is missing. a) Find the missing frequency. b) Find the relative frequency of that interval as a percentage. c) Explain how the total number of observations provides a check on all histogram frequencies.

Hints

- Add the frequencies that are visible. - Compare that subtotal with the stated number of observations. - Convert the missing count to a fraction of the full data set.

Solution

1. The four visible frequencies total \(7+11+9+8=35\). 2. The missing frequency is \(40-35=5\). 3. Its relative frequency is \(\frac{5}{40}=0.125=12.5\%\). 4. Every observation belongs to exactly one interval, so all bar frequencies must add to the data-set size \(40\).

Answer

a) \(5\) b) \(12.5\%\) c) The sum of all interval frequencies must equal \(40\).
5440359
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.
5440389
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.
5440399
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.
5440439
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.
5440879
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.
5441169
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.
5441839
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.
5441849
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.
5513859
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 \(Q_3\). The student's box plot is intended to use its whiskers for the actual minimum and maximum, with no separate outliers. A student 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: \((2, 6, 10, 18, 22)\).
5549529
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.
5549539
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 \([7,15]\).
5550609
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 \(7\).
5139259
The function \(T(x)=1.5x^2-4.5\) is given. a) Find \(T(-2)\), \(T(-1.5)\), \(T(-0.5)\), \(T(0.5)\), \(T(1.5)\), and \(T(2)\). b) Identify the pairs of opposite inputs that have equal outputs. Explain why this happens from the function rule.

Hints

- Evaluate the function at each listed input before comparing outputs. - Compare inputs that have the same absolute value. - What happens to the sign when a number is squared?

Solution

1. Evaluate the function: \(T(-2)=1.5\), \(T(-1.5)=-1.125\), \(T(-0.5)=-4.125\), \(T(0.5)=-4.125\), \(T(1.5)=-1.125\), and \(T(2)=1.5\). 2. The equal-output pairs are \(-2\) and \(2\), \(-1.5\) and \(1.5\), and \(-0.5\) and \(0.5\). 3. Opposite inputs have the same square, so \(T(-x)=1.5(-x)^2-4.5=1.5x^2-4.5=T(x)\).

Answer

a) \(T(-2)=1.5\), \(T(-1.5)=-1.125\), \(T(-0.5)=-4.125\), \(T(0.5)=-4.125\), \(T(1.5)=-1.125\), and \(T(2)=1.5\) b) The pairs are \((-2,2)\), \((-1.5,1.5)\), and \((-0.5,0.5)\). Opposite inputs give equal outputs because \((-x)^2=x^2\).
5351319
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

- Read each bar height as a percent. - Multiply the relative frequency by the total number of students to find a count. - For the estimated mean, multiply each interval midpoint by its relative frequency written as a decimal, then add the products. - Find the number halfway between the endpoints of each interval.

Solution

1. The relative frequencies are \(20\%\), \(35\%\), \(25\%\), \(15\%\), and \(5\%\) for the intervals from \(0\) to \(10\), \(10\) to \(20\), \(20\) to \(30\), \(30\) to \(40\), and \(40\) to \(50\) minutes, respectively. 2. The number in the \(10\)- to \(20\)-minute interval is \(0.35 \cdot 200 = 70\) students. 3. The interval midpoints are \(5, 15, 25, 35,\) and \(45\) minutes. The estimated mean 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) \(0\)–\(10\) minutes: \(20\%\); \(10\)–\(20\) minutes: \(35\%\); \(20\)–\(30\) minutes: \(25\%\); \(30\)–\(40\) minutes: \(15\%\); \(40\)–\(50\) minutes: \(5\%\) b) \(70\) students c) Approximately \(20\) minutes
5439959
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.
5439989
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.
5440009
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.
5440029
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 \(30\) visitors. Explain why simply averaging \(7.5\) and \(10.5\) would not be correct.

Hints

- Recover the total viewing time represented by each group mean. - Combine the totals before dividing by the combined number of visitors. - Think about whether the two means should contribute equally when the group sizes differ.

Solution

1. The first group contributed \(18\cdot 7.5=135\) total visitor-minutes. 2. The second group contributed \(12\cdot 10.5=126\) total visitor-minutes. 3. The combined total is \(135+126=261\) visitor-minutes for \(30\) visitors. 4. The combined mean is \(261\div 30=8.7\) minutes. 5. Averaging the two group means directly would give the groups equal weight even though their sizes differ.

Answer

The combined mean viewing time is \(8.7\) minutes. The two group means cannot be averaged directly because the groups contain different numbers of visitors.
5440059
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.
5440069
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: \(10.8^\circ\text{F}\)
5440119
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: \(8\)
5440179
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.
5440199
The bar chart shows counts for three equal-width commute-time intervals. The bars from left to right represent intervals \(1\), \(2\), and \(3\). The truncated axis makes interval \(3\) appear to have about three times as many observations as interval \(1\). Compare the actual counts and explain how the axis choice distorts the display.
Figure for problem 544019

Hints

- Read the numerical bar heights from the vertical scale. - Compare the full values rather than only the visible lengths above the baseline. - Identify what portion of each bar is omitted when the axis begins at \(40\).

Solution

1. The bar heights are \(42\), \(44\), and \(46\). The third interval has only \(46-42=4\) more observations than the first. 2. The ratio is \(\frac{46}{42}\approx 1.10\), not \(3\). 3. The vertical axis begins at \(40\), making the visible heights above the baseline \(2\) and \(6\). Their visible ratio is \(3\), which exaggerates the actual difference by omitting the common height from \(0\) to \(40\).

Answer

The claim is false. Interval \(3\) has \(46\) observations compared with \(42\), only about \(1.10\) times as many. The truncated vertical axis exaggerates the difference because it shows only the portions above \(40\).
5440219
Five prototype scores are \(62, 68, 71, 74, 80\). The score \(62\) is replaced by a corrected value so that the new mean is \(75\). a) Find the corrected value. b) Find the new median. c) Explain why the correction raises the mean but leaves \(74\) in the middle position.

Hints

- Convert the target mean into a required total. - Keep the four unchanged values together when finding the missing amount. - Reorder the completed data before locating the median.

Solution

1. A mean of \(75\) for \(5\) scores requires a total of \(5\cdot 75=375\). 2. The four unchanged scores total \(68+71+74+80=293\), so the corrected score is \(375-293=82\). 3. The ordered corrected data are \(68, 71, 74, 80, 82\), so the median is \(74\). 4. Replacing a low value with \(82\) increases the total used in the mean, but the corrected value moves to the upper end and does not change the third ordered value.

Answer

a) The corrected score is \(82\). b) The new median is \(74\). c) The total increases, so the mean rises, while \(74\) remains the third value in order.
5440229
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.
5440239
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 \(\text{IQR}=2\).
5440249
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.
5440269
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.
5440279
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.
5440319
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 \(b=6\).
5440339
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.
5440349
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 \(25.2\,\text{mg/L}\).
5440369
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 \(3.16\).
5440379
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: \(2,2,6,10,10\).
5440409
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.
5440429
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

\(x=12\)
5440449
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 \(10\).
5440829
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.
5549549
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.
5440209
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.
5440419
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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