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Boxplots and the five-number summary

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53973912
Read the five-number summary from the boxplot.
Figure for problem 539739

Hints

- Identify the whisker endpoints, box edges, and median line separately. - Record those five features in their left-to-right order.

Solution

1. The left whisker is the minimum \(4\). 2. The left box edge is \(Q_1=7\), the center line is the median \(11\), and the right box edge is \(Q_3=15\). 3. The right whisker is the maximum \(19\).

Answer

\((4,7,11,15,19)\).
53974112
Use the boxplot to find the range and IQR.
Figure for problem 539741

Hints

- Use the whisker endpoints for one spread and the box endpoints for the other. - Range uses maximum minus minimum; IQR uses \(Q_3-Q_1\).

Solution

1. The range is \(33-10=23\). 2. The IQR is \(25-16=9\).

Answer

Range: \(23\). IQR: \(9\).
53974512
Approximately what percentage of observations lie between \(Q_1=20\) and \(Q_3=36\)? Approximately what percentage lie from the median to the maximum?
Figure for problem 539745

Hints

- Use the four quartile sections represented by the boxplot. - Each quartile interval represents about one-fourth of the ordered observations regardless of its width.

Solution

1. The box from \(Q_1\) to \(Q_3\) contains the middle \(50\%\). 2. The interval from the median to the maximum contains the upper half, approximately \(50\%\).

Answer

About \(50\%\) lie between \(20\) and \(36\), and about \(50\%\) lie from the median to the maximum.
53974612
A five-number summary has \(Q_1=18\) and IQR \(=11\). Find \(Q_3\).

Hints

- Use the relationship between the two box edges and the box width. - Rearrange \(\text{IQR}=Q_3-Q_1\) to isolate the unknown quartile.

Solution

1. The IQR is \(Q_3-Q_1\). 2. \(Q_3=18+11=29\).

Answer

\(Q_3=29\).
53975212
A five-number summary has maximum \(84\) and range \(39\). Find the minimum.

Hints

- Work backward from the definition of range. - Rearrange \(\text{range}=\text{maximum}-\text{minimum}\) to solve for the minimum.

Solution

1. The range equals maximum minus minimum. 2. The minimum is \(84-39=45\).

Answer

The minimum is \(45\).
53975312
Which tuple could be a five-number summary? a) \((5,12,10,18,22)\) b) \((5,10,12,18,22)\) c) \((5,10,18,12,22)\) Explain.

Hints

- Recall the required order from minimum through maximum. - Reject any tuple in which a quartile or the median falls outside its neighboring landmarks.

Solution

1. A five-number summary must be nondecreasing from minimum through maximum. 2. Only \((5,10,12,18,22)\) satisfies the required order.

Answer

b) \((5,10,12,18,22)\).
53975712
What value is the \(75\)th percentile in the boxplot, and what does it mean?
Figure for problem 539757

Hints

- Match the percentile to its quartile landmark. - The \(75\)th percentile is the right edge of the box; interpret it as an approximate cumulative proportion.

Solution

1. The \(75\)th percentile is the third quartile. 2. The right edge of the box gives \(Q_3=31\). 3. Approximately \(75\%\) of observations are at or below \(31\).

Answer

The \(75\)th percentile is \(31\); approximately \(75\%\) of observations are at or below it.
54878712
A boxplot is shown without a sample size. Can the number of observations be determined from the boxplot alone? Explain why two samples of different sizes can produce the same boxplot.

Hints

- List the quantities represented directly by a boxplot. - Ask whether each observation receives its own visible mark. - Consider what information is discarded when data are reduced to five summary values.

Solution

1. A boxplot displays selected summary values and possibly outliers, not one mark for every observation. 2. Samples with different numbers of observations can share the same minimum, quartiles, median, maximum, and outlier pattern. 3. Therefore the sample size cannot be recovered unless it is stated separately.

Answer

No. A boxplot does not encode sample size, and samples of different sizes can have the same five-number summary.
54880712
A horizontal boxplot is redrawn vertically using the same numerical scale. Which statistical information changes, and which remains unchanged?

Hints

- Separate visual layout from numerical encoding. - Track the marked summary values rather than their page direction. - Rotation preserves distances along the value axis.

Solution

1. Changing orientation changes only the layout of the display. 2. The minimum, quartiles, median, maximum, outliers, and all numerical segment lengths remain unchanged. 3. No conclusion about center, spread, or skewness should change solely because the plot is rotated.

Answer

No statistical information changes. Only the orientation of the display changes.
53974012
Using the median-of-halves method, find the five-number summary for \(2,4,5,7,8,10,12,14\).

Hints

- Order the values and find the median of each half. - For an even sample, average the two middle values for the overall median and for each half.

Solution

1. The minimum and maximum are \(2\) and \(14\). 2. The median is \(\frac{7+8}{2}=7.5\). 3. The lower-half median is \(Q_1=4.5\), and the upper-half median is \(Q_3=11\).

Answer

The five-number summary is \((2,4.5,7.5,11,14)\).
53974212
The boxplot includes outlier symbols. State the whisker endpoints and the observed outliers. Are the whisker endpoints necessarily the data minimum and maximum?
Figure for problem 539742

Hints

- Distinguish symbols outside the whiskers from the whisker endpoints. - When outliers are plotted separately, whiskers stop at the most extreme nonoutlying observations.

Solution

1. The whiskers end at \(21\) and \(38\). 2. The plotted outliers are \(12\) and \(49\). 3. With outliers shown separately, whisker endpoints are the most extreme nonoutliers, not necessarily the data extrema.

Answer

Whiskers: \(21\) and \(38\). Outliers: \(12\) and \(49\). The whiskers are nonoutlier endpoints, not the full-data minimum and maximum.
53974312
What shape does the boxplot suggest, and what relationship between mean and median is likely?
Figure for problem 539743

Hints

- Compare the two whisker lengths and the median’s position inside the box. - A longer right side suggests a high-value tail; the mean tends to be pulled toward that tail.

Solution

1. The right whisker is much longer than the left whisker, suggesting a long right tail. 2. The distribution is likely skewed right. 3. For a right-skewed distribution, the mean is usually greater than the median.

Answer

The distribution is likely right-skewed, so the mean is likely greater than the median.
53974412
What shape does the boxplot suggest, and which center is likely smaller?
Figure for problem 539744

Hints

- Look for the longer tail and connect its direction to the mean. - A longer left side suggests a low-value tail; compare how that tail affects the mean.

Solution

1. The left whisker is much longer, suggesting a long left tail. 2. The distribution is likely skewed left. 3. The mean is usually smaller than the median in a left-skewed distribution.

Answer

The distribution is likely left-skewed, and the mean is likely smaller than the median.
53974812
A student says, “Because the median line is at \(14\), exactly one observation equals \(14\).” Explain why the boxplot does not support that conclusion.
Figure for problem 539748

Hints

- Separate a positional summary from an exact-value frequency. - The median describes position, not the number of observations equal to its numerical value.

Solution

1. The median gives a central ordered position or average of two central positions. 2. A boxplot does not show the frequency of any exact value. 3. There may be zero, one, or multiple observations equal to \(14\).

Answer

The boxplot shows the median value but not how many observations equal it.
53974912
The interval from the median to \(Q_3\) is much wider than the interval from \(Q_1\) to the median. Do the two intervals contain different percentages of the data? Explain.
Figure for problem 539749

Hints

- Recall the proportion represented by each adjacent quartile interval. - Use interval width to describe numerical spread, not the number of observations.

Solution

1. Each adjacent quartile interval contains approximately \(25\%\) of the observations. 2. The wider interval indicates that its observations are more spread out, not that it contains more observations.

Answer

No. Each interval contains about \(25\%\) of the data; the median-to-\(Q_3\) values are simply more spread out.
53975012
Every value represented by the boxplot is transformed by \(Y=2X+4\). Find the transformed five-number summary.
Figure for problem 539750

Hints

- Apply the increasing transformation to every landmark. - Check all five transformed values and preserve their order.

Solution

1. Apply the increasing linear transformation to each summary value. 2. The transformed values are \(2(3)+4=10\), \(2(5)+4=14\), \(2(8)+4=20\), \(2(11)+4=26\), and \(2(15)+4=34\).

Answer

The transformed five-number summary is \((10,14,20,26,34)\).
53975112
A boxplot in centimeters has five-number summary \((25.4,50.8,76.2,101.6,127.0)\). Convert the summary to inches using \(1\,\text{in}=2.54\,\text{cm}\).

Hints

- Use the same unit conversion on every landmark. - Check that the converted summary remains increasing and is labeled in inches.

Solution

1. Divide each value by \(2.54\). 2. The converted values are \(10,20,30,40,50\) inches.

Answer

\((10,20,30,40,50)\) inches.
53975412
Does the boxplot suggest approximate symmetry? Cite at least two visual relationships.
Figure for problem 539754

Hints

- Compare corresponding distances on the left and right sides. - Check whether the median bisects the box and whether the two whiskers have similar lengths.

Solution

1. The median is centered in the box because \(10-6=14-10=4\). 2. The whiskers have equal lengths because \(6-2=18-14=4\). 3. These relationships suggest approximate symmetry.

Answer

Yes. The median is centered in the box and the whiskers have equal lengths.
53975612
The boxplot marks the mean with an \(\times\). Compare the mean and median and state the suggested skew.
Figure for problem 539756

Hints

- Compare the mean marker with the median line. - Use the longer whisker or side of the box as supporting evidence for the skew direction.

Solution

1. The mean marker is at \(12\), while the median is \(9\). 2. The mean is greater than the median. 3. Together with the longer right side, this suggests right skew.

Answer

Mean \(=12\), median \(=9\); the distribution is likely right-skewed.
53975812
Can two different raw data sets have the same boxplot? Explain.

Hints

- Consider how much of the original data the representation retains. - A boxplot omits the exact values and frequencies within each quartile interval.

Solution

1. A boxplot records only the five-number summary and any displayed outliers. 2. Many different arrangements of values within the quartile intervals can share those same landmarks. 3. Therefore, distinct raw data sets can produce the same boxplot.

Answer

Yes. A boxplot does not preserve all individual values, so different data sets can share the same five-number summary and boxplot.
54878312
Let the ordered observations be \(x_1\leq x_2\leq\cdots\leq x_{12}\). Using the median-of-halves convention, identify the ordered positions used to calculate the minimum, \(Q_1\), median, \(Q_3\), and maximum.

Hints

- Locate the two central positions of the full ordered list. - Treat the first six and last six observations as separate ordered halves. - Each even-sized list has a center halfway between two positions.

Solution

1. The minimum and maximum are \(x_1\) and \(x_{12}\). 2. The lower half is \(x_1\) through \(x_6\), so \(Q_1=\frac{x_3+x_4}{2}\). 3. The overall median is \(\frac{x_6+x_7}{2}\). 4. The upper half is \(x_7\) through \(x_{12}\), so \(Q_3=\frac{x_9+x_{10}}{2}\).

Answer

Minimum: \(x_1\) \(Q_1=\frac{x_3+x_4}{2}\) Median: \(\frac{x_6+x_7}{2}\) \(Q_3=\frac{x_9+x_{10}}{2}\) Maximum: \(x_{12}\)
54880012
Use the median-of-halves convention for the data \(1,2,3,4,5,6,7,8\). Find the five-number summary. Which summary values are not observations in the data set, and why can a valid boxplot contain those values?

Hints

- Even-sized ordered lists have centers between two observations. - Apply the same idea to each four-observation half. - A summary statistic need not equal a recorded observation.

Solution

1. The minimum and maximum are \(1\) and \(8\). 2. The median is \(\frac{4+5}{2}=4.5\). 3. The lower-half median is \(Q_1=\frac{2+3}{2}=2.5\), and the upper-half median is \(Q_3=\frac{6+7}{2}=6.5\). 4. The values \(2.5\), \(4.5\), and \(6.5\) are not observations. They are averages of adjacent ordered observations used to locate quartile positions.

Answer

The five-number summary is \(1,2.5,4.5,6.5,8\). The quartiles and median are not observed data values; they are averages of paired middle observations.
54880112
A data set has five-number summary \(2,2,5,9,9\). Explain why repeated entries can form a valid five-number summary. Describe what the zero-length lower and upper whiskers mean in the corresponding standard boxplot.
Figure for problem 548801

Hints

- Check whether the required ordering permits equality. - Find each whisker length by subtracting its endpoints. - Interpret coincident summary values as overlapping parts of the display.

Solution

1. A five-number summary must be nondecreasing, not strictly increasing, and \(2\leq2\leq5\leq9\leq9\). 2. The lower whisker has length \(Q_1-\min=2-2=0\). 3. The upper whisker has length \(\max-Q_3=9-9=0\). 4. The minimum coincides with \(Q_1\), and the maximum coincides with \(Q_3\). Repeated values or concentrated tails can produce these equalities.

Answer

The summary is valid because equality is allowed. Both whiskers have length \(0\): the minimum equals \(Q_1\), and the maximum equals \(Q_3\).
54880412
A boxplot has a long box and short whiskers. A student says the raw data must have no gaps or clusters inside the box. Evaluate the statement.

Hints

- List the exact locations marked by a boxplot. - Ask what happens to individual positions between those marked values. - Different raw patterns can share the same quartiles.

Solution

1. The box shows only \(Q_1\), the median, and \(Q_3\), not the locations of individual observations within those intervals. 2. The same quartile values could arise from data spread smoothly through the box or from observations concentrated in clusters with gaps between them. 3. Therefore the boxplot cannot establish the absence of gaps or clusters inside the box.

Answer

The statement is not justified. A boxplot does not reveal within-quartile gaps or clusters.
54880512
A boxplot has minimum \(5\), \(Q_1=18\), median \(22\), \(Q_3=30\), and maximum \(55\). Find the range, interquartile range, and the interquartile range as a percentage of the full range.
Figure for problem 548805

Hints

- Use the outer endpoints for one spread measure and the box endpoints for the other. - Compare the two widths with a ratio. - Convert the resulting decimal to a percentage.

Solution

1. The range is \(55-5=50\). 2. The interquartile range is \(30-18=12\). 3. The ratio is \(\frac{12}{50}=0.24\), so the interquartile range is \(24\%\) of the full range.

Answer

Range: \(50\); interquartile range: \(12\); the interquartile range is \(24\%\) of the full range.
54880912
A nonnegative data set has five-number summary \(1,4,9,16,25\). Every observation is replaced by its square root. Find the transformed five-number summary.

Hints

- Check that the transformation is monotonic over the data's domain. - Apply it to each summary value. - Keep the results in increasing order.

Solution

1. The square-root function is increasing on nonnegative values, so order is preserved. 2. Taking square roots gives \(1,2,3,4,5\).

Answer

The transformed five-number summary is \(1,2,3,4,5\).
54881012
A boxplot is drawn on an axis whose equally spaced tick labels are \(0,10,20,100\). The designer intends the axis to be linear. Explain why interval lengths in this boxplot are misleading. State one valid way to repair the display.

Hints

- Compare the numerical change between each pair of adjacent labels. - A boxplot encodes spread through distance along its axis. - A repair must make the visual spacing consistent with the stated scale.

Solution

1. On a linear axis, equal physical distances must represent equal numerical differences. 2. The first two tick gaps each represent \(10\) units, but the final equal-sized gap represents \(80\) units. 3. Therefore box and whisker lengths cannot be interpreted consistently as numerical spread. 4. The display can be repaired by using proportional tick spacing on a continuous linear axis or by clearly using and labeling a different valid scale.

Answer

The axis is not linear because equal tick gaps represent differences of \(10,10,\) and \(80\). Redraw it with proportional linear spacing, or use a clearly labeled valid alternative scale.
53975512
A data set has \(Q_1=20\) and \(Q_3=32\). Under the \(1.5\cdot\text{IQR}\) rule, are values exactly equal to \(2\) or \(50\) outliers?

Hints

- Compute the IQR from the two quartiles. - Use the IQR to form the lower and upper fences. - Check whether equality with a fence satisfies the rule’s strict outlier condition.

Solution

1. The IQR is \(32-20=12\). 2. The fences are \(20-1.5(12)=2\) and \(32+1.5(12)=50\). 3. Outliers are values more extreme than the fences, so values exactly on the fences are not outliers.

Answer

No. Values \(2\) and \(50\) lie exactly on the fences and are not outliers by the stated rule.
54878212
The boxplot has five-number summary \(10,18,24,40,52\). Find the lengths of the four quartile intervals. Which interval is shortest, and what does that suggest about concentration? Explain why ties at quartile boundaries can prevent each closed interval from containing exactly \(25\%\) of the observations.
Figure for problem 548782

Hints

- Subtract consecutive values in the five-number summary. - Compare the numerical widths of the four quartile segments. - Distinguish quartile positions in ordered data from the number of tied observations included in a closed interval.

Solution

1. The quartile-interval lengths are \(18-10=8\), \(24-18=6\), \(40-24=16\), and \(52-40=12\). 2. The interval \([18,24]\) is shortest, so the observations represented between \(Q_1\) and the median are packed into the narrowest numerical span. 3. Quartiles divide ordered positions into quarters, but tied values can occur on both sides of a quartile position. Therefore, the number of observations whose values lie in a closed quartile interval need not be exactly \(25\%\).

Answer

The interval lengths are \(8,6,16,12\). The shortest interval is \([18,24]\), suggesting the greatest concentration between \(Q_1\) and the median. Quartiles divide ordered positions, but boundary ties can make a closed interval contain more or less than exactly \(25\%\) of the observations.
54878512
A data set has five-number summary \(10,20,30,50,70\). Every value is transformed using \(y=100-x\). Find the transformed five-number summary.
Figure for problem 548785

Hints

- Determine whether larger original values become larger or smaller transformed values. - The transformed minimum comes from the original maximum under a decreasing rule. - List the final summary from least to greatest.

Solution

1. The transformation is decreasing, so it reverses the order of the data. 2. Transform the original maximum, third quartile, median, first quartile, and minimum in that order: \(100-70=30\), \(100-50=50\), \(100-30=70\), \(100-20=80\), and \(100-10=90\).

Answer

The transformed five-number summary is \(30,50,70,80,90\).
54878612
A modified boxplot has \(Q_1=12\), median \(16\), and \(Q_3=20\). Its whiskers end at \(2\) and \(30\), and it marks \(-3\) and \(35\) as outliers. Calculate the outlier fences and verify whether the marked values are correctly classified.
Figure for problem 548786

Hints

- Find the middle-half spread first. - Extend the appropriate multiple of that spread below and above the quartiles. - Compare outliers and whisker endpoints with the resulting cutoffs.

Solution

1. The interquartile range is \(20-12=8\). 2. The lower fence is \(12-1.5\cdot8=0\), and the upper fence is \(20+1.5\cdot8=32\). 3. The values \(-3<0\) and \(35>32\), so both are outliers. 4. The whisker endpoints \(2\) and \(30\) lie within the fences, so they are plausible most-extreme nonoutliers.

Answer

The fences are \(0\) and \(32\). Both \(-3\) and \(35\) are correctly marked as outliers.
54878812
Boxplot A has \(Q_1=20\) and \(Q_3=40\) and is drawn with a scale of \(1\,\text{cm}\) for every \(5\) units. Boxplot B has \(Q_1=12\) and \(Q_3=24\) and is drawn with a scale of \(1\,\text{cm}\) for every \(2\) units. Which distribution has the larger interquartile range? Which box appears wider on the page? Explain why visual width alone gives the wrong comparison here.

Hints

- Compute each interquartile range in data units first. - Convert each numerical interval to its displayed length using that plot's scale. - Separate the size of a statistic from the physical size of its drawing.

Solution

1. Boxplot A has \(\text{IQR}=40-20=20\) units. 2. Boxplot B has \(\text{IQR}=24-12=12\) units. 3. A's box is drawn \(\frac{20}{5}=4\,\text{cm}\) wide, while B's box is drawn \(\frac{12}{2}=6\,\text{cm}\) wide. 4. A has the larger numerical IQR, but B appears wider because the plots use different scales.

Answer

Boxplot A has the larger IQR: \(20\) units versus \(12\) units. Boxplot B appears wider: \(6\,\text{cm}\) versus \(4\,\text{cm}\). Numerical spread must be compared from the axes, not raw page width.
54879012
A box has interquartile range \(24\), and its median is \(32\). The distance from \(Q_1\) to the median is one-third of the distance from the median to \(Q_3\). Find \(Q_1\) and \(Q_3\).

Hints

- Represent the two parts of the box with lengths in the stated ratio. - Their sum is the interquartile range. - Move left and right from the median after finding both lengths.

Solution

1. Let the lower half of the box have length \(a\). Then the upper half has length \(3a\). 2. The full box length is \(a+3a=24\), so \(a=6\). 3. Therefore, \(Q_1=32-6=26\) and \(Q_3=32+18=50\).

Answer

\(Q_1=26\) and \(Q_3=50\).
54879112
A positive data set has five-number summary \(1,2,4,5,10\). Every observation is transformed using \(y=\frac{1}{x}\). Find the five-number summary of the transformed data and explain why the original order of the five values must be reversed.

Hints

- Determine whether the transformation increases or decreases as the input grows. - Match the transformed minimum with the appropriate original endpoint. - List the final summary from least to greatest.

Solution

1. The reciprocal function is decreasing for positive values, so larger original observations become smaller transformed observations. 2. Transform the original maximum, third quartile, median, first quartile, and minimum in that order. 3. The transformed five-number summary is \(\frac{1}{10},\frac{1}{5},\frac{1}{4},\frac{1}{2},1\).

Answer

The transformed five-number summary is \(\left(\frac{1}{10},\frac{1}{5},\frac{1}{4},\frac{1}{2},1\right)\). The order reverses because taking reciprocals is decreasing on positive numbers.
54879212
The cumulative frequency table summarizes \(16\) observations. <table><thead><tr><th>Value</th><th>Cumulative frequency</th></tr></thead><tbody><tr><td>\(1\)</td><td>\(2\)</td></tr><tr><td>\(2\)</td><td>\(5\)</td></tr><tr><td>\(3\)</td><td>\(9\)</td></tr><tr><td>\(4\)</td><td>\(12\)</td></tr><tr><td>\(5\)</td><td>\(15\)</td></tr><tr><td>\(6\)</td><td>\(16\)</td></tr></tbody></table> Using the median-of-halves convention, find the five-number summary without expanding the full data list.

Hints

- Determine which ordered positions define each quartile for \(16\) observations. - Use cumulative counts to locate a position's value. - You do not need to write every repeated observation.

Solution

1. The minimum is \(1\), and the maximum is \(6\). 2. For \(16\) observations, \(Q_1\) is the average of positions \(4\) and \(5\). Both positions have value \(2\), so \(Q_1=2\). 3. The median is the average of positions \(8\) and \(9\). Both positions have value \(3\), so the median is \(3\). 4. \(Q_3\) is the average of positions \(12\) and \(13\). Those values are \(4\) and \(5\), so \(Q_3=4.5\).

Answer

The five-number summary is \(1,2,3,4.5,6\).
54879312
A boxplot begins at minimum \(4\). From left to right, the lower whisker has length \(3\), the lower half of the box has length \(5\), the upper half has length \(8\), and the upper whisker has length \(4\). Reconstruct the five-number summary.

Hints

- Move from left to right along the boxplot. - Add each stated segment length to the previous endpoint. - Check that the resulting summary values increase.

Solution

1. The first quartile is \(4+3=7\). 2. The median is \(7+5=12\). 3. The third quartile is \(12+8=20\). 4. The maximum is \(20+4=24\).

Answer

The five-number summary is \(4,7,12,20,24\).
54879512
A boxplot has median \(10\), and many observations are tied at \(10\). Is it necessarily true that exactly half of the observations are strictly less than \(10\)? Explain.

Hints

- Distinguish “at or below” from “strictly below.” - Consider how repeated values can occupy positions on both sides of the middle. - A boxplot does not display multiplicities at the median.

Solution

1. The median divides the ordered positions so that at least half the observations are at or below it and at least half are at or above it. 2. When many observations equal \(10\), some of the middle observations and possibly many others are tied at the median. 3. Therefore fewer than half may be strictly less than \(10\); the boxplot does not determine the exact strict-inequality count.

Answer

No. Ties at the median can make fewer than half of the observations strictly less than \(10\).
54879612
A boxplot is drawn on a base-10 logarithmic axis. Its five-number summary is \(1,10,100,1000,10{,}000\), so the four displayed segments have equal physical lengths. Are the four numerical differences equal? What common multiplicative relationship does each segment represent?

Hints

- Compare consecutive values by subtraction and by division. - Identify which comparison remains constant. - Interpret equal spacing according to the stated axis scale.

Solution

1. The numerical differences are \(10-1=9\), \(100-10=90\), \(1000-100=900\), and \(10{,}000-1000=9000\), so they are not equal. 2. Each right endpoint is \(10\) times its left endpoint. 3. Equal lengths on a base-10 logarithmic axis represent equal ratios, not equal additive differences.

Answer

The numerical differences are \(9,90,900,9000\), so they are not equal. Each segment represents multiplication by \(10\).
54879712
Use the median-of-halves convention for the ordered data \(5,5,5,5,5,5,5,5,5,5,8,9\). Find \(Q_1\), \(Q_3\), the interquartile range, and the outlier fences. Describe the resulting modified boxplot.

Hints

- Repeated values can make two quartiles equal. - Follow the fence rule even when the interquartile range is zero. - Distinguish the collapsed nonoutlier display from separately plotted values.

Solution

1. The lower half is six \(5\)s, so \(Q_1=5\). 2. The upper half is \(5,5,5,5,8,9\), whose middle two values are both \(5\), so \(Q_3=5\). 3. The interquartile range is \(0\), and both fences equal \(5\). 4. Every value different from \(5\) lies beyond a fence, so \(8\) and \(9\) are outliers. 5. The box and both whiskers collapse at \(5\), with separate outlier points at \(8\) and \(9\).

Answer

\(Q_1=5\), \(Q_3=5\), \(\text{IQR}=0\), and both fences are \(5\). The modified boxplot collapses at \(5\) and marks \(8\) and \(9\) as outliers.
54879812
After a new value \(-3\) is added and the quartiles are recomputed, a data set still has \(Q_1=12\) and \(Q_3=20\). Its smallest other observation is \(8\). Under the \(1.5\cdot\text{IQR}\) rule, how should the lower end of a modified boxplot display \(-3\) and \(8\)?

Hints

- Calculate the lower cutoff from the quartiles. - Compare each low value with that cutoff. - A modified whisker reaches the most extreme value that is not an outlier.

Solution

1. The interquartile range is \(20-12=8\). 2. The lower fence is \(12-1.5\cdot8=0\). 3. Since \(-3<0\), it is plotted as a separate outlier. 4. Since \(8\) is the smallest observation within the fence, the lower whisker ends at \(8\).

Answer

Plot \(-3\) as an outlier and end the lower whisker at \(8\).
54880212
A data set has \(Q_1=5\) and \(Q_3=9\). Every observation is multiplied by \(3\). Find the original and transformed outlier fences. Explain whether an observation's outlier status is preserved.

Hints

- Compute the original fences before transforming anything. - Track how quartiles and the interquartile range scale. - Compare a transformed observation with correspondingly transformed cutoffs.

Solution

1. The original interquartile range is \(9-5=4\), so the fences are \(5-1.5\cdot4=-1\) and \(9+1.5\cdot4=15\). 2. The transformed quartiles are \(15\) and \(27\), with interquartile range \(12\). 3. The transformed fences are \(15-1.5\cdot12=-3\) and \(27+1.5\cdot12=45\). 4. The fences and every observation are multiplied by the same positive factor, so outlier status is preserved.

Answer

Original fences: \(-1\) and \(15\). Transformed fences: \(-3\) and \(45\). Outlier status is preserved.
54880312
A standard boxplot has minimum \(6\), maximum \(20\), and interquartile range \(14\). Determine \(Q_1\), \(Q_3\), and both whisker lengths. Explain why these values are forced.

Hints

- Compare the middle-half width with the full data width. - Quartiles must lie between the minimum and maximum. - Ask when a contained interval can have the same length as the interval containing it.

Solution

1. The full range is \(20-6=14\), equal to the interquartile range. 2. Because \(Q_1\geq6\) and \(Q_3\leq20\), the difference \(Q_3-Q_1\) cannot equal the full range unless \(Q_1=6\) and \(Q_3=20\). 3. The lower whisker length is \(6-6=0\), and the upper whisker length is \(20-20=0\).

Answer

\(Q_1=6\), \(Q_3=20\), and both whisker lengths are \(0\). Equality of IQR and range forces the box to span the entire data range.
54880812
A data set has minimum \(8\), \(Q_1=20\), median \(25\), \(Q_3=30\), and maximum \(42\). Determine whether a standard boxplot and a modified boxplot would have the same whisker endpoints under the \(1.5\cdot\text{IQR}\) rule.
Figure for problem 548808

Hints

- Compute the outlier fences from the box endpoints. - Compare both extremes with those fences. - Modified and standard whiskers differ only when outliers are present.

Solution

1. The interquartile range is \(30-20=10\). 2. The fences are \(20-1.5\cdot10=5\) and \(30+1.5\cdot10=45\). 3. Both the minimum \(8\) and maximum \(42\) lie within the fences, so there are no outliers. 4. Both types of boxplot use whisker endpoints \(8\) and \(42\).

Answer

Yes. Both boxplots have whiskers at \(8\) and \(42\) because there are no outliers.
53974712
Use the median-of-halves method and the \(1.5\cdot\text{IQR}\) rule to determine the modified boxplot values for \(4,5,7,8,9,10,11,12,13,30\): \(Q_1\), median, \(Q_3\), whisker endpoints, and outliers.

Hints

- Find the quartiles and fences before deciding where the whiskers end. - After ordering the data, apply the specified median-of-halves convention consistently. - Use the fences to identify outliers; place whiskers at the most extreme remaining data values.

Solution

1. \(Q_1=7\), median \(=9.5\), and \(Q_3=12\). 2. The IQR is \(5\), giving fences \(-0.5\) and \(19.5\). 3. The nonoutlier endpoints are \(4\) and \(13\). 4. The value \(30\) is an outlier.

Answer

\(Q_1=7\), median \(=9.5\), \(Q_3=12\); whiskers at \(4\) and \(13\); outlier \(30\).
54878412
Using the median-of-halves convention, analyze the ordered data \(1,2,3,4,5,6,7,8,30\). a) Find the quartiles, outlier fences, and outliers. b) For a separate sensitivity analysis, temporarily remove the outlier and recompute the quartiles and fences. Explain why the primary analysis should still report the verified original value and why the sensitivity result differs.

Hints

- Complete the primary five-number analysis before changing the data set. - Treat the second calculation as a comparison, not as automatic deletion of a valid observation. - Recompute ordered positions from the beginning after the temporary removal.

Solution

1. For the nine values, the median is \(5\), \(Q_1=\frac{2+3}{2}=2.5\), and \(Q_3=\frac{7+8}{2}=7.5\). 2. The interquartile range is \(5\), so the fences are \(2.5-1.5\cdot5=-5\) and \(7.5+1.5\cdot5=15\). Thus \(30\) is an outlier. 3. After temporarily removing \(30\), the eight values have median \(4.5\), \(Q_1=2.5\), and \(Q_3=6.5\). 4. The new interquartile range is \(4\), so the new fences are \(2.5-1.5\cdot4=-3.5\) and \(6.5+1.5\cdot4=12.5\). 5. The verified value \(30\) should remain in the primary analysis. The sensitivity analysis differs because removing an observation changes ordered positions, quartiles, the interquartile range, and therefore the fences.

Answer

a) \(Q_1=2.5\), median \(=5\), \(Q_3=7.5\); fences: \(-5\) and \(15\); outlier: \(30\). b) In the sensitivity analysis, \(Q_1=2.5\), median \(=4.5\), \(Q_3=6.5\); fences: \(-3.5\) and \(12.5\). The verified value \(30\) remains part of the primary data set.
54878912
Use the median-of-halves convention for these two ordered data sets: A: \(0,2,2,4,6,6,8\) B: \(0,2,3.9,4,5.9,6,8\) Show that they have the same five-number summary but different means. Explain what this demonstrates about boxplots.

Hints

- Compute the five summary positions separately from the mean. - Internal values can move without changing a quartile boundary. - Compare what information a boxplot keeps with what it discards.

Solution

1. For both sets, the minimum is \(0\), the median is \(4\), and the maximum is \(8\). 2. In each set, the median of the lower three values is \(2\), and the median of the upper three values is \(6\). Thus both summaries are \(0,2,4,6,8\). 3. The mean of A is \(\frac{28}{7}=4\). 4. The mean of B is \(\frac{29.8}{7}\approx4.257\). 5. Identical boxplots can hide different within-quartile arrangements and different means.

Answer

Both five-number summaries are \(0,2,4,6,8\). The means are \(4\) for A and approximately \(4.257\) for B, so the same boxplot does not determine the raw distribution or mean.
54879412
A modified boxplot has \(Q_1=10\), \(Q_3=18\), an upper whisker ending at \(24\), and high outliers plotted at \(31\) and \(40\). a) Find the upper outlier fence. b) Could the data set contain an observation equal to \(27\)? Explain. c) State an interval above \(24\) that is guaranteed to contain no observations.

Hints

- Use the quartiles to locate the high-outlier cutoff. - Interpret the whisker as the most extreme observation still inside the fences. - Separate the nonoutlier interval from the region where outlier symbols may occur.

Solution

1. The interquartile range is \(18-10=8\). 2. The upper fence is \(18+1.5\cdot8=30\). 3. The upper whisker ends at the largest nonoutlier, \(24\). Since \(27\leq30\), a value of \(27\) would be a nonoutlier and would force the whisker to extend beyond \(24\). Therefore \(27\) cannot be present. 4. No observations can lie in \((24,30]\). Values greater than \(30\) may appear as plotted outliers.

Answer

a) The upper fence is \(30\). b) No. A value of \(27\) would be a nonoutlier larger than the displayed whisker endpoint. c) The interval \((24,30]\) contains no observations.
54879912
Group A has data \(0,0,0,0,100\), and Group B has data \(100,100,100,100,100\). Use the median-of-halves convention. a) Find each group's five-number summary. b) Pool all ten observations and find the pooled five-number summary. c) Show that averaging corresponding entries of the two group summaries does not produce the pooled summary.

Hints

- Compute each summary from its own ordered observations first. - Pooling requires reordering all observations together. - Compare the actual pooled positions with a componentwise numerical average.

Solution

1. Group A has summary \(0,0,0,50,100\). Group B has summary \(100,100,100,100,100\). 2. The pooled ordered data contain four \(0\)s followed by six \(100\)s. 3. The pooled median is \(100\), the lower-half median is \(0\), and the upper-half median is \(100\). Thus the pooled summary is \(0,0,100,100,100\). 4. Averaging corresponding entries gives \(50,50,50,75,100\), which differs from the pooled summary.

Answer

a) A: \(0,0,0,50,100\); B: \(100,100,100,100,100\) b) Pooled: \(0,0,100,100,100\) c) Componentwise averaging gives \(50,50,50,75,100\), so five-number summaries cannot be pooled by averaging their entries.
54880612
Using the median-of-halves convention, give one ordered data set of \(9\) observations with five-number summary \(0,2,5,8,10\).

Hints

- Place the minimum, median, and maximum first. - Build four lower values whose middle pair averages to the first quartile. - Build four upper values whose middle pair averages to the third quartile.

Solution

1. Choose minimum \(0\), median \(5\), and maximum \(10\). 2. Make the median of the four lower observations equal to \(2\), for example with \(0,2,2,4\). 3. Make the median of the four upper observations equal to \(8\), for example with \(6,8,8,10\). 4. One valid ordered data set is \(0,2,2,4,5,6,8,8,10\).

Answer

One possible data set is \(0,2,2,4,5,6,8,8,10\).

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