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Graphs for one quantitative variable

Learning goals

  • Constructs or interprets appropriate graphs for quantitative data.
  • Describes distribution shape, center, spread, and unusual features.
  • Chooses mean and standard deviation or median and IQR to summarize a data set.

Click problems to add them to your worksheet.

Graphs for one quantitative variable 1
The dotplot displays quiz scores. What score occurs most often, and how many observations are shown?
Figure for problem 539680

Hints

- Use the tallest stack to identify the most frequent score. - Count every dot separately to determine the sample size.

Solution

1. The tallest stack is at \(17\), with \(4\) dots. 2. Counting all dots gives \(13\) observations.

Answer

The mode is \(17\), and there are \(13\) observations.
Graphs for one quantitative variable 2
A stem-and-leaf plot uses the key \(4\mid 7=47\). \(3\mid 2\ 5\ 8\) \(4\mid 0\ 1\ 1\ 6\ 9\) \(5\mid 2\ 4\) List the original data values in increasing order.

Hints

- Use the key to interpret one stem–leaf pair before reading the rest. - Read stems as leading digits and leaves as trailing digits, preserving repeated leaves.

Solution

1. Combine each stem with each leaf using the key. 2. The values are \(32,35,38,40,41,41,46,49,52,54\).

Answer

\(32,35,38,40,41,41,46,49,52,54\).
Graphs for one quantitative variable 3
A stem-and-leaf plot uses the key \(12\mid 4=12.4\). \(11\mid 8\ 9\) \(12\mid 0\ 2\ 4\ 4\ 7\) \(13\mid 1\ 5\) What values are represented by the two leaves \(4\) on stem \(12\), and why are both leaves needed?

Hints

- Use the key to translate one stem–leaf pair into a data value. - Keep repeated leaves as separate observations rather than merging them.

Solution

1. Each \(12\mid 4\) represents \(12.4\). 2. Two leaves are needed because the value \(12.4\) occurred twice.

Answer

Both leaves represent \(12.4\); the repeated leaves show two observations at that value.
Graphs for one quantitative variable 4
The histogram uses equal-width bins. What is the bin width, and which interval contains the greatest frequency?
Figure for problem 539687

Hints

- Read adjacent interval boundaries and locate the tallest bar. - Bin width is the difference between consecutive boundaries, not the visual width of a bar.

Solution

1. Consecutive bin boundaries differ by \(20\), so the bin width is \(20\). 2. The tallest bar has frequency \(12\) for \([140,160)\).

Answer

Bin width: \(20\). Greatest-frequency interval: \([140,160)\).
Graphs for one quantitative variable 5
A stem-and-leaf plot has \(18\) observations. The displayed rows contain \(4\), \(6\), \(3\), and \(4\) leaves, plus one row with an unknown number of leaves. How many leaves must the unknown row contain?

Hints

- Use the one-to-one relationship between leaves and observations. - Subtract the known number of leaves from the stated number of observations.

Solution

1. The known rows contain \(4+6+3+4=17\) leaves. 2. Each leaf represents one observation. 3. The unknown row must contain \(18-17=1\) leaf.

Answer

The unknown row must contain \(1\) leaf.
Graphs for one quantitative variable 6
List the data values represented by the dotplot in increasing order.
Figure for problem 539695

Hints

- Translate each vertical stack into repeated observations at that x-value. - Work from left to right so the list stays in increasing order.

Solution

1. Read one value for each dot, repeating an x-value according to its stack height. 2. The data are \(1,1,2,3,3,3,4,5,5\).

Answer

\(1,1,2,3,3,3,4,5,5\).
Graphs for one quantitative variable 7
How many days had exactly \(3\) visitors?
Figure for problem 563395

Hints

- Each dot represents one observed day. - Count the dots directly above the target value.

Solution

1. Find the stack of dots at \(3\) on the horizontal axis. 2. The stack has \(4\) dots, so four days had that count.

Answer

\(4\) days.
Graphs for one quantitative variable 8
Compare the two number lines. Their ticks are equally spaced, but their labels follow different patterns. Which one uses a logarithmic scale? Explain what stays the same from one tick to the next on each.
Figure for problem 563420

Hints

- Look at how each label changes to the next. - Is the same number added at every step, or is the same factor used? - Test what happens at more than one pair of neighboring ticks.

Solution

1. Axis A increases by \(10\) at each step, so it is linear. 2. Axis B multiplies by \(10\) at each step, so it is logarithmic. 3. Equal distances mean equal differences on A but equal ratios on B.

Answer

Axis B is logarithmic: each step multiplies by \(10\). Axis A is linear: each step adds \(10\).
Graphs for one quantitative variable 9
The histogram groups bicycle commute times in minutes. How many commutes lasted at least \(10\) but less than \(15\) minutes?
Figure for problem 563516

Hints

- The interval includes \(10\) but not \(15\). - Read the height of that interval's bar on the frequency axis.

Solution

1. Find the histogram interval \([10,15)\). 2. Its bar has frequency \(5\).

Answer

\(5\) commutes.
Graphs for one quantitative variable 10
The histogram shows the time, in minutes, that \(30\) customers waited for service. How many customers waited at least \(10\) minutes but less than \(20\) minutes?
Figure for problem 539679

Hints

- Identify every interval included in the requested range, then combine their frequencies. - Use the interval endpoints carefully: include bins beginning at the lower bound and ending before the upper bound.

Solution

1. The relevant bins are \([10,15)\) and \([15,20)\). 2. Their frequencies are \(11\) and \(6\). 3. The total is \(11+6=17\).

Answer

\(17\) customers.
Graphs for one quantitative variable 11
The histogram displays \(30\) test scores using bins of width \(10\). What proportion of scores are in the interval \([80,90)\)?
Figure for problem 539682

Hints

- Read the frequency of the specified bin and compare it with the total. - A histogram bar gives the bin frequency; divide that count by the total number of scores.

Solution

1. The frequency in \([80,90)\) is \(10\). 2. The proportion is \(\frac{10}{30}=\frac{1}{3}\approx 0.333\).

Answer

\(\frac{1}{3}\), or approximately \(0.333\).
Graphs for one quantitative variable 12
Use the dotplot to determine the frequency of values greater than \(5\) and the relative frequency of the value \(4\).
Figure for problem 539684

Hints

- Count the dots with x-values greater than \(5\). - Separately count the dots at \(4\) and the total number of dots. - Form the relative frequency using the count at \(4\) over the total.

Solution

1. Values greater than \(5\) are \(6,8,8,9\), so the frequency is \(4\). 2. There are \(11\) observations and three observations equal to \(4\). 3. The relative frequency of \(4\) is \(\frac{3}{11}\approx0.273\).

Answer

Frequency greater than \(5\): \(4\). Relative frequency of \(4\): \(\frac{3}{11}\approx0.273\).
Graphs for one quantitative variable 13
The relative frequency histogram represents \(200\) observations. How many observations fall in the bin \([4,6)\)?
Figure for problem 539685

Hints

- Identify whether the bar height is a count or a relative frequency. - Convert the displayed share to the requested count using the stated sample size.

Solution

1. The bin has relative frequency \(0.35\). 2. Its frequency is \(0.35\cdot 200=70\).

Answer

\(70\) observations.
Graphs for one quantitative variable 14
A stem-and-leaf plot is intended to show the data \(21,23,23,27,31,34\), using tens digits as stems. Bohdan writes \(2\mid 1\ 3\ 7\) \(3\mid 1\ 4\) Identify and correct the omission.

Hints

- Compare the display with the raw list one observation at a time. - Repeated raw values must appear as repeated leaves in the correct stem row.

Solution

1. The data contain two occurrences of \(23\). 2. The stem \(2\) must have two leaves equal to \(3\). 3. The corrected first row is \(2\mid 1\ 3\ 3\ 7\).

Answer

The second \(3\) leaf is missing. The corrected row is \(2\mid 1\ 3\ 3\ 7\).
Graphs for one quantitative variable 15
The dotplot contains a gap. Identify the integer value in the gap and describe the groups of observations it separates.
Figure for problem 539688

Hints

- Look for an x-axis value with no dot between occupied values. - After finding the empty position, describe the observed values on each side of it.

Solution

1. There are no observations at \(9\). 2. The gap at \(9\) separates the lower group at values \(5\) through \(8\) from the upper group at \(10\).

Answer

The gap is at \(9\); it separates the observations at \(5\) through \(8\) from those at \(10\).
Graphs for one quantitative variable 16
A data set contains exact whole-number values with many repeats. A researcher wants a graph that preserves every observed value and makes repeated values visible. Which is more suitable: a dotplot or a histogram? Explain.

Hints

- Decide whether preserving each exact observed value is essential. - Compare how the two displays represent repeated values and grouped intervals.

Solution

1. A dotplot places one dot for every observation and stacks repeated values. 2. A histogram groups values into intervals and may not preserve individual values. 3. Therefore, a dotplot is more suitable.

Answer

A dotplot, because it preserves individual values and displays repetitions directly.
Graphs for one quantitative variable 17
A data set has \(5000\) continuous measurements spanning a wide range. Which is generally more practical for displaying the overall distribution: a stem-and-leaf plot or a histogram? Explain.

Hints

- Consider how the number and type of observations affect the readability of each representation. - A useful graph should remain readable when thousands of continuous observations are included.

Solution

1. A stem-and-leaf plot would require a leaf for every observation and would be unwieldy. 2. A histogram summarizes many continuous measurements in intervals. 3. Therefore, a histogram is more practical.

Answer

A histogram, because it summarizes a large continuous data set using intervals.
Graphs for one quantitative variable 18
The histogram uses bins \([0,10)\), \([10,20)\), \([20,30)\), and \([30,40)\). If the value \(20\) is added, which bin frequency changes under the usual left-inclusive, right-exclusive convention?
Figure for problem 539691

Hints

- Use the endpoint convention stated in the problem. - At a shared boundary, the left-inclusive convention places the value in the interval that starts there.

Solution

1. Under the convention, \(20\) belongs to \([20,30)\), not \([10,20)\). 2. The frequency of \([20,30)\) increases from \(10\) to \(11\).

Answer

The \([20,30)\) bin increases to frequency \(11\).
Graphs for one quantitative variable 19
Constanza says, “The histogram proves that the exact value \(12\) occurred three times.” Explain why the statement is not justified.
Figure for problem 539694

Hints

- Distinguish information about an interval from information about a specific value. - Every value inside a bin contributes to the same bar, so the bar cannot isolate one exact value.

Solution

1. The histogram shows only that \(7\) observations are in the interval \([10,15)\). 2. It does not show the individual values within that bin. 3. Therefore, the exact frequency of \(12\) cannot be determined.

Answer

Constanza's claim is not justified because the histogram groups all values in \([10,15)\); it does not preserve exact values.
Graphs for one quantitative variable 20
Verify that the displayed relative frequency histogram represents a complete distribution. Then find the combined relative frequency for values in \([1,3)\).
Figure for problem 539696

Hints

- Check the total area represented by all equal-width bars, then combine the requested bins. - For equal-width bins, the relative frequencies should sum to \(1\); then select only bins contained in the requested interval.

Solution

1. The bar heights sum to \(0.15+0.30+0.40+0.15=1.00\), so the distribution is complete. 2. The interval \([1,3)\) includes the middle two bins. 3. Their combined relative frequency is \(0.30+0.40=0.70\).

Answer

The heights sum to \(1\), and the relative frequency in \([1,3)\) is \(0.70\).
Graphs for one quantitative variable 21
From the histogram, what can be concluded about the minimum and maximum data values? Give the most precise intervals supported by the graph.
Figure for problem 539697

Hints

- Use the first and last nonempty intervals without assuming locations inside them. - Report interval bounds rather than inventing exact values inside the end bins.

Solution

1. The first nonempty bin is \([10,20)\), so the minimum lies in \([10,20)\). 2. The last nonempty bin is \([40,50)\), so the maximum lies in \([40,50)\). 3. Exact extrema cannot be recovered from the histogram.

Answer

Minimum: somewhere in \([10,20)\). Maximum: somewhere in \([40,50)\). The exact values cannot be determined.
Graphs for one quantitative variable 22
A stem-and-leaf plot and a histogram are made from the same \(24\) values. State one numerical question the stem-and-leaf plot can answer exactly that the histogram generally cannot.

Hints

- Focus on the different level of detail retained by the two displays. - Ask a question whose answer depends on an individual observation, not merely an interval count.

Solution

1. A stem-and-leaf plot preserves each original value. 2. Therefore, it can answer an exact-value question such as “What is the maximum value?” or “How many observations equal a specified value?” 3. A histogram usually provides only interval counts.

Answer

For example, the stem-and-leaf plot can give the exact maximum value, while the histogram generally cannot.
Graphs for one quantitative variable 23
A histogram is intended to use the equal-width intervals \([0,10)\), \([10,20)\), and \([20,30)\). The draft graph is shown. Identify the placement error and give the correct horizontal center for every bar.
Figure for problem 548695

Hints

- Relate each rectangle's horizontal position to the interval it represents. - Find the point halfway between each pair of interval endpoints. - Check whether the misplaced bar is centered or aligned with an endpoint.

Solution

1. A histogram bar is centered at the midpoint of its interval. 2. The interval midpoints are \(\frac{0+10}{2}=5\), \(\frac{10+20}{2}=15\), and \(\frac{20+30}{2}=25\). 3. The second bar was placed at its right endpoint instead of its midpoint.

Answer

The correct centers are \(5\), \(15\), and \(25\). The second bar should be moved from \(20\) to \(15\).
Graphs for one quantitative variable 24
A frequency histogram has four equal-width intervals with counts \(6\), \(12\), \(14\), and \(8\). Convert it to a relative-frequency histogram. Give the new bar heights and state which features of the histogram remain unchanged.

Hints

- First find the total number of observations. - Use that same total to rescale every bar. - A common vertical rescaling does not alter horizontal structure.

Solution

1. The sample size is \(6+12+14+8=40\). 2. The relative-frequency heights are \(\frac{6}{40}=0.15\), \(\frac{12}{40}=0.30\), \(\frac{14}{40}=0.35\), and \(\frac{8}{40}=0.20\). 3. The bin locations, ordering of heights, and overall shape remain unchanged because all bars are divided by the same total.

Answer

The new heights are \(0.15,0.30,0.35,0.20\). The bin positions and shape remain the same; only the vertical scale changes.
Graphs for one quantitative variable 25
A histogram with equal-width bins has frequencies \(3,5,7,9,6,4\) from left to right. Each pair of adjacent bins will be merged to make bins twice as wide. Find the three new frequencies and describe one detail that the wider bins hide.

Hints

- A wider bin combines all observations from the smaller intervals it covers. - Work from left to right without changing the total count. - Consider what comparisons are no longer possible after two bars become one.

Solution

1. The first merged frequency is \(3+5=8\). 2. The second merged frequency is \(7+9=16\). 3. The third merged frequency is \(6+4=10\). 4. The wider bins hide how observations were divided between the two original subintervals within each merged interval.

Answer

The new frequencies are \(8,16,10\). The graph no longer shows the within-pair differences between the original narrower bins.
Graphs for one quantitative variable 26
A stem-and-leaf plot uses the key \(2\mid4=2.4\). <table><tbody><tr><td>\(1\)</td><td>\(2\;5\;8\)</td></tr><tr><td>\(2\)</td><td>\(0\;0\;4\;9\)</td></tr><tr><td>\(3\)</td><td>\(1\)</td></tr></tbody></table> Find the sample size, minimum, maximum, and range.

Hints

- Use the key to translate one stem-leaf pair into a data value. - Each leaf represents one observation. - Read the smallest and largest entries before finding their difference.

Solution

1. There are \(3+4+1=8\) leaves, so the sample size is \(8\). 2. The minimum is \(1.2\), and the maximum is \(3.1\). 3. The range is \(3.1-1.2=1.9\).

Answer

The sample size is \(8\), the minimum is \(1.2\), the maximum is \(3.1\), and the range is \(1.9\).
Graphs for one quantitative variable 27
Construct a split stem-and-leaf plot for the data \(42,44,45,47,48,51,53,55,57,59\). Use each tens digit twice: the first row for leaves \(0\) through \(4\), and the second row for leaves \(5\) through \(9\). Include a key.

Hints

- Separate the tens digit from the ones digit in each observation. - Place leaves \(0\) through \(4\) on the first occurrence of a stem. - Keep the leaves in increasing order within each row.

Solution

1. For stem \(4\), the low leaves are \(2,4\), and the high leaves are \(5,7,8\). 2. For stem \(5\), the low leaves are \(1,3\), and the high leaves are \(5,7,9\). 3. A suitable key is \(5\mid7=57\).

Answer

\(4\mid2\;4\) \(4\mid5\;7\;8\) \(5\mid1\;3\) \(5\mid5\;7\;9\) Key: \(5\mid7=57\).
Graphs for one quantitative variable 28
Every value in a dotplot is increased by \(3\). Describe exactly how the new dotplot differs from the original and which distribution features remain unchanged.

Hints

- Track what happens to one plotted value first. - Compare differences between pairs of values before and after the change. - A common translation changes location but not relative spacing.

Solution

1. Each dot moves \(3\) units to the right because every observation is increased by \(3\). 2. The center increases by \(3\). 3. Distances between observations do not change, so the spread, shape, gaps, clusters, and multiplicities remain unchanged.

Answer

The entire dotplot shifts \(3\) units right. Its center increases by \(3\), while its shape and spread remain unchanged.
Graphs for one quantitative variable 29
A stem-and-leaf display was entered with unsorted leaves: <table><tbody><tr><td>\(3\)</td><td>\(8\;2\;5\)</td></tr><tr><td>\(4\)</td><td>\(1\;9\;3\)</td></tr></tbody></table> The key is \(3\mid2=32\). Rewrite the display correctly and find the median.

Hints

- Leaves should increase from left to right within each stem. - Translate the corrected display into one ordered list. - Locate the middle position based on the number of observations.

Solution

1. Sort the leaves within each stem to obtain \(3\mid2\;5\;8\) and \(4\mid1\;3\;9\). 2. The ordered data are \(32,35,38,41,43,49\). 3. With \(6\) observations, the median is the average of the third and fourth values: \(\frac{38+41}{2}=39.5\).

Answer

Corrected display: \(3\mid2\;5\;8\) \(4\mid1\;3\;9\) The median is \(39.5\).
Graphs for one quantitative variable 30
The histogram uses intervals of width \(5\): \([0,5)\), \([5,10)\), \([10,15)\), \([15,20)\), and \([20,25)\). Identify the gap in the distribution and find the sample size.
Figure for problem 548710

Hints

- Look for consecutive intervals with no bar height. - Translate the empty intervals back into a numerical range. - Add all interval frequencies, including zeros.

Solution

1. The bars for \([10,15)\) and \([15,20)\) have frequency \(0\), so no observations lie from \(10\) up to but not including \(20\). 2. The nonzero frequencies are \(4\), \(6\), and \(5\). 3. The sample size is \(4+6+5=15\).

Answer

The gap is \([10,20)\), and the sample size is \(15\).
Graphs for one quantitative variable 31
The histogram shown uses left-closed, right-open intervals. Give one possible raw data set that produces this histogram, and explain why the histogram does not determine a unique raw data set.
Figure for problem 548711

Hints

- Read the required number of observations in each displayed interval. - Respect the stated left-closed, right-open endpoint convention. - Choose values inside each bin without assuming the graph fixes their exact positions.

Solution

1. The histogram shows frequency \(2\) in \([0,10)\) and frequency \(3\) in \([10,20)\). 2. One valid choice is \(1,9\) for the first bin and \(10,15,19\) for the second bin. 3. Thus \(1,9,10,15,19\) is one possible data set. 4. Many other values within the same intervals would produce the same histogram because exact positions inside bins are not recorded.

Answer

One possible data set is \(1,9,10,15,19\). The answer is not unique because the histogram specifies only interval membership, not exact within-bin values.
Graphs for one quantitative variable 32
A stem-and-leaf plot has the rows \(1\mid2\;5\) and \(2\mid0\;3\), with key \(1\mid2=1.2\). Every data value is multiplied by \(10\). Write the new data values and explain how the same written stems and leaves could be reused with a different key.

Hints

- Decode the original plot before applying the transformation. - Apply the scale factor to every observation. - A stem-and-leaf key determines place value, so it may need to change even when the digits do not.

Solution

1. The original values are \(1.2,1.5,2.0,2.3\). 2. Multiplying by \(10\) gives \(12,15,20,23\). 3. The written rows can remain \(1\mid2\;5\) and \(2\mid0\;3\), but the key must change to \(1\mid2=12\).

Answer

The new values are \(12,15,20,23\). The same rows may be used if the key is changed to \(1\mid2=12\).
Graphs for one quantitative variable 33
Three observations were all recorded as \(7\), but the displayed dotplot spreads their dots horizontally to prevent overlap. Explain why this changes the represented data and describe the correct way to display the repeats.
Figure for problem 548714

Hints

- Identify which coordinate carries the quantitative data value in a standard dotplot. - Compare the intended repeated value with the horizontal positions actually shown. - Preserve one x-position and use stacking to show multiplicity.

Solution

1. Horizontal position in a dotplot encodes the numerical value, so the separated dots are shown at different values rather than all at \(7\). 2. Vertical position in a standard dotplot is used only to stack repeats and does not represent a second measured variable. 3. The three observations should therefore be stacked vertically at x-value \(7\).

Answer

Horizontal spreading falsely changes the represented values. Stack all three dots vertically at \(7\).
Graphs for one quantitative variable 34
The histogram shown uses left-closed, right-open bins. Every observation is multiplied by \(2\). State the new intervals and their frequencies.
Figure for problem 548715

Hints

- Read the original class boundaries and bar heights from the histogram. - Apply the positive scale factor to the class boundaries as well as to the observations. - Decide whether scaling changes how many observations correspond to each transformed bin.

Solution

1. The original intervals are \([0,5)\), \([5,10)\), and \([10,15)\), with frequencies \(4,9,7\). 2. Multiplying every observation by \(2\) doubles each interval endpoint. 3. The new intervals are \([0,10)\), \([10,20)\), and \([20,30)\). 4. The transformation does not add or remove observations, so the corresponding frequencies remain \(4,9,7\).

Answer

The new intervals are \([0,10)\), \([10,20)\), and \([20,30)\), with frequencies \(4,9,7\).
Graphs for one quantitative variable 35
A stem-and-leaf plot with key \(1\mid2=12\) is written as <table><tbody><tr><td>\(1\)</td><td>\(2\;8\)</td></tr><tr><td>\(3\)</td><td>\(1\;4\)</td></tr></tbody></table> The writer omitted stems with no leaves. Rewrite the plot so its numerical scale is explicit, and describe the gap shown by the completed display.

Hints

- Preserve every stem needed for a continuous tens-place scale. - Translate the leaves into full data values. - Use the empty row to identify the interval containing no observations.

Solution

1. Stem \(2\) must appear between stems \(1\) and \(3\), even though it has no leaves. 2. The corrected plot is \(1\mid2\;8\), \(2\mid\), and \(3\mid1\;4\). 3. The data are \(12,18,31,34\), so there are no observations from \(20\) through \(29\).

Answer

<table><tbody><tr><td>\(1\)</td><td>\(2\;8\)</td></tr><tr><td>\(2\)</td><td></td></tr><tr><td>\(3\)</td><td>\(1\;4\)</td></tr></tbody></table> The empty stem shows a gap from \(20\) through \(29\).
Graphs for one quantitative variable 36
The displayed bars in a relative-frequency histogram sum to \(0.95\). The intervals shown are mutually exclusive, but the horizontal axis ends before the largest observations. What proportion of the data lies outside the displayed range, and what graphing correction is needed?

Hints

- Compare the displayed total with the proportion of a complete data set. - Use the description of the horizontal axis to locate the missing observations conceptually. - A complete graph must cover the full data range.

Solution

1. A complete relative-frequency histogram must account for a total proportion of \(1\). 2. The omitted proportion is \(1-0.95=0.05\), or \(5\%\). 3. The horizontal range must be extended and one or more intervals added to include the omitted largest observations.

Answer

The missing proportion is \(0.05\), or \(5\%\). Extend the axis and include bins for those observations.
Graphs for one quantitative variable 37
A data set contains one temperature for each of \(30\) consecutive days. An analyst makes a histogram and claims it shows that temperatures rose steadily during the month. Explain why a histogram cannot support that claim and name the information that the graph discards.

Hints

- Identify what the horizontal axis of a histogram represents. - Ask whether observations retain their original sequence after grouping. - A trend requires information about when values occurred.

Solution

1. A histogram groups temperatures by value and counts how many fall in each interval. 2. It does not preserve which temperature occurred on which day. 3. A steady rise is a statement about chronological order, so the original day sequence or a time plot is needed to evaluate it.

Answer

The histogram cannot show a steady rise because it discards the order of the \(30\) days. The chronological sequence is required.
Graphs for one quantitative variable 38
The waiting times, in minutes, are \(2,3,3,4,4,4,5,5,6,7,7,9\). Find the number of dots a dot plot would place above each integer from \(2\) through \(9\). Identify the mode and any gap inside the displayed range.

Hints

- Count every repeated value before comparing stack heights. - The mode is the value with the greatest dot count. - Look for an interior integer with no data while both sides contain observations.

Solution

1. Count the values at each integer from \(2\) through \(9\): \(1,2,3,2,1,2,0,1\). 2. The tallest stack would be at \(4\), so the mode is \(4\). 3. There are observations on both sides of \(8\) but none at \(8\), so \(8\) is a gap.

Answer

Dot counts for \(2\) through \(9\): \(1,2,3,2,1,2,0,1\). Mode: \(4\). Gap: \(8\).
Graphs for one quantitative variable 39
Package masses, in kilograms, are \(1.1,1.7,2.0,2.2,2.4,2.8,3.1,3.6,3.9,4.4,4.9,5.3\). A histogram uses bins \([1,2)\), \([2,3)\), \([3,4)\), \([4,5)\), and \([5,6)\). What would the five bar heights be? Which bin is tallest?

Hints

- Place each mass into exactly one interval. - Pay attention to values that lie exactly on a left endpoint. - Compare the bin counts after every observation has been counted.

Solution

1. Each bin includes its left endpoint and excludes its right endpoint. 2. Count the values in the five bins to get frequencies \(2,4,3,2,1\). 3. The tallest bar is for \([2,3)\), with frequency \(4\).

Answer

Bar heights: \(2,4,3,2,1\). Tallest bin: \([2,3)\).
Graphs for one quantitative variable 40
The table groups measured values into equal-width intervals. <table><thead><tr><th>Interval</th><th>Frequency</th></tr></thead><tbody><tr><td>\([0,10)\)</td><td>\(2\)</td></tr><tr><td>\([10,20)\)</td><td>\(5\)</td></tr><tr><td>\([20,30)\)</td><td>\(9\)</td></tr><tr><td>\([30,40)\)</td><td>\(6\)</td></tr><tr><td>\([40,50)\)</td><td>\(3\)</td></tr></tbody></table> A frequency polygon uses each interval's midpoint as the x-coordinate and its frequency as the y-coordinate. List the five points in increasing x-order.

Hints

- Find the midpoint of each interval before using its frequency. - Keep the midpoint paired with the count from the same row. - Order the points by their midpoint values.

Solution

1. The interval midpoints are \(5,15,25,35,45\). 2. Pair each midpoint with its own interval frequency. 3. The points are \((5, 2)\), \((15, 5)\), \((25, 9)\), \((35, 6)\), and \((45, 3)\).

Answer

\((5, 2)\), \((15, 5)\), \((25, 9)\), \((35, 6)\), \((45, 3)\).
Graphs for one quantitative variable 41
The table summarizes \(20\) observations. <table><thead><tr><th>Interval</th><th>Frequency</th></tr></thead><tbody><tr><td>\([0,10)\)</td><td>\(4\)</td></tr><tr><td>\([10,20)\)</td><td>\(6\)</td></tr><tr><td>\([20,30)\)</td><td>\(8\)</td></tr><tr><td>\([30,40)\)</td><td>\(2\)</td></tr></tbody></table> A cumulative relative-frequency graph starts at the lowest boundary with proportion \(0\) and uses each later upper boundary for the running proportion. List its five points from left to right.

Hints

- Add frequencies as you move from left to right. - Change each running count into a share of all \(20\) observations. - Use the ends of intervals for the x-coordinates, not the midpoints.

Solution

1. The cumulative counts at upper boundaries \(10,20,30,40\) are \(4,10,18,20\). 2. Divide each by \(20\) to get \(0.20,0.50,0.90,1.00\). 3. Include the starting point \((0, 0)\). 4. The points are \((0, 0)\), \((10, 0.20)\), \((20, 0.50)\), \((30, 0.90)\), and \((40, 1.00)\).

Answer

\((0, 0)\), \((10, 0.20)\), \((20, 0.50)\), \((30, 0.90)\), \((40, 1.00)\).
Graphs for one quantitative variable 42
Use the ogive. a) Between which two displayed thresholds does the median lie? b) What proportion of observations are greater than \(30\)? c) Between which two displayed thresholds does the \(80\)th percentile lie? d) What proportion of observations are greater than \(20\) and at most \(40\)?
Figure for problem 556216

Hints

- Read each ogive height as the proportion at or below that threshold. - Locate a percentile by finding where its cumulative proportion falls between displayed points. - A proportion between two thresholds is found by subtracting cumulative proportions.

Solution

1. The cumulative proportion is \(0.35\) at \(20\) and \(0.70\) at \(30\), so the median lies in \((20,30]\). 2. The cumulative proportion at \(30\) is \(0.70\), so the proportion greater than \(30\) is \(1-0.70=0.30\). 3. The cumulative proportion is \(0.70\) at \(30\) and \(0.90\) at \(40\), so the \(80\)th percentile lies in \((30,40]\). 4. The proportion greater than \(20\) and at most \(40\) is \(0.90-0.35=0.55\).

Answer

a) \((20,30]\) b) \(0.30\) c) \((30,40]\) d) \(0.55\)
Graphs for one quantitative variable 43
A graph has one bar for each weekday: Monday, Tuesday, Wednesday, Thursday, and Friday. The bars touch, and its title says “Histogram of Daily Ticket Sales.” Is it really a histogram? Explain how the graph should be displayed for the categories shown.

Hints

- Check whether the x-axis labels are number intervals or category names. - Touching bars represent neighboring numerical intervals in a histogram. - Think about what spacing suits weekday categories.

Solution

1. The horizontal labels are days of the week, which are named categories. 2. A histogram uses intervals of a quantitative variable, not a category for each weekday. 3. The graph should be a categorical bar chart with separated bars and a frequency scale.

Answer

No. With weekdays as categories, it is a bar chart; the bars should be separated.
Graphs for one quantitative variable 44
The histogram uses intervals that include the left endpoint and exclude the right. A new observation is exactly \(10\). Which bin would contain it? How many earlier observations are in that bin?
Figure for problem 563517

Hints

- The new value sits on a border. Check the stated endpoint rule. - What are the two bins touching that border? - Read the height of the bin that contains the value.

Solution

1. Under the given rule, \(10\) belongs to \([10,15)\), not \([5,10)\). 2. The histogram shows frequency \(4\) in \([10,15)\).

Answer

Bin \([10,15)\); it currently contains \(4\) observations.
Graphs for one quantitative variable 45
A stem-and-leaf display has no key. <table><tbody><tr><td>\(1\)</td><td>\(2\;5\)</td></tr><tr><td>\(2\)</td><td>\(0\;4\)</td></tr></tbody></table> Give two different data sets that this display could represent. Then state what a key must communicate to make the display unambiguous.

Hints

- Try reading the same stem-leaf pair with different decimal-place conventions. - Check whether both interpretations preserve the order shown in the display. - Think about what information a reader needs to turn one row entry into an actual number.

Solution

1. If the stem is the tens digit and the leaf is the ones digit, the data are \(12,15,20,24\). 2. If the stem is the ones digit and the leaf is the tenths digit, the data are \(1.2,1.5,2.0,2.4\). 3. A key must show the place value represented by the stem and leaf, using one complete example such as \(1\mid2=12\) or \(1\mid2=1.2\).

Answer

Two possible data sets are \(12,15,20,24\) and \(1.2,1.5,2.0,2.4\). A key must identify the place values of the stem and leaf by interpreting one stem-leaf pair.
Graphs for one quantitative variable 46
A draft histogram contains nonempty bars for \([0,10)\) and \([30,40)\). The designer omits the zero-frequency intervals \([10,20)\) and \([20,30)\) from the horizontal axis and places the two nonempty bars next to each other. Explain why the graph is misleading and describe the correction.

Hints

- Compare the numerical endpoints of the occupied intervals. - Decide what physical distance between bars should represent on a quantitative axis. - Preserve zero-frequency intervals when they carry information about gaps.

Solution

1. The omitted intervals represent a genuine \(20\)-unit gap in the data. 2. Placing the nonempty bars next to each other falsely suggests that the occupied intervals are adjacent. 3. The horizontal scale must retain the two zero-frequency intervals as empty spaces so distance and bin adjacency remain accurate.

Answer

The graph hides the gap from \(10\) to \(30\). Keep the full numerical axis and display empty bins for \([10,20)\) and \([20,30)\).
Graphs for one quantitative variable 47
The data are \(1.9,2.1,3.9,4.1,5.9,6.1,7.9,8.1\). a) Find the histogram frequencies for \([0,2)\), \([2,4)\), \([4,6)\), \([6,8)\), and \([8,10)\). b) Find the frequencies for \([1,3)\), \([3,5)\), \([5,7)\), and \([7,9)\). c) Explain what the comparison shows about bin placement.

Hints

- Apply the endpoint convention separately to each set of intervals. - Count every observation exactly once in each histogram. - Compare the resulting bar patterns rather than only their totals.

Solution

1. For part a), the frequencies are \(1,2,2,2,1\). 2. For part b), the frequencies are \(2,2,2,2\). 3. The same raw data can produce visibly different bar patterns when bin boundaries shift, so histogram appearance depends partly on bin placement.

Answer

a) \(1,2,2,2,1\) b) \(2,2,2,2\) c) Shifting the bin boundaries changes the visible pattern even though the data do not change.
Graphs for one quantitative variable 48
The dotplot is shown below. One additional observation will be added so that the completed dotplot is symmetric about \(10\). What value must be added?
Figure for problem 548705

Hints

- Use the stated center of symmetry to pair points at equal horizontal distances. - Check which plotted point currently lacks a reflected partner. - Place the missing value the same distance on the opposite side of the center.

Solution

1. Symmetry about \(10\) pairs values that are the same distance from \(10\). 2. The pairs \(8\) and \(12\), and \(9\) and \(11\), are already present; \(10\) is at the center. 3. The point at \(6\) is \(4\) units below \(10\), so its missing partner is \(4\) units above \(10\), at \(14\).

Answer

Add the value \(14\).
Graphs for one quantitative variable 49
A relative-frequency histogram has four bar heights \(0.15\), \(0.25\), \(0.35\), and \(0.25\). The second interval contains \(15\) observations. Find the sample size and the frequency in every interval.

Hints

- Match the known interval count to its relative-frequency height. - Use that relationship to recover the total number of observations. - Apply the total to each remaining bar height.

Solution

1. The second interval gives \(0.25n=15\), so \(n=60\). 2. The interval frequencies are \(0.15\cdot60=9\), \(0.25\cdot60=15\), \(0.35\cdot60=21\), and \(0.25\cdot60=15\). 3. The frequencies total \(9+15+21+15=60\), matching the sample size.

Answer

The sample size is \(60\), and the interval frequencies are \(9,15,21,15\).
Graphs for one quantitative variable 50
A new dotplot is made by replacing every observation \(x\) with \(-x\). Describe how the new graph relates to the original, including what happens to its center, shape, and spread.

Hints

- Track the location of a point on each side of zero. - Compare distances before and after changing signs. - Consider what reflection does to the direction of a tail.

Solution

1. Each point at \(x\) moves to the point at \(-x\), so the dotplot is reflected across \(0\). 2. Measures of center change sign. 3. Distances between values are preserved, so the range, interquartile range, and standard deviation remain unchanged. 4. The shape is mirrored: right skew becomes left skew and left skew becomes right skew.

Answer

The dotplot is reflected across \(0\). Its center changes sign, its spread stays the same, and its shape is reversed left to right.
Graphs for one quantitative variable 51
The ogive shows cumulative relative frequency at the upper boundaries of four consecutive intervals. The sample size is \(80\). Find each interval's relative frequency and frequency.
Figure for problem 548713

Hints

- Read cumulative proportions at successive upper class boundaries. - Subtract adjacent cumulative values to isolate each later interval. - Convert each interval proportion to a count using the stated sample size.

Solution

1. Read the cumulative relative frequencies as \(0.10,0.35,0.75,1.00\). 2. Consecutive differences give interval relative frequencies \(0.10\), \(0.35-0.10=0.25\), \(0.75-0.35=0.40\), and \(1.00-0.75=0.25\). 3. Multiplying by \(80\) gives frequencies \(8,20,32,20\). 4. The relative frequencies sum to \(1\), and the frequencies sum to \(80\).

Answer

The interval relative frequencies are \(0.10,0.25,0.40,0.25\), and the frequencies are \(8,20,32,20\).
Graphs for one quantitative variable 52
Waiting times are recorded to the nearest whole minute. An analyst wants histogram groups for recorded values \(0\) through \(4\), \(5\) through \(9\), and \(10\) through \(14\). Give interval boundaries that place every recorded integer unambiguously in the intended group.

Hints

- The data values lie on whole-number marks. - Put boundaries between neighboring possible recorded values. - Check the placement of the boundary values \(4\), \(5\), \(9\), and \(10\).

Solution

1. Place boundaries halfway between consecutive whole-number values. 2. Suitable intervals are \([-0.5,4.5)\), \([4.5,9.5)\), and \([9.5,14.5)\). 3. Each recorded integer lies inside exactly one interval, and the group labels match the intended integer ranges.

Answer

Use \([-0.5,4.5)\), \([4.5,9.5)\), and \([9.5,14.5)\).
Graphs for one quantitative variable 53
A histogram gives only the interval frequencies for a quantitative data set; the raw observations are unavailable. Can the exact mean always be calculated from the histogram? Explain what can be calculated instead and why it is only approximate.

Hints

- Ask what information about individual observations is lost during grouping. - Consider whether two values in the same interval must contribute equally to the mean. - Identify a representative value that could stand in for each interval.

Solution

1. A histogram records how many observations fall in each interval but not their exact values within the intervals. 2. Different raw data sets can have the same interval frequencies and different means. 3. An approximate mean can be found by treating every observation in an interval as if it were located at that interval's midpoint and computing a weighted average.

Answer

The exact mean cannot always be determined. A midpoint-weighted mean can be calculated, but it is only an estimate because positions within bins are unknown.
Graphs for one quantitative variable 54
The frequency polygon shown uses class midpoints on the x-axis and class frequencies on the y-axis. Find the sample size. Then explain why the connecting line at \(x=20\) does not show observations with value \(20\).
Figure for problem 548718

Hints

- Read the y-coordinate at each plotted class midpoint. - Add the class frequencies to obtain the sample size. - Distinguish the polygon's connecting segments from the underlying observations.

Solution

1. The plotted class frequencies are \(4,7,5,2\), so the sample size is \(4+7+5+2=18\). 2. The x-coordinates of the plotted points are class midpoints, not individual observations. 3. The line segments connect class-frequency points to show the distribution's pattern; interpolated points on a segment are not additional data values.

Answer

The sample size is \(18\). The line at \(x=20\) is only part of the visual connection between class midpoints and does not represent observations equal to \(20\).
Graphs for one quantitative variable 55
The histogram shown uses left-closed, right-open intervals. Determine the interval that must contain the median. Can the exact median be found from the histogram?
Figure for problem 548720

Hints

- Read the bar frequencies and use them to locate the middle ordered positions cumulatively. - Identify which displayed interval contains both middle observations. - Remember that a histogram does not preserve exact positions within a bin.

Solution

1. The displayed frequencies are \(4,6,2\), so the sample size is \(12\) and the median is the average of the sixth and seventh ordered observations. 2. Positions \(1\) through \(4\) lie in \([0,10)\), and positions \(5\) through \(10\) lie in \([10,20)\). 3. Both middle observations lie in \([10,20)\), so their average also lies in \([10,20)\). 4. Their exact values within the bin are unknown, so the exact median cannot be determined.

Answer

The median must lie in \([10,20)\), but its exact value cannot be found from the histogram.
Graphs for one quantitative variable 56
The histogram below uses frequency as the height of each unequal-width class. Gonçalo compares the bar areas as if they directly represent frequency. Explain why that is misleading, and state how the bar heights should be defined so that area represents frequency.
Figure for problem 539692

Hints

- Compare the widths of the classes before interpreting bar area. - Ask which geometric quantity of a histogram bar should represent frequency when widths differ. - Determine what height would make each bar's area match its class frequency.

Solution

1. The class widths are \(10\), \(10\), and \(30\), so the bars do not have equal widths. 2. If frequency is used as height, the bar areas are \(10\cdot8=80\), \(10\cdot12=120\), and \(30\cdot15=450\). Those areas are not proportional to the frequencies. 3. Use frequency density, defined as \(\frac{\text{frequency}}{\text{class width}}\), for bar height. 4. The appropriate heights are \(0.8\), \(1.2\), and \(0.5\), so each bar's area equals its frequency.

Answer

The unequal class widths make frequency heights produce misleading areas. Use frequency density as the height: \(0.8\), \(1.2\), and \(0.5\) for the three classes.
Graphs for one quantitative variable 57
A stem-and-leaf plot uses the key \(-2\mid4=-2.4\). The negative-stem rows are written as <table><tbody><tr><td>\(-2\)</td><td>\(4\;8\)</td></tr><tr><td>\(-1\)</td><td>\(2\;9\)</td></tr></tbody></table> Explain why the leaves are not ordered from least to greatest numerical value. Rewrite the two rows so the represented data are ordered within each stem.

Hints

- Translate each stem-and-leaf entry into its full decimal value. - Compare negative numbers rather than sorting only the leaf digits. - Check the row by reading its represented values from left to right.

Solution

1. On stem \(-2\), the values are \(-2.4\) and \(-2.8\); \(-2.8<-2.4\), so leaf \(8\) must come first. 2. On stem \(-1\), the values are \(-1.2\) and \(-1.9\); \(-1.9<-1.2\), so leaf \(9\) must come first. 3. The corrected rows are \(-2\mid8\;4\) and \(-1\mid9\;2\).

Answer

Corrected rows: <table><tbody><tr><td>\(-2\)</td><td>\(8\;4\)</td></tr><tr><td>\(-1\)</td><td>\(9\;2\)</td></tr></tbody></table> For negative stems, larger leaf digits produce more negative values.
Graphs for one quantitative variable 58
A designer replaces each stack in a dotplot with one circle whose diameter equals the frequency at that value. One value has frequency \(1\), and another has frequency \(3\). Explain why the circle display exaggerates the second frequency when viewers compare areas, and state the conventional dotplot correction.

Hints

- Compare the intended frequency ratio with the visual area ratio. - Recall how a circle’s area changes when its diameter changes. - Identify the basic observation-to-mark rule of a dotplot.

Solution

1. The intended frequency ratio is \(3:1\). 2. Circle area is proportional to the square of diameter, so diameters \(3:1\) create an area ratio of \(9:1\). 3. A conventional dotplot should use equal-sized dots, one dot for each observation, stacked at the recorded value.

Answer

The second circle has \(9\) times the area of the first even though its frequency is only \(3\) times as large. Use equal-sized, one-observation-per-dot stacks instead.

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