Dot plots and histograms

Learning goals

Constructs the requested data display with an appropriate scale.
Reads values and frequencies from dot plots or histogram intervals.
Describes center, spread, shape, and unusual features when appropriate.
Compares representations or distributions using evidence from the data.

Difficulty: 1 – 5

Click problems to add them to your worksheet.

Dot plots and histograms 1

Each dot stands for one student. How many students scored \(3\) points?

Hints

- Find the requested value on the horizontal scale. - Count only the dots directly above it.

Solution

1. Find the stack of dots above \(3\). 2. Count the four dots in that stack.

Answer

\(4\) students.

Dot plots and histograms 2

The dot plot records the number of goals scored in separate soccer games. How many games were recorded in all?

Hints

- A dot is one observation, even when several dots sit over the same number. - Count all dots, not just the different goal values.

Solution

1. Every dot represents one game. 2. The stacks contain \(1+3+2+4=10\) dots, so ten games were recorded.

Answer

\(10\) games.

Dot plots and histograms 3

Which interval contains the most students' walking times to school?

Hints

- The tallest bar represents the interval with the largest count. - Use the interval endpoints, not a single middle value, to name your answer.

Solution

1. Compare the bar heights in the histogram. 2. The bar from \(10\) to less than \(15\) minutes is highest.

Answer

\(10\) to less than \(15\) minutes.

Dot plots and histograms 4

The school records each student's favorite lunch: pizza, salad, or sandwich. Should the class make a bar graph or a histogram? Why?

Hints

- Are the lunch choices number intervals or named groups? - Which display keeps different categories in separate bars?

Solution

1. The lunch choices are categories, not measurements on a number scale. 2. A bar graph is suitable for comparing category counts.

Answer

A bar graph, because the data are categories.

Dot plots and histograms 5

The histogram shows puzzle-completion times. Identify the gap and the two clusters. How many observations are in each cluster?

Hints

- Look for an empty interval between occupied intervals. - Add the frequencies of nearby bars on each side. - Check both cluster counts against the total.

Solution

1. No time falls in the interval from \(10\) to less than \(15\) minutes, so that interval is a gap. 2. The cluster from \(0\) to less than \(10\) minutes has \(2+5=7\) observations. 3. The cluster from \(15\) to less than \(25\) minutes has \(4+1=5\) observations. 4. Together the clusters account for all \(12\) observations.

Answer

Gap: \(10\) to less than \(15\) minutes. First cluster: \(0\) to less than \(10\) minutes, with \(7\) times. Second cluster: \(15\) to less than \(25\) minutes, with \(5\) times.

Dot plots and histograms 6

The dot plot shows measurements rounded to the nearest \(0.5\) unit. a) How many observations are shown? b) What is the mode? c) What fraction of the observations are from \(1\) through \(2\), inclusive?

Hints

- Count every dot exactly once. - Compare stack heights to find the most frequent value. - Include both endpoints when combining the requested stacks.

Solution

1. Counting all dots gives \(9\) observations. 2. The tallest stack is at \(1.5\), with \(3\) dots, so the mode is \(1.5\). 3. The values from \(1\) through \(2\), inclusive, have frequencies \(2+3+2=7\). 4. The requested fraction is \(\frac{7}{9}\).

Answer

a) \(9\) b) \(1.5\) c) \(\frac{7}{9}\)

Dot plots and histograms 7

The histogram groups delivery times into equal-width intervals. a) How many observations are shown? b) What fraction lie in the middle three intervals? c) Is the pattern of bar heights symmetric? Explain.

Hints

- Add every interval frequency to get the total. - Count only the three middle intervals for part b). - Compare pairs of bars the same distance from the center.

Solution

1. The total is \(2+4+7+4+2=19\) observations. 2. The three middle intervals contain \(4+7+4=15\) observations, a fraction of \(\frac{15}{19}\). 3. The first and fifth bars have the same height, as do the second and fourth. Thus the grouped bar pattern is symmetric around the middle interval; this does not prove the individual times are symmetric within bins.

Answer

a) \(19\) b) \(\frac{15}{19}\) c) Yes, the histogram's bar pattern is symmetric. That alone does not prove the exact times form a symmetric list.

Dot plots and histograms 8

The numbers of correct answers in ten puzzle rounds were \(6,8,7,9,8,10,6,8,7,11\). Organize the data as dot-plot frequencies. Then state the mode, range, and number of rounds with at most \(7\) correct answers.

Hints

- Tally repeated values before interpreting the display. - Use the tallest planned stack for the mode. - Include both values below and equal to the stated cutoff.

Solution

1. The frequencies are: \(6\) occurs \(2\) times, \(7\) occurs \(2\) times, \(8\) occurs \(3\) times, and \(9,10,11\) each occur once. 2. The greatest frequency is \(3\) at \(8\), so the mode is \(8\). 3. The range is \(11-6=5\). 4. Values at most \(7\) occur \(2+2=4\) times.

Answer

Frequencies: \(6\) occurs \(2\) times, \(7\) occurs \(2\) times, \(8\) occurs \(3\) times, and \(9,10,11\) each occur once. Mode: \(8\) Range: \(5\) Rounds with at most \(7\): \(4\)

Dot plots and histograms 9

Which data set could produce the histogram? Explain why the other choices cannot. A) \(1,4,5,8,9,12\) B) \(1,5,5,8,9,15\) C) \(1,4,7,9,12,13\)

Hints

- Sort each possible data set into the histogram's intervals. - Check carefully what happens to a value on an interval boundary. - Compare the three bar frequencies with your counts.

Solution

1. Set A gives counts \(2,3,1\) in the three intervals from \(0\) to less than \(5\), \(5\) to less than \(10\), and \(10\) to less than \(15\). 2. Set B contains \(15\), which is outside these three intervals because the last interval stops before \(15\). 3. Set C gives interval frequencies \(2,2,2\), which differ from the plotted bar heights.

Answer

A) \(1,4,5,8,9,12\).

Dot plots and histograms 10

The dot plot shows a data distribution. Describe the cluster, gap, and separated value. Find the range with all observations and the range after removing the observation at \(15\).

Hints

- Look for a consecutive region containing most of the dots. - Identify empty values between occupied positions. - Recheck the maximum after removing the separated observation.

Solution

1. Fourteen observations form a cluster from \(5\) through \(9\). 2. There are no observations from \(10\) through \(14\), creating a gap. 3. The value \(15\) is separated from the main cluster. 4. With all observations, the range is \(15-5=10\). 5. After removing \(15\), the maximum is \(9\), so the range is \(9-5=4\).

Answer

Cluster: \(5\) through \(9\) Gap: \(10\) through \(14\) Separated value: \(15\) Range with \(15\): \(10\) Range without \(15\): \(4\)

Dot plots and histograms 11

A histogram has three intervals. The first bar has frequency \(4\), the third has frequency \(5\), and the histogram represents \(15\) observations total. Find the second bar’s frequency. Then identify which interval contains the median observation.

Hints

- Use the total frequency to fill the missing bar. - Locate the middle position for an odd number of observations. - Add interval frequencies from left to right until reaching that position.

Solution

1. The second frequency is \(15-4-5=6\). 2. With \(15\) observations, the median is the \(8\)th observation. 3. The first interval contains positions \(1\) through \(4\). 4. The second interval contains positions \(5\) through \(10\), so the \(8\)th observation lies in the second interval.

Answer

Second-bar frequency: \(6\) Median interval: the second interval

Dot plots and histograms 12

The histogram shows \(24\) observations in four equal-width intervals. a) Give the frequency of each interval. b) Verify whether both statements are true: the first two intervals contain \(14\) observations in all, and the middle two intervals contain \(13\) observations in all. c) Identify the modal interval.

Hints

- Read each bar height from the common frequency scale. - For each claim, add only the bars named in that claim. - The modal interval is represented by the tallest bar.

Solution

1. Reading the four bar heights gives frequencies \(6,8,5,5\). 2. The first two intervals contain \(6+8=14\) observations, so the first statement is true. 3. The middle two intervals contain \(8+5=13\) observations, so the second statement is true. 4. The interval from \(10\) up to \(20\) has the greatest frequency, \(8\), so it is the modal interval.

Answer

a) \(6,8,5,5\) b) Both statements are true. c) The interval from \(10\) up to \(20\) is modal.

Dot plots and histograms 13

The histogram shows trail lengths. Find the total number of trails, the fraction from \(2\) to less than \(6\) miles, and the cumulative frequency below \(6\) miles.

Hints

- Add every interval frequency for the total. - Combine only the two intervals fully inside the stated range. - For a cumulative count, include every interval below the cutoff.

Solution

1. The total frequency is \(3+7+8+2=20\). 2. The intervals from \(2\) to less than \(6\) contain \(7+8=15\) trails, so the fraction is \(\frac{15}{20}=\frac{3}{4}\). 3. Lengths below \(6\) miles include the first three intervals, with cumulative frequency \(3+7+8=18\).

Answer

Total trails: \(20\) Fraction from \(2\) to less than \(6\) miles: \(\frac{3}{4}\) Cumulative frequency below \(6\) miles: \(18\)

Dot plots and histograms 14

The histogram shows times needed to complete a puzzle. a) How many puzzle times were recorded? b) Which interval has the greatest frequency, and how many more times does it contain than the first interval?

Hints

- Add the bar frequencies to find the number of recorded times. - Read the numerical boundaries below the tallest bar. - Compare that frequency with the first interval's frequency.

Solution

1. The four intervals contain \(3,6,9,6\) times, so the total is \(3+6+9+6=24\). 2. The \(20\)-to-less-than-\(30\)-minute interval is the largest at \(9\) times. 3. The first interval contains \(3\), so the difference is \(9-3=6\) times.

Answer

a) \(24\) times b) From \(20\) to less than \(30\) minutes; \(6\) more times than the first interval.

Dot plots and histograms 15

A data set should have frequency \(2\) at value \(1\), frequency \(5\) at value \(2\), frequency \(4\) at value \(3\), and frequency \(1\) at value \(4\). The dot plot shows Renzo's attempt. Renzo made exactly one frequency error. Identify the incorrect stack, describe how to correct it, and state how many dots the completed dot plot should contain.

Hints

- Compare each visible stack with the required frequency for that value. - One dot represents one observation. - After correcting the one wrong stack, check the total number of observations.

Solution

1. Compare the number of dots above each value with the required frequency. 2. The stacks at \(1\), \(3\), and \(4\) have frequencies \(2\), \(4\), and \(1\), so those stacks are correct. 3. There is only one dot above \(2\), but the required frequency is \(5\). Add four more dots to that stack. 4. The completed plot contains \(2+5+4+1=12\) dots.

Answer

The incorrect stack is above \(2\). It has \(1\) dot but should have \(5\), so add \(4\) dots. The completed dot plot should contain \(12\) dots.

Dot plots and histograms 16

The histogram shows how long students spent making paper sculptures. Then \(3\) more students are added to each time interval. Find the new frequencies and the new total. Does the tallest interval change? What happens to the difference between the tallest and shortest frequencies?

Hints

- Add the same amount to all three plotted counts. - Compare the first and last bars before and after the additions. - Find a new total from the updated frequencies.

Solution

1. The original counts are \(1,4,7\), totaling \(12\). 2. Adding \(3\) to each gives \(4,7,10\), totaling \(21\). 3. The last interval is still tallest. 4. The difference was \(7-1=6\) and becomes \(10-4=6\), so it does not change.

Answer

New frequencies: \(4,7,10\). New total: \(21\). The last interval remains tallest. The frequency difference stays \(6\).

Dot plots and histograms 17

The histogram shows four equal-width intervals. a) Give the total number of observations. b) Describe the numerical pattern in the four bar heights. c) Identify the modal interval and state how many more observations it contains than the first interval.

Hints

- Read each height from the frequency axis. - Add the four frequencies for the total. - Compare successive bar heights and then compare the tallest bar with the first.

Solution

1. The bar heights are \(3,4,5,6\). 2. The total number of observations is \(3+4+5+6=18\). 3. The frequencies are consecutive whole numbers increasing by \(1\). 4. The fourth interval, from \(30\) up to \(40\), is modal because its frequency is \(6\). 5. It contains \(6-3=3\) more observations than the first interval.

Answer

a) \(18\) b) The bar heights are consecutive whole numbers increasing by \(1\): \(3,4,5,6\). c) The interval from \(30\) up to \(40\) is modal and contains \(3\) more observations than the first interval.

Dot plots and histograms 18

Use the dot plot to find the median and range. Identify the gap, and explain how the median can be a value where no dot appears.

Hints

- Put the four plotted values in order. - Use both middle observations for an even-sized data set. - Separate a calculated center from an observed dot position.

Solution

1. The ordered data are \(0,1,3,4\). 2. The median is the average of the two middle values: \((1+3)\div2=2\). 3. The range is \(4-0=4\). 4. There is a gap at \(2\), because no observation has that value. 5. For an even number of observations, the median can be an average of two data values and does not have to be an observed value.

Answer

Median: \(2\) Range: \(4\) Gap: \(2\) The median lies in the gap because it is the average of \(1\) and \(3\).

Dot plots and histograms 19

Use the dot plot to find the total number of observations, mode, median, range, and number of observations at least \(22\).

Hints

- Add all stack heights before locating middle positions. - Use the tallest stack for the mode. - Combine all stacks at and above the stated cutoff.

Solution

1. The total frequency is \(2+1+4+3=10\). 2. The tallest stack is at \(22\), so the mode is \(22\). 3. For \(10\) observations, the median is the average of the \(5\)th and \(6\)th values. Both positions are in the stack at \(22\), so the median is \(22\). 4. The range is \(23-20=3\). 5. Values at least \(22\) occur \(4+3=7\) times.

Answer

Total: \(10\) Mode: \(22\) Median: \(22\) Range: \(3\) At least \(22\): \(7\)

Dot plots and histograms 20

The histogram shows four equal-width intervals. Kertu makes two claims about the histogram: - The last two intervals together contain more than half of all observations. - The interval from \(20\) up to \(30\) is the modal interval. Check both claims using the graph.

Hints

- Read all four frequencies from the graph before checking either claim. - Compare the combined frequency of the last two bars with half of the total frequency. - The modal interval corresponds to the greatest bar height.

Solution

1. Reading the graph gives bar frequencies \(4, 5, 8, 7\). 2. The histogram contains \(4+5+8+7=24\) observations in all, so half is \(12\). 3. The last two intervals contain \(8+7=15\) observations, which is more than \(12\), so the first claim is correct. 4. The interval from \(20\) up to \(30\) has frequency \(8\), the greatest of the four frequencies, so the second claim is also correct.

Answer

Both claims are correct. The last two intervals contain \(15\) of the \(24\) observations, and the interval from \(20\) up to \(30\) is modal.

Dot plots and histograms 21

The travel times, in minutes, for \(12\) students are \(4,7,10,12,15,19,20,21,27,29,30,34\). A histogram will use these equal-width intervals: \(0\) to less than \(10\), \(10\) to less than \(20\), \(20\) to less than \(30\), and \(30\) to less than \(40\). Give the height of each histogram bar. Which interval contains a time of exactly \(20\) minutes?

Hints

- Sort each value by the lower and upper boundary of its interval. - “Less than \(20\)” does not include \(20\). - Check that the four frequencies add to all \(12\) observations.

Solution

1. From \(0\) to less than \(10\): \(4,7\), so the frequency is \(2\). 2. From \(10\) to less than \(20\): \(10,12,15,19\), so the frequency is \(4\). 3. From \(20\) to less than \(30\): \(20,21,27,29\), so the frequency is \(4\). 4. From \(30\) to less than \(40\): \(30,34\), so the frequency is \(2\). 5. A value of exactly \(20\) belongs in the interval that starts at \(20\), because the preceding interval stops before \(20\).

Answer

Bar heights: \(2,4,4,2\) Exactly \(20\) minutes belongs in the interval \(20\) to less than \(30\).

Dot plots and histograms 22

A class measured the heights of sunflower seedlings. A student says, “Six seedlings were exactly \(7\,\text{cm}\) tall.” Does the histogram support this claim? Explain what we can tell from the graph.

Hints

- Look at the boundaries of the bar containing \(7\,\text{cm}\). - Could different sets of measurements produce the same histogram?

Solution

1. The bar from \(5\,\text{cm}\) to less than \(10\,\text{cm}\) has a frequency of \(6\). 2. Those \(6\) seedlings could have different heights within that interval. 3. The histogram does not show how many were exactly \(7\,\text{cm}\) tall.

Answer

We cannot tell whether the claim is true. The graph shows that \(6\) seedlings were at least \(5\,\text{cm}\) but less than \(10\,\text{cm}\) tall. It does not show their exact heights.

Dot plots and histograms 23

Eight paper airplanes flew these distances in feet: \(2,5,9,13,16,19,20,22\). Make a frequency table for the intervals \(0\) to less than \(10\), \(10\) to less than \(20\), and \(20\) to less than \(30\) feet.

Hints

- Sort each measurement into one interval only. - The endpoint at \(20\) belongs in the interval starting at \(20\). - Add your table counts to check the total.

Solution

1. The first interval contains \(2,5,9\), so its count is \(3\). 2. The second contains \(13,16,19\), so its count is \(3\). 3. The third contains \(20,22\), so its count is \(2\). The counts add to eight.

Answer

<table><tr><th>Distance (ft)</th><th>Frequency</th></tr><tr><td>0 to less than 10</td><td>3</td></tr><tr><td>10 to less than 20</td><td>3</td></tr><tr><td>20 to less than 30</td><td>2</td></tr></table>

Dot plots and histograms 24

The dot plot records points scored on different quiz attempts. Make a frequency table for scores \(1,2,3,4\), then say how many attempts there were.

Hints

- One dot is one quiz attempt. - Make one table row for each number on the horizontal scale. - Check that the frequencies add to the number of dots.

Solution

1. Count the dots at each score: \(1\) point occurs twice, \(2\) points once, \(3\) points three times, and \(4\) points twice. 2. The total number of attempts is \(2+1+3+2=8\).

Answer

<table><tr><th>Score</th><th>Frequency</th></tr><tr><td>1</td><td>2</td></tr><tr><td>2</td><td>1</td></tr><tr><td>3</td><td>3</td></tr><tr><td>4</td><td>2</td></tr><tr><td>Total</td><td>8</td></tr></table>

Dot plots and histograms 25

A class made two histograms from the same set of seedling heights. Choose one bar in graph a). Find the bars in graph b) that cover the same heights. Show how their counts add up to the count in graph a). How many seedlings are shown in each graph?

Hints

- Find the endpoints of one interval in graph a). - Which consecutive intervals in graph b) cover exactly that same range? - Compare their combined frequency with the one bar in graph a).

Solution

1. In graph a), the \(0\)-to-less-than-\(4\,\text{cm}\) interval contains \(5\) seedlings. In graph b), its two smaller intervals contain \(2+3=5\) seedlings. 2. Likewise, the \(4\)-to-less-than-\(8\,\text{cm}\) interval in a) contains \(3\) seedlings, and the matching intervals in b) contain \(1+2=3\). 3. Both graphs show \(5+3=8\) seedlings. Only the interval widths change.

Answer

For example, graph a)'s \(0\)-to-less-than-\(4\,\text{cm}\) bar has \(5\) seedlings. The matching bars in b) have \(2\) and \(3\), and \(2+3=5\). Both graphs show \(8\) seedlings.

Dot plots and histograms 26

The dot plot shows how many target rings each of several foam gliders passed through. a) How many gliders were tested? b) What is the mode of the data? c) One result shown at \(4\) rings was entered incorrectly. It should be \(6\) rings. After the correction, which values are tied for the mode?

Hints

- Each dot represents one tested glider. - Compare the heights of the stacks to identify the most frequent result. - For the correction, decrease one stack and increase another before comparing again.

Solution

1. Count the dots: \(1+2+4+1+2+1=11\) gliders were tested. 2. The value \(4\) has \(4\) dots, more than any other value, so the original mode is \(4\) rings. 3. Moving one dot from \(4\) to \(6\) changes both frequencies to \(3\). No other value has more than \(3\) dots, so \(4\) and \(6\) are tied for the mode.

Answer

a) \(11\) gliders b) \(4\) rings c) \(4\) rings and \(6\) rings

Dot plots and histograms 27

The dot plot shows how many minutes \(12\) readers worked before asking for their first hint. The values will be grouped into equal-width histogram intervals. Complete the frequency table. <table> <tr><th>Time interval in minutes</th><th>Frequency</th></tr> <tr><td>\(1\leq t<3\)</td><td>?</td></tr> <tr><td>\(3\leq t<5\)</td><td>?</td></tr> <tr><td>\(5\leq t<7\)</td><td>?</td></tr> <tr><td>\(7\leq t<9\)</td><td>?</td></tr> </table>

Hints

- Treat the left endpoint as included and the right endpoint as excluded. - Combine the dot stacks whose values fall in the same interval. - Use the stated total to check the completed table.

Solution

1. The values \(1\) and \(2\) occur \(1+2=3\) times, so \(1\leq t<3\) has frequency \(3\). 2. The values \(3\) and \(4\) occur \(1+3=4\) times, so \(3\leq t<5\) has frequency \(4\). 3. The values \(5\) and \(6\) occur \(2+1=3\) times, so \(5\leq t<7\) has frequency \(3\). 4. The value \(7\) occurs \(2\) times, so \(7\leq t<9\) has frequency \(2\). 5. The frequencies check because \(3+4+3+2=12\).

Answer

<table> <tr><th>Time interval in minutes</th><th>Frequency</th></tr> <tr><td>\(1\leq t<3\)</td><td>\(3\)</td></tr> <tr><td>\(3\leq t<5\)</td><td>\(4\)</td></tr> <tr><td>\(5\leq t<7\)</td><td>\(3\)</td></tr> <tr><td>\(7\leq t<9\)</td><td>\(2\)</td></tr> </table>

Dot plots and histograms 28

The same \(12\) data values are grouped into two different histograms: \(10, 10, 10, 11, 12, 13, 14, 14, 15, 16, 17, 17\). a) Find the frequencies for intervals \(10\leq x<12\), \(12\leq x<14\), \(14\leq x<16\), and \(16\leq x<18\). b) Find the frequencies for intervals \(10\leq x<14\) and \(14\leq x<18\). c) Explain why the two histograms can look different even though they represent the same data.

Hints

- Count every value in exactly one interval for each grouping. - Notice how each wider interval is made from two neighboring narrower intervals. - Compare what detail is preserved or hidden by each bin width.

Solution

1. For width-\(2\) intervals, the frequencies are \(4, 2, 3, 3\). 2. For width-\(4\) intervals, the first two smaller intervals combine to \(4+2=6\), and the last two combine to \(3+3=6\). 3. The wider intervals combine neighboring values, so they hide some of the variation visible with narrower intervals. 4. Both histograms still represent all \(12\) observations.

Answer

a) \(4, 2, 3, 3\) b) \(6, 6\) c) Different interval widths combine the same observations in different ways, changing the visible shape without changing the underlying data.

Dot plots and histograms 29

Ten shell thicknesses, in millimeters, are \(1.5, 1.8, 2.0, 2.1, 2.4, 2.5, 2.9, 3.0, 3.2, 3.7\). Complete the histogram frequencies for these intervals: \(1.5\leq x<2.0\), \(2.0\leq x<2.5\), \(2.5\leq x<3.0\), \(3.0\leq x<3.5\), and \(3.5\leq x<4.0\). Explain why a thickness of exactly \(2.5\,\text{mm}\) belongs in the third interval rather than the second.

Hints

- Read the inequality symbols at both ends of every interval. - Place each decimal measurement in exactly one group. - Add the completed frequencies to check that all observations were counted.

Solution

1. The interval \(1.5\leq x<2.0\) contains \(1.5, 1.8\), so its frequency is \(2\). 2. The interval \(2.0\leq x<2.5\) contains \(2.0, 2.1, 2.4\), so its frequency is \(3\). 3. The interval \(2.5\leq x<3.0\) contains \(2.5, 2.9\), so its frequency is \(2\). 4. The interval \(3.0\leq x<3.5\) contains \(3.0, 3.2\), so its frequency is \(2\). 5. The interval \(3.5\leq x<4.0\) contains \(3.7\), so its frequency is \(1\). 6. The second interval excludes its right endpoint, while the third includes its left endpoint, so \(2.5\) belongs only in the third interval.

Answer

The frequencies are \(2, 3, 2, 2, 1\). A value of \(2.5\,\text{mm}\) belongs in \(2.5\leq x<3.0\) because the left endpoint is included and the previous interval’s right endpoint is excluded.

Dot plots and histograms 30

Panel a) is a dot plot and panel b) is a histogram of the same eight signal-flash counts. a) How many trials had exactly \(6\) flashes? b) Can the histogram alone answer part a)? Explain. c) What information does the histogram give that agrees with the dot plot?

Hints

- Use individual dot stacks for exact-value frequencies. - Identify what information is lost when several values are grouped into one bar. - Compare the total number represented in the stated interval.

Solution

1. The dot stack at \(6\) has \(3\) dots, so \(3\) trials had exactly \(6\) flashes. 2. The histogram alone cannot give that exact count because it combines every value from \(4\) through \(8\) into one interval. 3. Both displays show that there are \(8\) observations with values in \(4\leq x<9\).

Answer

a) \(3\) trials b) No. The histogram groups all values from \(4\) through \(8\) together. c) Both displays show \(8\) total trials in the interval \(4\leq x<9\).

Dot plots and histograms 31

The dot plot shows scores on a short pattern-recognition task. Every score is increased by \(2\) points because two bonus points are added. a) List the new data values in order. b) Give the new mode and range. c) Describe how the dot plot changes.

Hints

- Apply the same change to every plotted value. - Track horizontal positions separately from stack heights. - Compare the distance between the new extremes with the old distance.

Solution

1. The original values are \(1, 1, 2, 4, 4, 4, 5\). 2. Adding \(2\) to each gives \(3, 3, 4, 6, 6, 6, 7\). 3. The new mode is \(6\), because it occurs \(3\) times. 4. The new range is \(7-3=4\), the same as the original range \(5-1=4\). 5. Every dot moves \(2\) units to the right, while the stack heights and horizontal spread stay unchanged.

Answer

a) \(3, 3, 4, 6, 6, 6, 7\) b) Mode \(6\); range \(4\) c) Every dot shifts \(2\) units right, with the same stack pattern.

Dot plots and histograms 32

A dot plot shows \(7\) observations. A second dot plot is made by duplicating every observation, so every original dot is copied once at the same value. a) How many observations are in the second plot? b) Find the original and new modes and ranges. c) Describe which features of the dot plot change and which stay the same.

Hints

- Track what happens to each individual frequency. - Separate vertical stack height from horizontal location. - Check whether the minimum and maximum move.

Solution

1. Duplicating all \(7\) observations gives \(14\) observations. 2. The original mode is \(5\), and doubling every frequency keeps \(5\) as the mode. 3. The original range is \(5-1=4\), and the minimum and maximum are unchanged, so the new range is also \(4\). 4. Every stack doubles in height, but the occupied x-values, mode, range, and overall horizontal pattern remain the same.

Answer

a) \(14\) observations b) Original and new mode: \(5\); original and new range: \(4\) c) Stack heights and total frequency double, while horizontal positions and relative shape stay the same.

Dot plots and histograms 33

The histogram shows package-sorting times. One time recorded as \(18\) seconds should have been \(28\) seconds. Give the corrected frequency for every interval. Which interval or intervals now have the greatest frequency?

Hints

- Find the bin for the incorrect time and the bin for the corrected time. - One observation moves; the total stays unchanged. - Compare the new bar counts rather than the original ones.

Solution

1. The incorrect value \(18\) lies in the interval from \(10\) to less than \(20\) seconds. Its count falls from \(7\) to \(6\). 2. The corrected value \(28\) lies in the interval from \(20\) to less than \(30\) seconds. Its count rises from \(5\) to \(6\). 3. The outer interval counts remain \(4\) and \(2\). 4. The two middle intervals now tie for greatest frequency at \(6\).

Answer

Corrected frequencies, left to right: \(4,6,6,2\). The intervals from \(10\) to less than \(20\) seconds and from \(20\) to less than \(30\) seconds tie.

Dot plots and histograms 34

The data are \(3, 4, 7, 8, 9, 12, 13, 13, 17, 19, 21, 22\). Rihards wants equal-width histogram intervals that do not overlap and that cover all the data. Choose the valid interval plan, then find its frequencies. A) \(0\) to less than \(5\), \(5\) to less than \(10\), \(10\) to less than \(15\), \(15\) to less than \(20\), \(20\) to less than \(25\) B) \(0\) through \(5\), \(5\) through \(10\), \(10\) through \(15\), \(15\) through \(20\), \(20\) through \(25\) C) \(0\) to less than \(5\), \(5\) to less than \(12\), \(12\) to less than \(20\), \(20\) to less than \(25\)

Hints

- Compare the numerical width of every proposed interval. - Check whether a boundary value could belong to two intervals. - Sort each data value into exactly one interval in the valid plan.

Solution

1. Plan A has equal width \(5\), has no overlap, and covers every value. 2. Plan B is ambiguous at shared endpoints such as \(5\), because each endpoint is included in two intervals. 3. Plan C uses unequal widths. 4. For Plan A, the frequencies are \(2\) in \(0\) to less than \(5\), \(3\) in \(5\) to less than \(10\), \(3\) in \(10\) to less than \(15\), \(2\) in \(15\) to less than \(20\), and \(2\) in \(20\) to less than \(25\).

Answer

Plan A is valid. Its frequencies are \(2, 3, 3, 2, 2\), in interval order.

Dot plots and histograms 35

One dot in the plot is moved from \(0\) to \(4\). Compare the original and changed distributions. State the range and mode of each. Was the original pattern symmetric, and is the changed pattern symmetric?

Hints

- Read the plotted stacks rather than making a new data list first. - Only two stacks change after the move. - Compare heights on opposite sides of the original center.

Solution

1. The original minimum is \(0\), the maximum is \(4\), and the range is \(4-0=4\). The mode is \(2\), with three dots. 2. The original stack heights are \(1,2,3,2,1\), so they mirror each other around \(2\). 3. Moving the dot leaves a minimum of \(1\), a maximum of \(4\), and a range of \(4-1=3\). The mode remains \(2\). 4. The new stack heights at \(1,2,3,4\) are \(2,3,2,2\), which are not symmetric.

Answer

Original: range \(4\), mode \(2\), symmetric. Changed: range \(3\), mode \(2\), not symmetric.

Dot plots and histograms 36

The whole-number data set is \(4, 5, 6, 7, 8, 9, 10, 11\). Plan A groups the data into \(4\) to less than \(8\) and \(8\) to less than \(12\). Plan B groups the data into \(3\) to less than \(7\), \(7\) to less than \(11\), and \(11\) to less than \(15\). Find the frequencies for both plans. Explain why the histogram from Plan A would have two equal tallest bars while the histogram from Plan B would have one tallest bar, even though the data are identical.

Hints

- Assign each value to exactly one interval in each plan. - Compare the interval widths and then compare their starting points. - Separate a change in grouping from a change in the data.

Solution

1. Histogram A has \(4\) values in \(4\) to less than \(8\) and \(4\) values in \(8\) to less than \(12\). 2. Histogram B has \(3\) values in \(3\) to less than \(7\), \(4\) values in \(7\) to less than \(11\), and \(1\) value in \(11\) to less than \(15\). 3. The intervals have the same width in both displays, but their starting boundaries differ. 4. Changing boundaries changes which values are grouped together, so the visible tallest bars can change without any change to the raw data.

Answer

Histogram A frequencies: \(4,4\). Histogram B frequencies: \(3,4,1\). Different starting boundaries group the same values differently.

Dot plots and histograms 37

New observations may be added to the histogram, but none may be removed. What is the smallest number of observations that must be added to make all three bars equally tall? Which intervals receive them?

Hints

- Which bar is already tallest? - Because observations cannot be removed, what must the final height be at least? - Find how many observations each shorter interval needs.

Solution

1. No bar can be lowered, so all bars must reach at least the current greatest frequency, \(5\). 2. The first and last intervals each have frequency \(4\), so each needs one more observation. 3. Two additions, one in each outer interval, make all frequencies \(5\). Fewer additions cannot do this.

Answer

Add \(2\) observations: one from \(0\) to less than \(10\), and one from \(20\) to less than \(30\). All frequencies become \(5\).

Dot plots and histograms 38

The histogram shows song-clip lengths, in seconds. Which interval contains the median? Can you find the exact median from the histogram? Explain.

Hints

- Find the middle positions for an even number of clips. - Count observations across the intervals from left to right. - Consider what a histogram hides about values inside each interval.

Solution

1. The frequencies sum to \(2+5+6+3=16\), so the median is halfway between the eighth and ninth ordered values. 2. The first two intervals contain \(2+5=7\) clips. Both middle observations therefore lie in the third interval, from \(10\) to less than \(15\) seconds. 3. The histogram shows only interval counts, not exact clip lengths, so the exact median is unknown.

Answer

The median is in the \(10\)-to-less-than-\(15\)-second interval. Its exact value cannot be determined.

Dot plots and histograms 39

A histogram of whole-number scores has frequency \(8\) in the interval \(0\) to less than \(5\), which is the tallest bar. Aisyah says, “The mode must be \(2\), the middle whole number in that interval.” Explain why Aisyah's claim is not justified. What can the histogram show about the most frequent interval, and what can it not show about an exact mode?

Hints

- Separate the frequency of an interval from the frequency of one value. - List the possible whole numbers hidden inside the bar. - Decide whether the bar reveals how its observations are divided among those values.

Solution

1. The histogram shows that the interval \(0\) to less than \(5\) contains more observations than any other interval. 2. It does not show how the \(8\) observations are distributed among \(0, 1, 2, 3, 4\). 3. The value \(2\) might occur often, once, or not at all. 4. Therefore, the display identifies the most frequent interval but does not necessarily identify an exact most frequent value.

Answer

The interval \(0\) to less than \(5\) is the most frequent interval, but the exact mode cannot be determined from the histogram.

Dot plots and histograms 40

The dot plot shows \(9\) observations. One more observation will be added so that the completed \(10\)-observation data set has mean \(2.3\). At which value must the new dot be placed? After the dot is added, what is the mode?

Hints

- Read each stack from the dot plot before doing any arithmetic. - Convert the target mean and final number of observations into a required total. - The difference between the required total and the plotted total is the new observation.

Solution

1. Reading the plot gives frequencies \(2\) at \(1\), \(3\) at \(2\), \(3\) at \(3\), and \(1\) at \(4\). 2. The current total of the data values is \(2\times1+3\times2+3\times3+1\times4=21\). 3. A mean of \(2.3\) for \(10\) observations requires a total of \(10\times2.3=23\). 4. The added observation must therefore be \(23-21=2\). 5. The stack at \(2\) then has \(4\) dots, so the mode is \(2\).

Answer

Place the new dot at \(2\). The completed data set has mode \(2\).

Dot plots and histograms 41

The measurements in the dot plot are rounded to the nearest whole number and replotted. Give the rounded data and compare the modes before and after rounding. Explain how rounding changes the stacks.

Hints

- Round each observation separately before counting frequencies. - Compare exact repeats in the original list with repeats in the rounded list. - Think about what happens when nearby values collapse onto one plotted position.

Solution

1. The rounded values are \(1,1,2,2,2\). 2. Before rounding, every value occurs once, so there is no single mode. 3. After rounding, \(2\) occurs \(3\) times and \(1\) occurs \(2\) times, so the mode is \(2\). 4. Rounding combines several distinct nearby measurements into the same horizontal positions, creating taller stacks.

Answer

Rounded data: \(1,1,2,2,2\). The original data have no single mode; the rounded data have mode \(2\).

Dot plots and histograms 42

The histogram shows a set of measurements. a) Which intervals contain the two middle observations? b) One observation is removed from the first interval. Which interval then contains the middle observation? c) Why can neither exact median be found from this graph?

Hints

- Find the middle positions from the total frequency. - Use running totals from the leftmost bar. - After removal, repeat with the new sample size.

Solution

1. The original total is \(3+4+5+2=14\), so the middle observations are the seventh and eighth. 2. The first two bins hold \(3+4=7\) observations. The seventh lies in the second interval, from \(5\) to less than \(10\); the eighth lies in the third, from \(10\) to less than \(15\). 3. Removing one from the first bin gives counts \(2,4,5,2\), total \(13\). The seventh observation is now in the third interval. 4. A histogram gives grouped counts, not exact values within the intervals.

Answer

a) The seventh observation is from \(5\) to less than \(10\), and the eighth from \(10\) to less than \(15\). b) From \(10\) to less than \(15\). c) Individual values within intervals are not shown.

Dot plots and histograms 43

The histogram shows puzzle times in minutes. Identify the intervals containing the observations needed to find \(Q_1\) and \(Q_3\). Explain why the exact quartiles cannot be found from this histogram.

Hints

- Find the sample size before locating the quartile positions. - Use running frequencies to locate each position. - What can you tell about the observations inside one bar, and what can you not tell?

Solution

1. There are \(5+7+6+2=20\) observations. 2. Using the median-of-halves method, \(Q_1\) is halfway between observations \(5\) and \(6\). The fifth is in the first interval and the sixth in the second. 3. \(Q_3\) is halfway between observations \(15\) and \(16\). Both are in the third interval because the first two hold \(12\) observations and the third reaches \(18\). 4. Exact values are unknown within each interval, so the quartile numbers cannot be calculated exactly.

Answer

\(Q_1\) uses one observation from \(0\) to less than \(10\) and one from \(10\) to less than \(20\). \(Q_3\) uses two observations from \(20\) to less than \(30\). The exact quartiles are unavailable.

Dot plots and histograms 44

The histogram shows lengths of ribbon pieces cut for an art project. How many pieces were measured? Which interval has the greatest frequency? Can you tell which exact length occurs most often? Explain.

Hints

- Read the frequency of each whole interval. - Look for any bar taller than all the others. - Does one bar tell you which individual numbers occurred repeatedly inside it?

Solution

1. All four intervals have \(6\) observations, so the total is \(4\times6=24\). 2. All four intervals tie for greatest frequency. There is no single modal interval. 3. The bars do not tell how often each exact value appears, so an exact-value mode might exist and cannot be found from this histogram alone.

Answer

There are \(24\) observations. All four intervals tie for greatest frequency. The exact-value mode of the raw data cannot be determined.

Dot plots and histograms 45

The dot plot shows a data distribution. Find the total number of observations, median, range, and mode. Describe the symmetry of the plot.

Hints

- Add all stack heights before locating the middle observation. - Find the range as the distance on the horizontal number line from the minimum to the maximum. - Compare frequencies at values the same distance to the left and right of zero.

Solution

1. The total frequency is \(1+2+1+3+1+2+1=11\). 2. The middle observation is the \(6\)th. The cumulative frequency reaches \(4\) by \(-1\) and \(7\) at \(0\), so the median is \(0\). 3. The minimum is \(-3\) and the maximum is \(3\). On the number line, the distance from \(-3\) to \(0\) is \(3\) units and from \(0\) to \(3\) is \(3\) units, so the range is \(6\). 4. The tallest stack has height \(3\) at \(0\), so the mode is \(0\). 5. Values the same distance to the left and right of \(0\) have equal frequencies, so the plot is symmetric around \(0\).

Answer

Total: \(11\) Median: \(0\) Range: \(6\) Mode: \(0\) The plot is symmetric around \(0\).

Dot plots and histograms 46

The two dot plots each contain \(10\) observations and have range \(3\). Compare where the observations are concentrated. Which plot has more observations at the two endpoints, and which has more at the two middle values?

Hints

- Combine the frequencies at the minimum and maximum for each plot. - Combine the frequencies at the two interior values. - Compare stack placement, not only total count and range.

Solution

1. Plot A has \(1+1=2\) endpoint observations and \(4+4=8\) middle observations. 2. Plot B has \(3+3=6\) endpoint observations and \(2+2=4\) middle observations. 3. Plot A is concentrated in the middle, while Plot B places more observations at the extremes. 4. Equal sample size and range do not make the frequency patterns identical.

Answer

Plot B has more endpoint observations: \(6\) versus \(2\). Plot A has more middle observations: \(8\) versus \(4\).

Dot plots and histograms 47

A histogram has a positive-height bar in every interval from \(0\) to less than \(5\) and from \(5\) to less than \(10\). Jón says, “There are no gaps in the exact data because no histogram bar is empty.” Explain why Jón's statement is not justified. Give a two-value data set that makes both bars positive but has many missing whole-number values.

Hints

- Separate an occupied interval from occupied exact positions inside it. - Use one value near each outer end of the two intervals. - List the exact whole numbers hidden between those observations.

Solution

1. A nonempty histogram bar shows only that at least one value lies somewhere in its interval. 2. It does not show whether every exact value in that interval occurs. 3. The data set \(0, 9\) gives one observation in each interval. 4. The exact whole-number values \(1\) through \(8\) are all absent, even though neither histogram bar is empty.

Answer

Jón's statement is false. For example, \(0, 9\) makes both bars positive while all whole numbers \(1\) through \(8\) are missing.

Dot plots and histograms 48

The histogram shows paper-folding times in three equal \(10\)-minute intervals. Each interval will be split into two equal \(5\)-minute intervals. There are \(3\) observations from \(0\) to less than \(5\) minutes, \(5\) from \(10\) to less than \(15\), and \(4\) from \(20\) to less than \(25\). Read the three original interval frequencies from the histogram. Then find the six new frequencies, the total number of observations, and the new modal interval.

Hints

- Read each original bar height from the frequency scale before splitting that interval. - Within each original interval, the two new frequencies must add to the old bar height. - Splitting intervals changes the grouping but not the total number of observations.

Solution

1. The histogram shows original frequencies \(8,12,7\). 2. From \(0\) to less than \(10\), the second new frequency is \(8-3=5\). 3. From \(10\) to less than \(20\), the second new frequency is \(12-5=7\). 4. From \(20\) to less than \(30\), the second new frequency is \(7-4=3\). 5. The six new frequencies are \(3,5,5,7,4,3\), which total \(27\). 6. The greatest new frequency is \(7\), so the new modal interval is \(15\) to less than \(20\) minutes.

Answer

Original frequencies: \(8,12,7\) New frequencies, from \(0\) to less than \(5\) through \(25\) to less than \(30\): \(3,5,5,7,4,3\) Total observations: \(27\) Modal interval: \(15\) to less than \(20\) minutes

Dot plots and histograms 49

The dot plot shows the recorded data. The value \(1\) was later found to be an error and was changed to \(5\). Compare the mean, range, and mode or modes before and after the correction. Which features change?

Hints

- Replace the incorrect observation before recomputing any summaries. - Compare the endpoint distance in the two dot plots. - Check whether the tallest stacks change after one dot moves.

Solution

1. Before the correction, the mean is \(\frac{1+2+2+3+3+4}{6}=\frac{15}{6}=2.5\), the range is \(4-1=3\), and the modes are \(2\) and \(3\). 2. After the correction, the data are \(2,2,3,3,4,5\). 3. The new mean is \(\frac{19}{6}\approx3.17\), the new range is \(5-2=3\), and the modes remain \(2\) and \(3\). 4. Only the mean changes; the range and modes stay the same.

Answer

Before: mean \(2.5\), range \(3\), modes \(2\) and \(3\) After: mean \(\frac{19}{6}\approx3.17\), range \(3\), modes \(2\) and \(3\) Only the mean changes.

Dot plots and histograms 50

A symmetric dot plot has possible values \(2,3,4,5,\) and \(6\), with \(9\) observations total. There are \(2\) dots at \(2\), \(1\) dot at \(3\), \(1\) dot at \(5\), and \(2\) dots at \(6\). All remaining dots are at \(4\). Find the number of dots at \(4\), the mode, and the median. Explain how the frequencies confirm symmetry.

Hints

- Subtract all known stack heights from the total number of dots. - Locate the middle position after completing the frequency pattern. - Compare stacks at values equally far from the center.

Solution

1. The known frequencies total \(2+1+1+2=6\), so the frequency at \(4\) is \(9-6=3\). 2. The largest frequency is \(3\), so the mode is \(4\). 3. With \(9\) observations, the median is the fifth value. The ordered data place the fifth value at \(4\). 4. Values equally far from \(4\) have matching frequencies: \(2\) and \(6\) each have \(2\) dots, while \(3\) and \(5\) each have \(1\) dot.

Answer

Dots at \(4\): \(3\) Mode: \(4\) Median: \(4\) Matching frequencies at equal distances from \(4\) confirm symmetry.

Dot plots and histograms 51

The waiting times, in minutes, are \(1,3,4,4,6,7,8,8,12,13\). Two equal-width histogram plans are proposed. Plan A: \(0\) to less than \(3\), \(3\) to less than \(6\), \(6\) to less than \(9\), \(9\) to less than \(12\), \(12\) to less than \(15\). Plan B: \(0\) to less than \(5\), \(5\) to less than \(10\), \(10\) to less than \(15\). Find the bar heights for both plans. Which plan reveals an empty interval that the other plan hides?

Hints

- Tally the same raw observations separately under each set of interval boundaries. - A boundary value belongs to the interval that begins with that value. - After finding both frequency lists, compare whether either plan contains a zero-frequency bin.

Solution

1. For Plan A, the frequencies are \(1,3,4,0,2\). The interval from \(9\) to less than \(12\) is empty. 2. For Plan B, the frequencies are \(4,4,2\). None of its intervals is empty. 3. Plan A therefore reveals the empty interval, while the wider bins in Plan B group values on either side of that gap into occupied bars.

Answer

Plan A bar heights: \(1,3,4,0,2\) Plan B bar heights: \(4,4,2\) Plan A reveals the empty interval \(9\) to less than \(12\) minutes.

Dot plots and histograms 52

The histogram shows \(9\) observations. Aoife claims that the exact mean can be found from the histogram. Show that Aoife's claim is false by giving two different data sets that match the histogram but have different means.

Hints

- Keep the required number of values inside each interval. - Move values within their intervals without changing the bar frequencies. - Compare a data set using low values in each interval with one using high values.

Solution

1. One matching data set is \(1, 1, 1, 1, 1, 11, 11, 11, 11\). Its mean is \(49\div9=\frac{49}{9}\). 2. Another matching data set is \(9, 9, 9, 9, 9, 19, 19, 19, 19\). Its mean is \(121\div9=\frac{121}{9}\). 3. Both data sets place \(5\) values in the first interval and \(4\) in the second, but their means differ. 4. The histogram does not show the exact positions of values inside each interval, so the exact mean is not determined.

Answer

One valid pair is \(1, 1, 1, 1, 1, 11, 11, 11, 11\) and \(9, 9, 9, 9, 9, 19, 19, 19, 19\). Their means are \(\frac{49}{9}\) and \(\frac{121}{9}\), so the exact mean cannot be found from the histogram alone.

Dot plots and histograms 53

A dot plot should show \(8\) observations whose values have a total of \(31\). One dot was accidentally omitted from the displayed plot. a) At what value should the missing dot be placed? b) After adding it, what is the mode or set of modes?

Hints

- Read the visible dots as a list and find their total. - Compare the displayed total with the required total. - Update the stack frequencies after restoring the missing observation.

Solution

1. The displayed values are \(2, 2, 3, 4, 4, 5, 6\), with total \(26\). 2. The missing value is \(31-26=5\). 3. After adding a dot at \(5\), the values \(2\), \(4\), and \(5\) each occur twice. 4. Therefore, all three values are modes.

Answer

a) Place the missing dot at \(5\). b) The modes are \(2\), \(4\), and \(5\).

Dot plots and histograms 54

Whole-number measurements are grouped into the intervals \(10\) to less than \(15\), \(15\) to less than \(20\), \(20\) to less than \(25\), and \(25\) to less than \(30\). The first and last intervals each contain at least one observation. What is the smallest possible range of the data? What is the largest possible range? Explain why the exact range cannot be determined from the grouped data.

Hints

- Use the fact that the measurements are whole numbers. - For the smallest range, move the extreme observations toward each other within their intervals. - For the largest range, move them toward the outer ends of the displayed intervals.

Solution

1. For the smallest range, make the minimum as large as possible in the first interval, \(14\), and the maximum as small as possible in the last interval, \(25\). The range is \(25-14=11\). 2. For the largest range, make the minimum \(10\) and the maximum \(29\). The range is \(29-10=19\). 3. The histogram gives interval membership but not the exact values within the first and last intervals, so several ranges are possible.

Answer

Smallest possible range: \(11\) Largest possible range: \(19\) The exact range is not determined because the exact minimum and maximum are hidden inside their intervals.

Dot plots and histograms 55

What is the smallest possible number of observations that are at least \(25\)? What is the largest possible number? Explain the uncertainty.

Hints

- Identify intervals entirely above the cutoff. - Identify the interval split by the cutoff. - Consider both extreme ways the observations in that split interval could be placed.

Solution

1. The \(2\) observations in \(30\) to less than \(40\) are definitely at least \(25\). 2. The \(5\) observations in \(20\) to less than \(30\) could all be below \(25\), so the smallest possible count is \(2\). 3. Those \(5\) observations could all be from \(25\) to less than \(30\), so the largest possible count is \(2+5=7\). 4. The histogram does not show where values fall inside the \(20\)-to-less-than-\(30\) interval.

Answer

Smallest possible count: \(2\) Largest possible count: \(7\)

Dot plots and histograms 56

Create one dot-plot data set of \(12\) whole numbers that meets all these conditions: - The minimum is \(2\) and the maximum is \(7\). - Exactly \(3\) observations are below \(4\). - The unique mode is \(4\), with frequency \(4\). Give the stack height at every value from \(2\) through \(7\).

Hints

- Reserve the required number of observations below the mode first. - Make the modal stack taller than every other stack. - Use both endpoint values and check the total frequency last.

Solution

1. One valid choice below \(4\) is one observation at \(2\) and two at \(3\), giving exactly \(3\) observations. 2. Place four observations at \(4\) to create the required mode. 3. Five observations remain. One valid distribution is two at \(5\), two at \(6\), and one at \(7\). 4. The frequencies \(1,2,4,2,2,1\) total \(12\), use both required extremes, and leave \(4\) as the only stack of height \(4\).

Answer

One possible set of stack heights is \(1\) at \(2\), \(2\) at \(3\), \(4\) at \(4\), \(2\) at \(5\), \(2\) at \(6\), and \(1\) at \(7\).

Dot plots and histograms 57

In the dot plot, two dots are moved to \(4\). Find every pair of original values whose dots can be moved so that the new plot has mean \(4\), range \(6\), and unique mode \(4\) with frequency \(3\).

Hints

- Use the unchanged mean to determine the required sum of the moved values. - List symmetric pairs around the target value. - Eliminate any pair whose removal changes an endpoint of the range.

Solution

1. The original mean is \(4\), so replacing two values \(a,b\) by \(4,4\) preserves the mean only if \(a+b=8\). 2. Possible distinct pairs from the plot are \((1,7)\), \((2,6)\), and \((3,5)\). 3. Moving \(1\) and \(7\) removes both extremes, reducing the range, so that pair fails. 4. Moving \(2\) and \(6\) leaves extremes \(1\) and \(7\), giving range \(6\), and creates three dots at \(4\). 5. Moving \(3\) and \(5\) also leaves the extremes and creates the required unique mode.

Answer

The valid pairs are \(2\) and \(6\), or \(3\) and \(5\).

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Dot plots and histograms

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