Free math worksheets

- Look for an interval with no observations between occupied intervals. - Combine neighboring occupied bars on each side of that empty interval. - Check the cluster totals against the full number of observations.

1. The interval \(10\leq t<15\) has frequency \(0\), so it is a gap. 2. The first cluster occupies \(0\leq t<10\) and contains \(2+5=7\) observations. 3. The second cluster occupies \(15\leq t<25\) and contains \(4+1=5\) observations. 4. The cluster counts check because \(7+5=12\).

Gap: \(10\leq t<15\) First cluster: \(0\leq t<10\), with \(7\) observations Second cluster: \(15\leq t<25\), with \(5\) observations

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

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 \(\frac{7}{9}\).

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

- Add every bar height for the total frequency. - Combine only the three intervals named in the question. - Compare bars the same distance from the center.

1. The total number of observations is \(2+4+7+4+2=19\). 2. The middle three intervals contain \(4+7+4=15\) observations. 3. The requested fraction is \(\frac{15}{19}\). 4. The first and fifth bars match, and the second and fourth bars match, so the bar pattern is symmetric around the middle interval.

a) \(19\) b) \(\frac{15}{19}\) c) The bar pattern is symmetric around the middle interval.

- 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.

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.

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 \(7\): \(4\)

- Sort each candidate value into the stated intervals. - Pay close attention to which endpoint is excluded. - Compare the resulting three frequencies with the target bar heights.

1. Choice A has \(1,4\) in the first interval, \(5,8,9\) in the second, and \(12\) in the third, giving frequencies \(2,3,1\). 2. Choice B has a value \(15\), which is not in the \(10\)-to-less-than-\(15\) interval and would require another interval. 3. Choice C gives frequencies \(2,2,2\), not \(2,3,1\).

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

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

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 \(9-5=4\).

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

- 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.

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.

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

- 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.

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 \(3+7+8=18\).

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

- Add the ratio parts before using the total frequency. - Determine the number of observations represented by one part. - Scale every ratio entry by the same amount.

1. The ratio contains \(1+2+3+2=8\) equal parts. 2. Each part represents \(24\div8=3\) observations. 3. The interval frequencies are \(3,6,9,6\). 4. The third interval has the greatest frequency, \(9\), so it is the modal interval.

Frequencies: \(3,6,9,6\). The third interval is the modal interval.

- Identify what one dot represents in this kind of display. - Separate the horizontal data value from the vertical stacking used to avoid overlap. - Make the number of visible marks equal the frequency.

1. In a dot plot, each dot represents one observation and is placed above its data value on the horizontal axis. 2. A point at \((2,5)\) can look like one observation with two coordinates rather than five repeated observations. 3. The correct display has five separate dots vertically stacked above \(2\), at stack levels \(1\) through \(5\). 4. The complete plot has \(2+5+4+1=12\) dots.

Draw \(5\) separate dots stacked above the value \(2\), not one point at \((2,5)\). The full dot plot contains \(12\) dots.

- Apply the same addition to every bar height. - Recalculate the total from all updated frequencies. - Compare differences between bars rather than only their new heights.

1. Adding \(3\) to each frequency gives \(4,7,10\). 2. The original total is \(1+4+7=12\), and the new total is \(4+7+10=21\). 3. The third interval remains the modal interval. 4. Originally, the tallest-shortest difference is \(7-1=6\). 5. Afterward, the difference is \(10-4=6\), so adding the same amount to every bar preserves the frequency differences.

New frequencies: \(4,7,10\) New total: \(21\) The third interval remains modal, and the tallest-shortest difference remains \(6\).

- Convert each stated share into a number of dots. - Use the total to find the unassigned stack. - Compare the completed stack heights.

1. The stack at \(6\) has \(0.40\times15=6\) dots. 2. The stack at \(7\) has \(\frac{1}{3}\times15=5\) dots. 3. The remaining stack has \(15-6-5=4\) dots at \(8\). 4. The tallest stack is at \(6\), so the mode is \(6\).

Stack heights: \(6\) dots at value \(6\), \(5\) dots at value \(7\), and \(4\) dots at value \(8\). Mode: \(6\).

- 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.

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.

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

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

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.

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

- 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.

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.

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

- 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.

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 \(3+4+3+2=12\).

<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>

- 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.

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.

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.

- 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.

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.

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.

- 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.

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 \(4\leq x<9\).

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 \(4\leq x<9\).

- 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.

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.

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.

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

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.

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.

- Identify the interval containing the incorrect value before changing any frequency. - Move one observation rather than changing the total number of observations. - Compare the corrected bar heights to find the new greatest frequency.

1. The incorrect value \(18\) belongs to the \(10\)-to-less-than-\(20\) interval, so that frequency decreases from \(7\) to \(6\). 2. The corrected value \(28\) belongs to the \(20\)-to-less-than-\(30\) interval, so that frequency increases from \(5\) to \(6\). 3. The other interval frequencies remain \(4\) and \(2\). 4. Originally, only the \(10\)-to-less-than-\(20\) interval had the greatest frequency. After correction, the two middle intervals tie with frequency \(6\).

Corrected frequencies: \(4, 6, 6, 2\), in interval order. The greatest frequency changes from one interval to a tie between the \(10\)-to-less-than-\(20\) and \(20\)-to-less-than-\(30\) intervals.

- 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.

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 \(25\).

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

- Find the total represented by all bars. - Compare each interval’s count with that same total. - Think about whether dividing every bar height by one common number changes the shape.

1. The total frequency is \(3+5+8+4=20\). 2. The relative-frequency heights are \(\frac{3}{20}=0.15\), \(\frac{5}{20}=0.25\), \(\frac{8}{20}=0.40\), and \(\frac{4}{20}=0.20\). 3. The intervals and the relative pattern of bar heights stay the same. 4. The vertical scale changes from numbers of observations to parts of the total.

Relative-frequency heights: \(0.15, 0.25, 0.40, 0.20\). The histogram keeps the same intervals and shape, but its y-axis changes from frequency to relative frequency.

- Update only the two affected stacks. - Recheck the lowest occupied value after moving the dot. - Compare stack heights at equal distances on opposite sides of the center.

1. The original distribution has minimum \(0\), maximum \(4\), and range \(4\). Its only mode is \(2\), which occurs \(3\) times. 2. The original frequencies \(1,2,3,2,1\) are symmetric around \(2\). 3. After the move, the data are \(1,1,2,2,2,3,3,4,4\). 4. The changed range is \(4-1=3\), and the only mode remains \(2\). 5. The changed frequencies are not mirror images around \(2\), so the distribution is no longer symmetric.

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

- 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.

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.

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

- The tallest existing bar sets the smallest possible common height. - Find how far each shorter bar is below that height. - Remember that one observation changes only one interval’s frequency.

1. The common final height cannot be below the current greatest frequency, \(5\). 2. Each outer bar has frequency \(4\), so each needs one new observation to reach \(5\). 3. Adding only one observation would raise only one outer bar, leaving the other at \(4\). 4. Therefore, two observations are required: one in the first interval and one in the third interval.

Add \(2\) observations: one in the first interval and one in the third interval. The frequencies become \(5,5,5\).

- Find the total number of observations from all bar heights. - Locate the middle position or positions in the ordered data. - Use cumulative frequencies to see which interval contains those positions.

1. The histogram represents \(2+5+6+3=16\) observations. 2. The median lies between the \(8\)th and \(9\)th observations in order. 3. The cumulative frequency after the first interval is \(2\), and after the second it is \(7\). 4. The third interval contains positions \(8\) through \(13\), so both middle observations lie there. 5. The exact values of those observations are hidden within the interval, so the exact median cannot be determined.

The median is in the third interval. Its exact value cannot be determined from the histogram.

- 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.

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.

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

- Convert the stated percentage of the total into a count. - Use all bar frequencies to check the total. - Combine the requested bars before comparing with the full data set.

1. The missing frequency is \(35\%\) of \(40\): \(0.35\times40=14\). 2. The known frequencies total \(6+9+7+4=26\), and \(26+14=40\), confirming the result. 3. The outermost intervals have frequencies \(6\) and \(4\), for a combined frequency of \(10\). 4. Their combined relative frequency is \(\frac{10}{40}=0.25=25\%\).

Missing frequency: \(14\) Relative frequency of the outermost intervals combined: \(0.25=25\%\)

- Subtract the known stack totals from the total number of dots. - Use the completed stack heights to identify the most frequent value. - Treat each stack height as the number of times its horizontal value contributes to the total.

1. The known stacks contain \(2+3+1=6\) dots, so the missing stack has \(10-6=4\) dots. 2. The tallest stack is at \(2\), so the mode is \(2\). 3. The total of the data values is \(2\times1+4\times2+3\times3+1\times4=23\). 4. The mean is \(23\div10=2.3\).

Missing stack height: \(4\) Mode: \(2\) Mean: \(2.3\)

- 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.

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.

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

- Find the middle position or positions from the total frequency. - Build cumulative totals from left to right. - Distinguish locating an interval from knowing a precise value inside it.

1. Originally there are \(3+4+5+2=14\) observations, so the middle positions are the \(7\)th and \(8\)th. 2. The first two intervals contain positions \(1\) through \(7\), so the \(7\)th observation is in the second interval. The third interval contains positions \(8\) through \(12\), so the \(8\)th is in the third interval. 3. After removing one observation from the first interval, the frequencies are \(2,4,5,2\), totaling \(13\). 4. The middle position is the \(7\)th. The first two intervals contain \(6\) observations, so the \(7\)th lies in the third interval. 5. A histogram hides exact values within each interval, so it does not give the exact median.

a) The \(7\)th observation is in the second interval, and the \(8\)th is in the third. b) The new middle observation is in the third interval. c) Exact values within intervals are not shown.

- Use the total frequency to identify quartile positions. - Build cumulative positions for each interval. - Distinguish knowing an observation’s interval from knowing its exact value.

1. There are \(5+7+6+2=20\) observations. 2. The lower half contains the first \(10\) observations, so \(Q_1\) is the average of the \(5\)th and \(6\)th observations. 3. The \(5\)th observation is in the first interval, and the \(6\)th is in the second interval. 4. The upper half contains positions \(11\) through \(20\), so \(Q_3\) is the average of the \(15\)th and \(16\)th observations. 5. The third interval contains positions \(13\) through \(18\), so both observations used for \(Q_3\) are in the third interval. 6. Exact positions inside the intervals are not shown, so exact quartile values are unavailable.

\(Q_1\) uses one observation from the first interval and one from the second. \(Q_3\) uses two observations from the third interval. The exact quartile values cannot be determined.

- Compare bar heights at the interval level. - Add all interval frequencies for the total. - Separate interval frequency from the hidden frequencies of exact values.

1. All four intervals tie for greatest frequency, so there is no single modal interval. 2. The total number of observations is \(4\times6=24\). 3. The histogram does not show frequencies of individual values within each interval. 4. One exact value could still occur more often than every other exact value, so the raw data may have a mode.

All four intervals tie as modal intervals, and the histogram contains \(24\) observations. The exact-value mode cannot be determined.

- Add all stack heights before locating the middle observation. - Treat subtraction of a negative minimum carefully. - Compare frequencies at opposite values the same distance from zero.

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 range is \(3-(-3)=6\). 4. The tallest stack has height \(3\) at \(0\), so the mode is \(0\). 5. Matching values \(-1\) and \(1\), \(-2\) and \(2\), and \(-3\) and \(3\) have equal frequencies, so the plot is symmetric around \(0\).

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

- Multiply each bar height by the proposed total. - Check that each set of counts adds to its sample size. - Consider what information is removed when counts are divided by one common total.

1. For \(10\) observations, the counts are \(0.20\times10=2\), \(0.50\times10=5\), and \(0.30\times10=3\). 2. For \(20\) observations, the counts are \(4,10,6\). 3. Both count sets have the same proportions, so they produce the same relative-frequency bar heights. 4. A relative-frequency scale shows shares of a total but not the total itself.

For \(10\) observations: \(2,5,3\). For \(20\) observations: \(4,10,6\). The sample size cannot be determined from relative frequencies alone.

- Use each combined-frequency statement in order. - Subtract known parts from their stated group totals. - Check that all four frequencies add to the full sample size.

1. The second frequency is \(14-6=8\). 2. The middle two frequencies total \(13\), so the third frequency is \(13-8=5\). 3. The first three frequencies total \(6+8+5=19\), so the fourth frequency is \(24-19=5\). 4. The frequencies are \(6,8,5,5\), and the second interval is the unique modal interval.

Frequencies: \(6,8,5,5\). The second interval is the modal interval.

- Compare each middle bar with its own histogram total. - Do not rank concentration by raw bar height when sample sizes differ. - Express both shares in the same form.

1. Histogram A’s middle relative frequency is \(\frac{10}{20}=0.50=50\%\). 2. Histogram B’s middle relative frequency is \(\frac{16}{40}=0.40=40\%\). 3. Although B has more observations in the middle interval, A has the greater proportion there.

Histogram A has the greater middle-interval concentration. A: \(50\%\) B: \(40\%\)

- 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.

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.

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

- 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.

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.

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

- Represent the four consecutive heights from one starting value. - Add the four expressions and set the sum equal to the total. - Check that the final heights increase by exactly one.

1. Let the first frequency be \(n\). The frequencies are \(n,n+1,n+2,n+3\). 2. Their total is \(4n+6=18\). 3. Thus \(4n=12\), so \(n=3\). 4. The frequencies are \(3,4,5,6\), and the fourth interval is modal.

Bar heights: \(3,4,5,6\). The fourth interval is the modal interval.

- The two smaller intervals must contain all observations from the original interval. - Check that splitting a bar does not change the total number of data values. - Compare the completed bar heights to identify the modal interval.

1. The original middle frequency \(12\) is divided between the two new intervals. 2. The frequency from \(15\) to less than \(20\) minutes is \(12-5=7\). 3. The four new frequencies are \(8,5,7,7\). 4. Their total is \(8+5+7+7=27\). 5. The greatest frequency is \(8\), so the new modal interval is from \(0\) to less than \(10\) minutes.

New frequencies: \(8,5,7,7\) Total observations: \(27\) Modal interval: from \(0\) to less than \(10\) minutes

- 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.

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.

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.

- Use the number of bars and their average height to find the total frequency. - Subtract the known bar heights from the total. - Compare the completed frequency with every other bar height.

1. An average frequency of \(6\) across four bars gives a total frequency of \(4\times6=24\). 2. The three known frequencies total \(4+5+7=16\). 3. The missing frequency is \(24-16=8\). 4. Since \(8\) is greater than \(4,5,\) and \(7\), the missing bar is the unique modal interval.

Missing frequency: \(8\) Yes, the missing bar is the unique modal interval.

- 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.

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.

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

- 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.

1. One matching data set is \(0, 0, 0, 0, 0, 10, 10, 10, 10\). Its mean is \(40\div9=\frac{40}{9}\). 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.

One valid pair is \(0, 0, 0, 0, 0, 10, 10, 10, 10\) and \(9, 9, 9, 9, 9, 19, 19, 19, 19\). Their means are \(\frac{40}{9}\) and \(\frac{121}{9}\), so the exact mean cannot be found from the histogram alone.

- 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.

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.

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

- 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.

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.

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.

- 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.

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.

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

- 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.

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 \(4\).

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 \(7\).

- 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.

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.

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