Free math worksheets

- Use the tallest stack to identify the most frequent score. - Use a separate count of every dot to determine the sample size.

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

The mode is \(17\), and there are \(13\) observations.

- 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.

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

\(32,35,38,40,41,41,46,49,52,54\).

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

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

Both leaves represent \(12.4\); the repeated leaves show two observations at that value.

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

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

Bin width: \(20\). Greatest-frequency interval: \([140,160)\).

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

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.

The unknown row must contain \(1\) leaf.

- Translate each stack into repeated observations. - Proceed from left to right so the resulting list remains in increasing order.

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

\(1,1,2,3,3,3,4,5,5\).

- 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.

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

\(17\) customers.

- 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.

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

\(\frac{1}{3}\), or approximately \(0.333\).

- Count dots satisfying each condition, then use the total for the relative frequency. - Use separate counts for the condition \(x>5\), the exact value \(x=4\), and the total number of dots.

1. Values greater than \(5\) are \(6,8,8,9\), so the frequency is \(4\). 2. There are \(11\) observations and three equal \(4\). 3. The relative frequency of \(4\) is \(\frac{3}{11}\approx 0.273\).

Frequency greater than \(5\): \(4\). Relative frequency of \(4\): \(\frac{3}{11}\approx 0.273\).

- 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.

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

\(70\) observations.

- 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.

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 \(2\mid 1\ 3\ 3\ 7\).

The second \(3\) leaf is missing. The corrected row is \(2\mid 1\ 3\ 3\ 7\).

- Look for an x-axis value with no plotted dot between groups of observations. - Describe both clusters by their x-values after locating the empty position between them.

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 \(10\).

The gap is at \(9\); it separates the observations at \(5\) through \(8\) from those at \(10\).

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

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.

A dotplot, because it preserves individual values and displays repetitions directly.

- 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.

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.

A histogram, because it summarizes a large continuous data set using intervals.

- 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.

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

The \([20,30)\) bin increases to frequency \(11\).

- 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.

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.

The claim is not justified because the histogram groups all values in \([10,15)\); it does not preserve exact values.

- 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.

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 \(0.30+0.40=0.70\).

The heights sum to \(1\), and the relative frequency in \([1,3)\) is \(0.70\).

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

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.

Minimum: somewhere in \([10,20)\). Maximum: somewhere in \([40,50)\). The exact values cannot be determined.

- 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.

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.

For example, the stem-and-leaf plot can give the exact maximum value, while the histogram generally cannot.

- 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.

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.

The correct centers are \(5\), \(15\), and \(25\). The second bar should be moved from \(20\) to \(15\).

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

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.

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.

- 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.

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.

The new frequencies are \(8,16,10\). The graph no longer shows the within-pair differences between the original narrower bins.

- 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.

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 \(3.1-1.2=1.9\).

The sample size is \(8\), the minimum is \(1.2\), the maximum is \(3.1\), and the range is \(1.9\).

- 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.

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 \(5\mid7=57\).

\(4\mid2\;4\) \(4\mid5\;7\;8\) \(5\mid1\;3\) \(5\mid5\;7\;9\) Key: \(5\mid7=57\).

- 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.

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.

The entire dotplot shifts \(3\) units right. Its center increases by \(3\), while its shape and spread remain unchanged.

- 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.

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: \(\frac{38+41}{2}=39.5\).

Corrected display: \(3\mid2\;5\;8\) \(4\mid1\;3\;9\) The median is \(39.5\).

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

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 \(4+6+5=15\).

The gap is \([10,20)\), and the sample size is \(15\).

- Respect the included and excluded endpoints of each interval. - Supply exactly the number of observations required in each bin. - Remember that a histogram does not fix locations within an interval.

1. Choose any two values in \([0,10)\), such as \(1\) and \(9\). 2. Choose any three values in \([10,20)\), such as \(10\), \(15\), and \(19\). 3. The data set \(1,9,10,15,19\) produces the required frequencies. 4. Many other values within the same intervals would produce the same histogram because exact positions inside bins are not recorded.

One possible data set is \(1,9,10,15,19\). The answer is not unique because the histogram specifies only interval membership.

- 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.

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 \(1\mid2=12\).

The new values are \(12,15,20,23\). The same rows may be used if the key is changed to \(1\mid2=12\).

- Identify which coordinate represents the data value. - Decide whether vertical position carries numerical meaning in a standard dotplot. - Preserve the recorded value while making multiplicity visible.

1. Horizontal position in a dotplot encodes the numerical value, so dots at \(6.8\), \(7.0\), and \(7.2\) represent three different values. 2. The recorded observations are all \(7\), so they must share the same horizontal position. 3. The correct display stacks the three dots vertically above \(7\).

Horizontal spreading falsely changes the values. Stack all three dots vertically at \(7\).

- Apply the transformation to the interval endpoints as well as to the data. - A positive scale factor preserves order. - Track whether any observations are added or removed.

1. Multiplying every observation by \(2\) doubles each interval endpoint. 2. The new intervals are \([0,10)\), \([10,20)\), and \([20,30)\). 3. No observations change their corresponding interval membership under the scaled boundaries, so the frequencies remain \(4\), \(9\), and \(7\).

The new intervals are \([0,10)\), \([10,20)\), and \([20,30)\), with frequencies \(4,9,7\).

- 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.

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 \(29\).

<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 \(29\).

- 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.

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.

The missing proportion is \(0.05\), or \(5\%\). Extend the axis and include bins for those observations.

- Read the unit attached to the axis label. - Convert both endpoints before subtracting. - State the final range in the requested unit.

1. The axis unit means each plotted unit represents \(1000\) people. 2. The minimum population is \(1.2\cdot1000=1200\), and the maximum is \(1.8\cdot1000=1800\). 3. The range is \(1800-1200=600\) people.

The populations are \(1200\) and \(1800\), and the range is \(600\) people.

- 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.

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.

The histogram cannot show a steady rise because it discards the order of the \(30\) days. The chronological sequence is required.

- 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.

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 \(1\mid2=1.2\).

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.

- 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.

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.

The graph hides the gap from \(10\) to \(30\). Keep the full numerical axis and display empty bins for \([10,20)\) and \([20,30)\).

- 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.

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.

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.

- Identify the proposed center of symmetry. - Match points that are equally far from that center. - Find the missing partner for the unpaired point.

1. Symmetry about \(10\) pairs values whose average is \(10\). 2. The pairs \(8\) and \(12\), and \(9\) and \(11\), are already present; \(10\) is at the center. 3. The value paired with \(6\) must satisfy \(\frac{6+x}{2}=10\), giving \(x=14\).

Add the value \(14\).

- 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.

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.

The sample size is \(60\), and the interval frequencies are \(9,15,21,15\).

- 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.

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.

The dotplot is reflected across \(0\). Its center changes sign, its spread stays the same, and its shape is reversed left to right.

- A cumulative height includes all earlier intervals. - Subtract adjacent cumulative heights to isolate one interval. - Use the sample size only after finding the interval proportions.

1. The interval relative frequencies are consecutive differences: \(0.10\), \(0.35-0.10=0.25\), \(0.75-0.35=0.40\), and \(1.00-0.75=0.25\). 2. Multiplying by \(80\) gives frequencies \(8\), \(20\), \(32\), and \(20\). 3. The relative frequencies sum to \(1\), and the frequencies sum to \(80\).

The relative frequencies are \(0.10,0.25,0.40,0.25\), and the frequencies are \(8,20,32,20\).

- 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 \(10\).

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.

Use \([-0.5,4.5)\), \([4.5,9.5)\), and \([9.5,14.5)\).

- 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.

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.

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.

- Interpret the second coordinate of each plotted point. - Add class frequencies rather than x-coordinates. - Distinguish the graph’s connecting line from the underlying observations.

1. The class frequencies are \(4,7,5,2\), so the sample size is \(4+7+5+2=18\). 2. The x-coordinates mark class midpoints rather than 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.

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 \(20\).

- Use cumulative frequencies to locate the middle positions. - Identify which interval contains each middle observation. - Remember that a histogram does not record exact within-bin values.

1. The sample size is \(4+6+2=12\), so 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.

The median must lie in \([10,20)\), but its exact value cannot be found from the histogram.

- Compare the class widths before interpreting the bar areas. - With unequal bins, area rather than height should encode frequency. - Use \(\text{height}=\frac{\text{frequency}}{\text{class width}}\).

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.

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.

- 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.

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 \(-1\mid9\;2\).

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.

- 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.

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.

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.