Free math worksheets

- Find only the least and greatest values. - The order of the other values does not affect the range. - Include the unit in your answer.

1. The least value is \(1.8\,\text{m}\). 2. The greatest value is \(3.0\,\text{m}\). 3. The range is \(3.0 - 1.8 = 1.2\,\text{m}\).

The range is \(1.2\,\text{m}\).

- Relate the minimum, maximum, and range. - Work backward from the given difference. - Check your result by subtraction.

1. The range equals the maximum minus the minimum. 2. Therefore, the maximum is \(7.5 + 12.8 = 20.3\).

The maximum is \(20.3\).

- Use the relationship between minimum, maximum, and range. - Ask what values can lie between two equal endpoints. - Consider the distance from a common value to the mean.

1. A range of \(0\) means the maximum and minimum are equal. 2. Every data value lies between that same minimum and maximum, so all values are identical. 3. The mean equals the common data value. 4. Every absolute distance from the mean is \(0\), so the mean absolute deviation is \(0\).

All values are equal. The mean is that common value, and the mean absolute deviation is \(0\).

- Identify the least and greatest measurements. - The range measures the distance from one extreme to the other. - Include centimeters in the interpretation.

1. The shortest seedling is \(11\,\text{cm}\), and the tallest is \(17\,\text{cm}\). 2. The range is \(17-11=6\,\text{cm}\). 3. The tallest seedling is \(6\,\text{cm}\) taller than the shortest seedling.

Range: \(6\,\text{cm}\). The tallest seedling is \(6\,\text{cm}\) taller than the shortest.

- Identify the greatest value in each version of the data set. - The least value does not change. - Compare the two ranges.

1. With the outlier, the range is \(42 - 11 = 31\). 2. Without the outlier, the greatest value is \(18\). 3. The range without the outlier is \(18 - 11 = 7\).

The range is \(31\) with the outlier and \(7\) without the outlier.

- List the individual distances before interpreting their average. - Distinguish a statement about an average from a statement about every observation. - Check the student’s claim against the actual distances.

1. The absolute distances from \(5\) are \(3, 1, 1, 5\). 2. Their mean is \((3+1+1+5)\div4=2.5\). 3. The individual distances are not all \(2.5\). 4. Mean absolute deviation describes the average distance of the data values from the mean, not the distance of every value.

The conclusion is false. The distances are \(3, 1, 1, 5\), whose average is \(2.5\). Mean absolute deviation is an average distance, not a common distance shared by all values.

- Use each frequency as the number of times its distance occurs. - Combine all distance contributions before averaging. - Use the total number of original observations as the divisor.

1. The total absolute distance is \(2\times0+4\times1+3\times3+1\times5=18\). 2. There are \(10\) observations. 3. The mean absolute deviation is \(18\div10=1.8\).

Total absolute distance: \(18\) Mean absolute deviation: \(1.8\)

- Relate an average to its total and number of items. - Treat the unknown number of observations as the divisor. - Check by dividing the given total distance by your result.

1. Mean absolute deviation equals total absolute distance divided by the number of observations. 2. Let \(n\) be the number of observations. Then \(21\div n=1.5\). 3. Solving gives \(n=21\div1.5=14\).

There are \(14\) observations.

- Turn each total distance into an average using its own observation count. - Compare averages rather than totals when group sizes differ. - Interpret the result as distance per observation.

1. Data Set A has mean absolute deviation \(12\div6=2\). 2. Data Set B has mean absolute deviation \(12\div8=1.5\). 3. Data Set A has the greater mean absolute deviation. 4. The same total distance is spread across fewer observations in A, producing a greater average distance. 5. A total alone does not account for different data-set sizes.

Data Set A: MAD \(2\) Data Set B: MAD \(1.5\) Data Set A is more variable by MAD.

- With two values, the mean is their midpoint. - Split the range equally around that midpoint. - Average the two equal distances from the center.

1. For two observations, the mean lies halfway between them. 2. A range of \(6\) places the observations \(3\) units below and above the mean \(4\). 3. The observations are \(4-3=1\) and \(4+3=7\). 4. Both observations are \(3\) units from the mean, so the mean absolute deviation is \((3+3)\div2=3\).

The observations are \(1\) and \(7\), and the mean absolute deviation is \(3\).

- Read one data value for each dot, including repeated values in a stack. - Find the mean before measuring distances from it. - Average the nonnegative distances, not signed differences.

1. The data values are \(3,5,5,7\). 2. The mean is \((3+5+5+7)\div4=20\div4=5\). 3. The distances from the mean are \(2,0,0,2\). 4. The mean absolute deviation is \((2+0+0+2)\div4=1\). 5. A data value is an average distance of \(1\) unit from the mean of \(5\).

Mean: \(5\) MAD: \(1\) The dots are an average distance of \(1\) unit from the mean.

- Use horizontal distance from each dot to the dashed mean line. - Distances are nonnegative even for dots to the left of the mean. - Average the five distances to obtain the MAD.

1. The plotted values are \(1,3,4,5,7\), and the marked mean is \(4\). 2. Their absolute distances from \(4\) are \(3,1,0,1,3\). 3. The MAD is \((3+1+0+1+3)\div5=8\div5=1.6\). 4. A plotted value is an average distance of \(1.6\) units from the mean of \(4\).

Absolute distances: \(3,1,0,1,3\) MAD: \(1.6\) The values are an average distance of \(1.6\) units from the mean.

- Find the original maximum and minimum. - Double the original range. - The colder Sunday would become the new minimum.

1. The greatest temperature is \(23\,^\circ\text{C}\), and the least is \(18\,^\circ\text{C}\). 2. The original range is \(23 - 18 = 5\,^\circ\text{C}\). 3. Twice the original range is \(2 \times 5 = 10\,^\circ\text{C}\). 4. A colder Sunday would become the new minimum while \(23\,^\circ\text{C}\) remains the maximum. Let the new Sunday temperature be \(T\). Then \(23 - T = 10\), so \(T = 13\). 5. Sunday would need to be \(13\,^\circ\text{C}\).

a) \(5\,^\circ\text{C}\) b) \(13\,^\circ\text{C}\)

- Compare the distances between the values, not only their centers. - Find a simple measure of spread for each data set. - Equal sums can be distributed very differently.

1. Each data set has a sum of \(40\), so each mean is \(40 \div 4 = 10\). 2. Data set A has a range of \(11 - 9 = 2\). 3. Data set B has a range of \(20 - 0 = 20\). 4. Data set A is tightly clustered around the mean, while data set B has values much farther from the mean. Therefore, the equal means are not enough to describe the data sets as similar.

No. Data set A has a range of \(2\), while data set B has a range of \(20\). Their spreads are very different despite the equal means.

- Work backward from an average to the total of the quantities being averaged. - Separate the known distances from the missing one. - Use the stated direction from the mean only after finding the distance.

1. A mean absolute deviation of \(2\,\text{cm}\) for \(5\) values requires a total absolute distance of \(5\times2=10\,\text{cm}\). 2. The known absolute distances total \(1+2+0+4=7\,\text{cm}\), so the unknown distance is \(10-7=3\,\text{cm}\). 3. Because the corresponding landing distance is above the mean, it is \(8+3=11\,\text{cm}\).

The unknown absolute distance is \(3\,\text{cm}\), and the landing distance is \(11\,\text{cm}\).

- Analyze the full list and the shortened list separately. - Recalculate the mean after removing the observation. - Consider whether the removed value was an extreme and how far it sat from the main group.

1. The original range is \(8-2=6\,\text{mm}\). 2. The original mean is \((2+3+3+4+8)\div5=4\,\text{mm}\). The absolute distances are \(2, 1, 1, 0, 4\), so the mean absolute deviation is \(8\div5=1.6\,\text{mm}\). 3. After removing \(8\), the new range is \(4-2=2\,\text{mm}\). 4. The new mean is \((2+3+3+4)\div4=3\,\text{mm}\). The absolute distances are \(1, 0, 0, 1\), so the new mean absolute deviation is \(2\div4=0.5\,\text{mm}\). 5. The range drops by \(4\,\text{mm}\), while the mean absolute deviation drops by \(1.1\,\text{mm}\). The removed value was the maximum and was separated from the cluster.

a) Range \(6\,\text{mm}\); mean absolute deviation \(1.6\,\text{mm}\) b) Range \(2\,\text{mm}\); mean absolute deviation \(0.5\,\text{mm}\) c) The range changes more sharply because the removed value was an isolated maximum.

- Check what the word “absolute” requires before averaging the deviations. - Distances cannot be negative. - Find the full spread separately from the average distance from the mean.

1. Mean absolute deviation uses absolute distances, so negative signed differences cannot be averaged directly. 2. The absolute distances from \(17\) are \(4, 1, 1, 4\). 3. The mean absolute deviation is \((4+1+1+4)\div4=2.5\). 4. The range is \(21-13=8\).

The student used signed differences instead of absolute distances. The correct mean absolute deviation is \(2.5\), and the range is \(8\).

- Check the extreme-value condition before doing every distance calculation. - Find each set’s own mean before measuring distances. - A choice must satisfy both conditions, not just one.

1. Set A has mean \(5\), range \(8-2=6\), and absolute distances \(3, 1, 1, 3\), so its mean absolute deviation is \(2\). 2. Set B has mean \(5\), range \(8-2=6\), and absolute distances \(3, 3, 3, 3\), so its mean absolute deviation is \(3\). 3. Set C has mean \(5\), range \(9-1=8\), and absolute distances \(4, 1, 1, 4\), so its mean absolute deviation is \(2.5\). 4. Only Set B has both required measures.

B: \(2, 2, 8, 8\)

- Find each set’s center before comparing distances. - Compare the average distance with the distance between the extremes. - Look for which observations contribute large distances in each set.

1. Data set A has mean \(4\), range \(8-0=8\), and absolute distances \(4, 0, 0, 4\), so its mean absolute deviation is \(2\). 2. Data set B has mean \(4\), range \(6-2=4\), and absolute distances \(2, 2, 2, 2\), so its mean absolute deviation is \(2\). 3. The sets have the same mean and mean absolute deviation but different ranges. 4. Equal mean absolute deviations do not require equal extreme-to-extreme spread.

Data set A: mean \(4\), range \(8\), mean absolute deviation \(2\) Data set B: mean \(4\), range \(4\), mean absolute deviation \(2\) The same mean absolute deviation can occur with different ranges.

- Complete the statistics in the original unit first. - Use one consistent conversion for every data value and every result. - Compare the physical spread with its numerical representation in the new unit.

1. In centimeters, the mean is \((3.2+3.4+3.6+3.8)\div4=3.5\,\text{cm}\). 2. The range is \(3.8-3.2=0.6\,\text{cm}\). The absolute distances from \(3.5\) are \(0.3, 0.1, 0.1, 0.3\), so the mean absolute deviation is \(0.8\div4=0.2\,\text{cm}\). 3. In millimeters, the data are \(32, 34, 36, 38\), with mean \(35\,\text{mm}\), range \(6\,\text{mm}\), and mean absolute deviation \(2\,\text{mm}\). 4. Each numerical statistic is multiplied by \(10\) because \(1\,\text{cm}=10\,\text{mm}\), while the physical measurements are unchanged.

a) Mean \(3.5\,\text{cm}\), range \(0.6\,\text{cm}\), mean absolute deviation \(0.2\,\text{cm}\) b) Mean \(35\,\text{mm}\), range \(6\,\text{mm}\), mean absolute deviation \(2\,\text{mm}\) c) All numerical values are multiplied by \(10\).

- Use the mean and number of observations to recover the required total. - Complete the data set before finding its spread. - List each distance from the stated center without using signs.

1. A mean of \(5\) for \(4\) values requires a total of \(5\times4=20\). 2. The known values total \(2+4+8=14\), so \(x=20-14=6\). 3. The completed data set is \(2, 4, 6, 8\), with range \(8-2=6\). 4. The distances from the mean \(5\) are \(3, 1, 1, 3\), totaling \(8\). 5. The mean absolute deviation is \(8\div4=2\).

\(x=6\), range \(6\), mean absolute deviation \(2\).

- Find the original center before deciding which observations change. - Write the complete changed data set before recomputing anything. - Compare both the extreme separation and the average distance from the new center.

1. The original mean is \((4+6+8+10)\div4=7\). 2. The original range is \(10-4=6\), and the absolute distances from \(7\) are \(3,1,1,3\), so the mean absolute deviation is \(8\div4=2\). 3. The values below \(7\) are \(4\) and \(6\); after adding \(2\), the data are \(6,8,8,10\). 4. The new mean is \(8\), the new range is \(10-6=4\), and the distances from \(8\) are \(2,0,0,2\). 5. The new mean absolute deviation is \(4\div4=1\). 6. Both spread measures decrease because the lower values move toward the center.

Before: mean \(7\), range \(6\), mean absolute deviation \(2\). After: mean \(8\), range \(4\), mean absolute deviation \(1\).

- Find the farthest any allowed value can be from the stated mean. - Think about the greatest possible average when every quantity has the same upper bound. - Compare that bound with the reported spread.

1. Every value is at most \(2\) units below or above the mean \(10\). 2. Therefore, every absolute distance from the mean is at most \(2\). 3. An average of numbers that are all at most \(2\) cannot be \(2.5\). 4. The greatest possible mean absolute deviation under these conditions is \(2\), so the report is impossible.

No. Every distance from the mean is at most \(2\), so the mean absolute deviation cannot exceed \(2\).

- Find the original center before creating the new values. - Ignore direction when recording each distance. - Compare the calculation used for the new mean with the meaning of the original spread measure.

1. The original mean is \((2+4+8+10)\div4=6\). 2. The absolute distances from \(6\) are \(4,2,2,4\), so the second data set is \(4,2,2,4\). 3. The mean of the second data set is \((4+2+2+4)\div4=3\). 4. Its range is \(4-2=2\). 5. The mean of the distance data is exactly the original mean absolute deviation, because MAD is defined as the average of those absolute distances.

Second data set: \(4,2,2,4\). Mean: \(3\) Range: \(2\) Its mean equals the original data set’s mean absolute deviation.

- Expand each value according to its frequency. - Find the center before measuring distances. - Include every repeated observation when averaging.

1. The full data set is \(0,2,2,2,4\). 2. The mean is \((0+2+2+2+4)\div5=10\div5=2\). 3. The range is \(4-0=4\). 4. The absolute distances from \(2\) are \(2,0,0,0,2\), totaling \(4\). 5. The mean absolute deviation is \(4\div5=0.8\).

Data: \(0,2,2,2,4\) Mean: \(2\) Range: \(4\) Mean absolute deviation: \(0.8\)

- Complete the original statistics before changing the data. - Round every observation using the same rule. - Compare each value’s distance from the center before and after rounding.

1. The original mean is \(2\). The original range is \(2.6-1.4=1.2\). 2. The original distances from \(2\) are \(0.6,0.6,0.6,0.6\), so the mean absolute deviation is \(0.6\). 3. The rounded data are \(1,1,3,3\), with mean \(2\). 4. The rounded range is \(3-1=2\). 5. The rounded distances from \(2\) are all \(1\), so the mean absolute deviation is \(1\). 6. Rounding moves both lower values farther down and both upper values farther up from the unchanged center.

Before rounding: range \(1.2\), MAD \(0.6\). After rounding: range \(2\), MAD \(1\).

- State what observations each spread measure uses. - Ask whether the two numbers were produced by the same calculation. - Decide what must be held consistent before ranking two distributions.

1. Range measures only the distance between the minimum and maximum. 2. Mean absolute deviation measures the average distance of all observations from the mean. 3. The numbers come from different definitions and scales of behavior within a distribution, so comparing \(12\) directly with \(4\) does not rank variability. 4. A fair comparison requires the same measure for both data sets, along with compatible units and context.

The conclusion is invalid because range and MAD measure spread differently. Compare both ranges or both MADs, using the same units and context.

- Use the common mean to list each player’s distances. - Compare both extreme spread and average distance. - In this context, greater consistency corresponds to less variability.

1. Player A’s range is \(14-6=8\). The distances from \(10\) are \(4,2,0,2,4\), totaling \(12\), so the MAD is \(12\div5=2.4\). 2. Player B’s range is \(12-8=4\). The distances from \(10\) are \(2,2,0,2,2\), totaling \(8\), so the MAD is \(8\div5=1.6\). 3. Player B has smaller values for both measures of variability, so Player B is more consistent.

Player A: range \(8\), MAD \(2.4\). Player B: range \(4\), MAD \(1.6\). Player B is more consistent.

- Compare totals and endpoints before calculating distances. - Count how many values lie at the mean in each set. - Track how moving values to existing extremes affects average distance.

1. Both data sets have total \(30\), so both means are \(30\div5=6\). 2. Both have minimum \(2\) and maximum \(10\), so both ranges are \(8\). 3. Data Set A has distances \(4,0,0,0,4\), totaling \(8\), so its MAD is \(8\div5=1.6\). 4. Data Set B has distances \(4,4,0,4,4\), totaling \(16\), so its MAD is \(16\div5=3.2\). 5. Moving interior values to the existing extremes does not change the endpoints, but it increases the number of values far from the mean.

Data Set A: mean \(6\), range \(8\), MAD \(1.6\). Data Set B: mean \(6\), range \(8\), MAD \(3.2\).

- Compare the values present before and after sorting. - Identify whether totals, endpoints, or distances depend on written order. - Calculate from the multiset of values, not their sequence on the page.

1. The mean is \((2+7+4+9+3)\div5=25\div5=5\). 2. The range is \(9-2=7\). 3. The absolute distances from \(5\) are \(3,2,1,4,2\), totaling \(12\), so the MAD is \(12\div5=2.4\). 4. Sorting changes only the positions in which the same values are written. The total, extremes, and distances remain unchanged.

Mean \(5\), range \(7\), mean absolute deviation \(2.4\). Reordering does not change any statistic because the data values and frequencies are unchanged.

- Find each midpoint between an observation and the mean. - Compare the extreme-to-extreme distance before and after. - Track how each absolute distance from the unchanged center is scaled.

1. The original range is \(8-0=8\). The distances from \(4\) are \(4,0,4\), so the original MAD is \(8\div3=\frac{8}{3}\). 2. Moving \(0\) halfway toward \(4\) gives \(2\); \(4\) stays \(4\); moving \(8\) halfway gives \(6\). 3. The new data set is \(2,4,6\), still with mean \(4\). 4. The new range is \(6-2=4\). 5. The new distances are \(2,0,2\), so the new MAD is \(4\div3=\frac{4}{3}\). 6. Both spread measures are halved because every distance from the mean is halved.

New data: \(2,4,6\). Range changes from \(8\) to \(4\). MAD changes from \(\frac{8}{3}\) to \(\frac{4}{3}\).

- Do not round the mean before finding distances. - Group the repeated zeros when totaling their distances. - Remember that an average of integer-based measurements can be non-integer.

1. The mean is \((0+0+0+5)\div4=1.25\). 2. The range is \(5-0=5\). 3. The three zeros are each \(1.25\) from the mean, and \(5\) is \(3.75\) from the mean. 4. The total absolute distance is \(3\times1.25+3.75=7.5\). 5. The mean absolute deviation is \(7.5\div4=1.875\). 6. Averaging distances from a non-integer mean can produce a non-integer MAD even when all observations are integers.

Mean \(1.25\), range \(5\), mean absolute deviation \(1.875\).

- Compare the total increase with the total decrease. - Write the changed data in order before finding its endpoints. - Measure how far every changed value lies from the unchanged center.

1. Originally, the mean is \(5\), the range is \(8-2=6\), and the distances \(3,1,1,3\) give MAD \(8\div4=2\). 2. The changed data are \(4,4,6,6\), with total \(20\), so the mean remains \(5\). 3. The new range is \(6-4=2\). 4. The new distances from \(5\) are \(1,1,1,1\), so the MAD is \(1\). 5. One value moved up by \(2\) and another moved down by \(2\), preserving the total while bringing both extremes toward the mean.

Before: mean \(5\), range \(6\), MAD \(2\). After: mean \(5\), range \(2\), MAD \(1\).

- Compare totals and endpoints before finding average distances. - Count how many values lie at the mean in each set. - Use MAD to detect differences that range cannot show.

1. Both sets have total \(20\), so both means are \(5\). 2. Both have minimum \(1\) and maximum \(9\), so both ranges are \(8\). 3. For \(1,1,9,9\), every value is \(4\) from the mean, so the MAD is \(4\). 4. For \(1,5,5,9\), the distances are \(4,0,0,4\), so the MAD is \(8\div4=2\). 5. The second set places half its values at the mean, making it more concentrated despite the same endpoints.

Both sets have mean \(5\) and range \(8\). \(1,1,9,9\): MAD \(4\). \(1,5,5,9\): MAD \(2\).

- Find the center before measuring distances from it. - Use the greatest and least temperatures for the range. - Treat each difference from the mean as a nonnegative distance.

1. The sum is \(37+39+40+42+42=200\), so the mean is \(200\div5=40\,^{\circ}\text{F}\). 2. The range is \(42-37=5\,^{\circ}\text{F}\). 3. The absolute distances from the mean are \(3,1,0,2,2\) degrees Fahrenheit. 4. Their sum is \(8\), so the mean absolute deviation is \(8\div5=1.6\,^{\circ}\text{F}\). 5. A test temperature was, on average, \(1.6\,^{\circ}\text{F}\) from the mean temperature of \(40\,^{\circ}\text{F}\).

Mean: \(40\,^{\circ}\text{F}\) Range: \(5\,^{\circ}\text{F}\) MAD: \(1.6\,^{\circ}\text{F}\) The test temperatures were an average distance of \(1.6\,^{\circ}\text{F}\) from the mean temperature.

- Read repeated dots as repeated observations. - The common mean is halfway between the two extreme values. - Compare the average distances from that center, not only the shared range.

1. Plot a) represents \(2,6,6,10\). Its mean is \(6\), its range is \(8\), and its distances from the mean are \(4,0,0,4\). Its MAD is \(8\div4=2\). 2. Plot b) represents \(2,2,10,10\). Its mean is also \(6\), its range is also \(8\), and all four distances from the mean are \(4\). Its MAD is \(16\div4=4\). 3. Plot a) is more concentrated around the mean because two observations are at the mean, while plot b) places every observation at an extreme.

a) MAD \(=2\) b) MAD \(=4\) Plot a) is more concentrated around the mean.

- Read repeated dots as repeated observations. - Range depends only on the least and greatest values. - For MAD, compare all four distances from the common mean of \(6\).

1. Plot a) represents \(4,4,8,8\). Its range is \(8-4=4\). Every value is \(2\) units from the mean, so its MAD is \(2\). 2. Plot b) represents \(2,6,6,10\). Its range is \(10-2=8\). Its distances from the mean are \(4,0,0,4\), so its MAD is \((4+0+0+4)\div4=2\). 3. The range uses only the two extremes, so moving the extremes farther apart changes the range. MAD averages all distances from the mean, and in these two plots those average distances are equal.

a) Range \(=4\), MAD \(=2\) b) Range \(=8\), MAD \(=2\) They have the same MAD because their average distance from the mean is the same, even though plot b) has more widely separated extremes.

- Write out the transformed data set before calculating its statistics. - Compare the distances from each data set’s own mean. - Look for a relationship between paired values in the two sets.

1. Data set A has mean \((3+5+7+9)\div4=6\), range \(9-3=6\), and absolute distances \(3, 1, 1, 3\), so its mean absolute deviation is \(8\div4=2\). 2. Data set B is \(21, 19, 17, 15\), with mean \((21+19+17+15)\div4=18\). 3. Data set B has range \(21-15=6\), and its absolute distances from \(18\) are \(3, 1, 1, 3\), so its mean absolute deviation is \(2\). 4. The transformation reflects every value across \(12\). It changes the center from \(6\) to \(18\) but preserves all distances between values and from the center.

Data set A: mean \(6\), range \(6\), mean absolute deviation \(2\) Data set B: mean \(18\), range \(6\), mean absolute deviation \(2\) The reflection changes location but preserves spread.

- Consider separately whether the unknown becomes a new minimum or a new maximum. - Compute each candidate data set from its own center. - Compare which statistics match and which differ between the two possibilities.

1. The current minimum is \(6\) and maximum is \(10\). To make the range \(6\), \(x\) can create a new minimum of \(4\) or a new maximum of \(12\). 2. If \(x=4\), the mean is \((6+8+8+10+4)\div5=7.2\). The absolute distances total \(8.8\), so the mean absolute deviation is \(8.8\div5=1.76\). 3. If \(x=12\), the mean is \((6+8+8+10+12)\div5=8.8\). The absolute distances also total \(8.8\), so the mean absolute deviation is \(1.76\). 4. The two different data sets have the same range and mean absolute deviation but different means, so those two spread measures do not determine the data set uniquely.

a) \(x=4\) or \(x=12\) b) If \(x=4\): mean \(7.2\), mean absolute deviation \(1.76\). If \(x=12\): mean \(8.8\), mean absolute deviation \(1.76\). c) Different data sets can have the same range and mean absolute deviation.

- Look for symmetry in the two added values around the original mean. - Update the extreme values for the range. - Recover the original total distance from the original average distance before adding the new distances.

1. The added values are equally far below and above \(10\), so they balance around the original mean and the mean remains \(10\). 2. The new minimum is \(6\) and maximum is \(14\), so the new range is \(14-6=8\). 3. The original total absolute distance is \(4\times2=8\). 4. The new values each add an absolute distance of \(4\), making the new total distance \(8+4+4=16\). 5. The new mean absolute deviation is \(16\div6=\frac{8}{3}\approx2.67\).

Mean \(10\); range \(8\); mean absolute deviation \(\frac{8}{3}\approx2.67\) The balanced additions keep the center fixed but place new observations farther from it and extend both extremes.

- Identify the removed value’s distance from the mean. - Check whether either extreme is removed. - Compare the same total distance averaged over different numbers of values.

1. The original range is \(13-5=8\). 2. The original absolute distances from \(9\) are \(4, 2, 0, 0, 2, 4\), so the mean absolute deviation is \(12\div6=2\). 3. After removing one \(9\), the mean remains \(9\) and the extremes remain \(5\) and \(13\), so the range remains \(8\). 4. The total absolute distance remains \(12\), but it is now averaged over \(5\) values. 5. The new mean absolute deviation is \(12\div5=2.4\).

Before: range \(8\), mean absolute deviation \(2\) After: range \(8\), mean absolute deviation \(2.4\) Removing a zero-distance value leaves the distance total unchanged but reduces the number of observations.

- Check the extreme values before recalculating the range. - Recalculate the mean after changing an interior observation. - Compare all distances from each data set’s own mean.

1. Originally, the mean is \(4\), the range is \(8-0=8\), and the absolute distances are \(4, 0, 0, 0, 4\). The mean absolute deviation is \(8\div5=1.6\). 2. The changed data are \(0, 4, 4, 6, 8\), with mean \((0+4+4+6+8)\div5=4.4\). 3. The new range remains \(8-0=8\). 4. The new absolute distances are \(4.4, 0.4, 0.4, 1.6, 3.6\), totaling \(10.4\). 5. The new mean absolute deviation is \(10.4\div5=2.08\). 6. The extremes did not change, but the interior values became less concentrated around the new mean.

Before: range \(8\), mean absolute deviation \(1.6\) After: range \(8\), mean absolute deviation \(2.08\) The range stays fixed because the minimum and maximum are unchanged; the mean absolute deviation increases because the values are more spread around the mean.

- Separate the rule into a multiplication and an addition. - Consider what each operation does to distances between data values. - A common shift affects location differently from spread.

1. Doubling every value doubles the mean, range, and every distance from the mean. 2. Adding \(5\) to every doubled value increases the mean by \(5\) but does not change distances between values or distances from the mean. 3. The new mean is \(2\times12+5=29\). 4. The new range is \(2\times10=20\). 5. The new mean absolute deviation is \(2\times2.5=5\).

Mean \(29\), range \(20\), mean absolute deviation \(5\). The factor \(2\) changes both measures of variability; the added \(5\) changes only the center.

- Use the mean to determine the total of all observations. - Start with one repeated value and track the total change caused by each replacement. - Once the frequencies are known, group equal distances from the mean.

1. A mean of \(7\) for \(8\) values requires a total of \(7\times8=56\). 2. If all \(8\) values were \(2\), their total would be \(16\). Replacing a \(2\) with a \(10\) increases the total by \(8\). 3. The required increase is \(56-16=40\), so \(40\div8=5\) values are \(10\). The other \(3\) values are \(2\). 4. The range is \(10-2=8\). 5. The five values of \(10\) are each \(3\) from the mean, and the three values of \(2\) are each \(5\) from the mean. 6. The mean absolute deviation is \((5\times3+3\times5)\div8=30\div8=3.75\).

The value \(10\) occurs \(5\) times, and the value \(2\) occurs \(3\) times. The range is \(8\), and the mean absolute deviation is \(3.75\).

- Compare how the total and the number of observations change together. - Check whether any new minimum or maximum is introduced. - Track what happens to both the total distance and the divisor in the spread calculation.

1. The original mean is \((1+3+6+10)\div4=5\), and the range is \(10-1=9\). 2. The original absolute distances from \(5\) are \(4, 2, 1, 5\), totaling \(12\), so the mean absolute deviation is \(12\div4=3\). 3. In the repeated list, both the total of the values and the number of values are multiplied by \(3\), so the mean remains \(5\). 4. The minimum and maximum remain \(1\) and \(10\), so the range remains \(9\). 5. The total absolute distance becomes \(36\), and \(36\div12=3\), so the mean absolute deviation remains \(3\).

Both lists have mean \(5\), range \(9\), and mean absolute deviation \(3\). Equal repetition changes frequencies but preserves the proportions of all values and distances.

- Convert each average distance back into a total distance. - Use the fact that both groups have the same center. - Combine totals and observation counts before forming a new average.

1. Because both sets have mean \(10\), the combined set also has mean \(10\), and all stated absolute distances use the same center. 2. Data Set A has total absolute distance \(4\times2=8\). 3. Data Set B has total absolute distance \(6\times3=18\). 4. The combined total absolute distance is \(8+18=26\) across \(4+6=10\) observations. 5. The combined mean absolute deviation is \(26\div10=2.6\). 6. Averaging \(2\) and \(3\) equally would ignore the different numbers of observations.

The combined mean absolute deviation is \(2.6\).

- A new whole-number value can create either a new maximum or a new minimum. - Test both ways to make the required extreme-to-extreme distance. - Recompute each completed data set's mean before averaging the distances from it.

1. The current minimum and maximum are \(11\) and \(15\). To make the range \(12\), \(x\) must create a new maximum of \(23\) or a new minimum of \(3\). 2. If \(x=23\), the mean is \((11+12+15+23)\div4=15.25\). The absolute distances are \(4.25,3.25,0.25,7.75\), totaling \(15.5\), so the mean absolute deviation is \(15.5\div4=3.875\). 3. If \(x=3\), the mean is \((3+11+12+15)\div4=10.25\). The absolute distances are \(7.25,0.75,1.75,4.75\), totaling \(14.5\), so the mean absolute deviation is \(14.5\div4=3.625\). 4. Both sets have range \(12\), but their mean absolute deviations differ.

\(x=23\) gives mean absolute deviation \(3.875\). \(x=3\) gives mean absolute deviation \(3.625\). The range alone does not determine the mean absolute deviation.

- Analyze each cluster before combining them. - Find the center of all eight values, not the center of either cluster alone. - Notice how separation between clusters contributes to distances from the combined center.

1. Data Set A has mean \(5\), range \(8-2=6\), and distances \(3,1,1,3\), so its mean absolute deviation is \(2\). 2. Data Set B is \(12,14,16,18\), with mean \(15\), range \(6\), and mean absolute deviation \(2\). 3. The combined set has mean \((20+60)\div8=10\). 4. Its range is \(18-2=16\). 5. The distances from \(10\) are \(8,6,4,2,2,4,6,8\), totaling \(40\), so the combined mean absolute deviation is \(40\div8=5\). 6. Each separate cluster is compact, but the clusters are far apart around the combined center.

Data Set A: mean \(5\), range \(6\), MAD \(2\). Data Set B: mean \(15\), range \(6\), MAD \(2\). Combined: mean \(10\), range \(16\), MAD \(5\).

- Think of the mean as a balance point on a number line. - Decide how the distances \(1,1,3,3\) can be placed on the two sides with equal total distance. - Once the values are known, use extreme distance for the range and average distance for the MAD.

1. To keep the mean at \(10\), the total distance below \(10\) must balance the total distance above \(10\). 2. The distances must therefore be paired as one distance of \(1\) and one distance of \(3\) on each side of \(10\). 3. The data values are \(10-3=7\), \(10-1=9\), \(10+1=11\), and \(10+3=13\). 4. The range is \(13-7=6\). 5. The mean absolute deviation is \((1+1+3+3)\div4=2\).

Data: \(7,9,11,13\) Range: \(6\) Mean absolute deviation: \(2\) The distances balance: the total distance below \(10\) equals the total distance above \(10\).

- Find the greatest possible distance from the stated mean within the allowed interval. - Ask when an average can equal the maximum of all quantities being averaged. - Use balance around the mean to compare the endpoint frequencies.

1. Because the mean is \(5\), every allowed value is at most \(5\) units from the mean. 2. The mean absolute deviation is \(5\), the greatest possible distance. 3. An average of distances can equal \(5\) only if every distance equals \(5\). 4. Therefore, every data value must be \(0\) or \(10\). 5. For the mean of those endpoint values to be \(5\), the data set must contain equal numbers of \(0\)s and \(10\)s.

Every value must be either \(0\) or \(10\), and the data set must contain equal numbers of the two values.

- Check whether removing one repeated maximum removes the maximum value itself. - Recalculate the mean after changing the sample size and total. - Measure every remaining value from the new mean.

1. Originally, the mean is \(4\), the range is \(8\), and the absolute distances total \(4+4+0+0+4+4=16\). The MAD is \(16\div6=\frac{8}{3}\approx2.67\). 2. After removing one \(8\), the data are \(0,0,4,4,8\), with mean \(16\div5=3.2\). 3. The range remains \(8-0=8\) because another \(8\) is still present as the maximum. 4. The new distances from \(3.2\) are \(3.2,3.2,0.8,0.8,4.8\), totaling \(12.8\). 5. The new MAD is \(12.8\div5=2.56\).

Before: mean \(4\), range \(8\), MAD \(\frac{8}{3}\approx2.67\). After: mean \(3.2\), range \(8\), MAD \(2.56\). The range stays \(8\) because both endpoint values \(0\) and \(8\) remain.

- Locate the mean relative to the minimum and maximum. - Compare any one distance from the mean with the full endpoint distance. - Consider when both spread measures vanish.

1. Every data value lies between the minimum and maximum, whose distance is the range. 2. In a nonconstant data set, the mean lies strictly between the minimum and maximum. 3. Therefore, every distance from a data value to the mean is strictly less than the full range \(6\). 4. The average of those distances must also be less than \(6\), so the MAD cannot equal the nonzero range. 5. Range and MAD can be equal only when all values are identical, giving both measures value \(0\).

The report is impossible. For a nonconstant data set, MAD is less than the range. They can be equal only when both are \(0\).