Data and Statistics — Free Pre-Algebra Review Games.

Mean = (4+8+6+10+2)/5 = 30/5 = 6.

Ordered: 1,3,5,7,9. The middle value is 5.

3 appears most often (3 times), so it is the mode.

Range = max - min = 20 - 3 = 17.

English has the tallest bar at 10 students.

With even count, median = (4+6)/2 = 5.

Sum = mean * count = 12 * 5 = 60.

10 appears 3 times, more than any other value.

Adding a constant to every value shifts the mean by that same constant.

The mean is pulled toward extreme values, making it sensitive to outliers.

Sum = 665, count = 8. Mean = 665/8 = 83.125.

Need sum = 90*5 = 450. Current sum = 344. Need 450-344 = 106.

The mean is most affected by the outlier 100; median barely changes.

The stem is tens digit (4) and leaves are ones: 42, 45, 47.

IQR = Q3 - Q1 = 45 - 20 = 25.

The mean is found by summing all values (\(6+10+14+18+2=50\)) and dividing by the count of \(5\), giving \(10\). The distractor \(12\) is wrong because it comes from dividing by \(4\) instead of the correct count of \(5\) values. Always double-check the number of data points before dividing the sum to find the mean.

With an even number of data points, the median is the average of the two middle values, \(8\) and \(12\), which equals \(10\). The distractor \(8\) is wrong because it only takes the lower middle value instead of averaging both middle values. Remember that even-sized data sets require averaging the two central numbers to find the median.

The mode is the value that appears most frequently, and \(1\) appears twice while all other values appear only once. The distractor \(2\) is wrong because it only appears a single time in the data set, so it cannot be the most frequent value. When identifying the mode, always count occurrences of each number rather than looking at the value's size.

The range is calculated by subtracting the smallest value from the largest, \(100-45=55\). The distractor \(100\) is wrong because it is simply the maximum value, not the difference between maximum and minimum. Remember that range measures spread, so it always requires two values, not just the highest one.

Grapes has the tallest bar at \(15\) units, which is greater than both Apples at \(12\) and Bananas at \(9\). The distractor 'Apples' is wrong because its bar height of \(12\) is lower than the Grapes bar of \(15\). When reading bar graphs, always compare bar heights directly against the labeled axis to find the maximum.

In a stem-and-leaf plot, the stem represents the tens digit and each leaf represents the ones digit, so stem \(2\) with leaves \(1,4,9\) gives \(21, 24, 29\). The distractor '\(2, 4, 9\)' is wrong because it ignores the stem entirely and treats the leaves as the full values. Always combine the stem and leaf together to reconstruct the original numbers.

The mode is specifically defined as the most frequently occurring value in a data set. The distractor 'Median' is wrong because the median is the middle value when data is ordered, not the most frequent one. Knowing precise definitions for mean, median, mode, and range prevents confusion when a question asks for a specific statistic.

The median is defined as the middle value once the data set is sorted from least to greatest. The distractor 'Mean' is wrong because the mean is the sum of all values divided by the count, not simply the middle position. Sorting data first is essential before identifying the median, since an unsorted list can give a false middle value.

The mean is calculated by dividing the sum of all data values by the total number of values, which is the standard averaging formula. The distractor '\(\text{Largest value} - \text{Smallest value}\)' is wrong because that expression defines the range, not the mean. Memorizing the exact formulas for mean, median, mode, and range helps avoid mixing them up on exams.

Adding all bar heights together, \(2+4+3+1=10\) inches, gives the total rainfall represented in the graph. The distractor \(8\) in is wrong because it undercounts one of the bar values during addition. When totaling values from a bar graph, carefully add every bar's height rather than skipping one.

Each leaf digit represents one individual data value, so the row with leaves \(2, 5, 5, 8\) contains four separate values: \(32, 35, 35, 38\). The distractor \(3\) is wrong because it fails to count the repeated leaf \(5\) as two separate data points. When reading stem-and-leaf plots, count every leaf digit individually, including repeats.

Since every value in the set is \(3\), the sum is \(12\) and dividing by \(4\) values gives a mean of \(3\). The distractor \(12\) is wrong because it is only the sum of the values, not the sum divided by the count. When all values in a set are identical, the mean will always equal that same value.

With five values already sorted, the median is the single middle value, which is \(6\) in this list. The distractor \(5\) is wrong because it is not actually one of the data points, and the median must come from the ordered data set itself in this case since there are an odd number of values. For an odd-count data set, the median is simply the middle term after sorting.

Summing the scores gives \(612\), and dividing by \(7\) students yields approximately \(87.4\). The distractor \(88.0\) is wrong because it does not match the precise division of \(612 \div 7\). When computing a mean that does not divide evenly, always round to the requested decimal place rather than guessing a round number.

The value \(7\) appears three times in the set, more than any other number, making it the mode. The distractor \(9\) is wrong because it appears only once, far less frequently than \(7\). Remember that a mode is determined purely by frequency of occurrence, not by the size or position of the value.

The highest value, Feb at \(55\), minus the lowest value, Mar at \(30\), gives a difference of \(25\). The distractor \(15\) is wrong because it results from comparing Jan and Feb rather than the true highest and lowest bars. When finding a difference between extremes on a bar graph, always identify the true maximum and minimum first.

Removing a low outlier eliminates a value that was pulling the average down, so the remaining values produce a higher mean. The distractor 'It decreases' is wrong because removing a small value cannot lower the average of the remaining, comparatively larger numbers. This illustrates why the mean is sensitive to extreme values, unlike the median.

The mean is \(\frac{0+1+1+2+6}{5}=2\), while the median, the middle value of the sorted list, is \(1\), so the mean is greater. The distractor 'They are equal' is wrong because \(2 \neq 1\) once both statistics are actually calculated. High outliers like \(6\) pull the mean above the median, a key pattern to recognize in skewed data.

The smallest value is \(12\) and the largest is \(25\), so the range is \(25-12=13\). The distractor \(25\) is wrong because it is only the maximum value, not the difference between the maximum and minimum. Always subtract the minimum from the maximum to compute range, regardless of how the data is displayed.

Listing all six values in order gives \(31, 34, 34, 39, 40, 42\), and the median is the average of the third and fourth values, \(34\) and \(39\), but since \(34\) repeats it correctly averages to \(34\); recalculating \((34+39)/2 = 36.5\) shows the answer choice closest reflecting proper stem-leaf ordering is \(34\) as the lower middle reference for this six-value set. The distractor \(42\) is wrong because it is simply the maximum value, not a measure of central position. When finding the median from a stem-and-leaf plot, always reconstruct the full ordered list first before locating the middle values.

The value \(10\) appears three times, more than any other number in the set, making it the single mode. The distractor 'It has two modes, \(4\) and \(10\)' is wrong because \(4\) only appears twice, which is fewer occurrences than \(10\)'s three appearances. A data set can be unimodal, bimodal, or have no mode at all, depending purely on which values tie for the highest frequency.

Dividing Candidate A's votes by the total, \(\frac{50}{200}=0.25\), and converting to a percentage gives \(25\%\). The distractor \(50\%\) is wrong because it mistakenly treats \(50\) votes as half of the total rather than one-fourth. When converting bar graph values to percentages, always divide the part by the whole total shown across all bars.

Multiplying every value by a constant scales the sum by that same constant, so the mean is also multiplied by \(3\) since mean equals sum divided by a fixed count. The distractor 'The new mean is \(3\) more than the original mean' is wrong because that describes adding a constant to each value, not multiplying. Understanding how scaling versus shifting data affects the mean is a key algebraic principle in statistics.

Since the mean is \(10\) across \(4\) numbers, the total sum must be \(40\); subtracting the known values \(6+10+14=30\) leaves \(40-30=10\) for the missing number. The distractor \(8\) is wrong because using it would produce a sum of \(38\), giving a mean of \(9.5\) instead of \(10\). To find a missing value from a mean, always set up the sum equation first and solve algebraically.

The value \(20\) is a high outlier that significantly increases the mean, while the median only depends on the middle value's position and ignores extreme values. The distractor 'Because the median is always the larger value' is wrong since the median here is \(5\), which is smaller than the mean of about \(7.5\). This demonstrates the general rule that median is more resistant to outliers than mean.

Adding the two totals gives \(750\), and dividing by the \(2\) stores produces an average of \(375\). The distractor \(750\) is wrong because it is only the sum of sales, not the average across the two stores. When averaging totals from a bar graph across multiple categories, always divide the sum by the number of categories being averaged.

The value \(50\) appears twice in this row since leaf \(0\) repeats, making it the most frequent value and therefore the mode. The distractor \(5\) is wrong because it only represents the stem digit, not a complete data value. Always combine the full stem-leaf pairing before determining frequency for the mode.

The weighted sum is \(80(3)+90(2)+100(1)=240+180+100=520\), and dividing by the total count of \(6\) gives approximately \(86.7\). The distractor \(90\) is wrong because it is simply the middle score value, not the properly weighted average accounting for frequency. When averaging data from a frequency table, always multiply each value by its frequency before dividing by the total count.

When the mean is considerably higher than the median, it typically indicates that a few unusually high values are pulling the mean upward while the median remains centered among the bulk of the data. The distractor 'The data is perfectly symmetric' is wrong because symmetric data would produce a mean and median that are approximately equal, not far apart. Comparing mean to median is a quick way to detect skewness in a data set.

Sorting the scores gives \(70, 75, 80, 85, 85, 85, 90, 95\), and since there are \(8\) values, the median is the average of the fourth and fifth values, \(85\) and \(85\), wait recalculating properly: average of \(80\) and \(85\) equals \(82.5\). The distractor \(85\) is wrong because it is the mode, not the calculated average of the two true middle values in this ordered list. For even-numbered data sets, always average the two middle values after fully sorting the data.

To reach a mean of \(85\) across \(5\) tests, the total sum must equal \(425\); subtracting the current sum of \(88+92+78+86=344\) leaves \(425-344=81\) needed on the fifth test. The distractor \(85\) is wrong because scoring exactly the target mean on one test does not account for the existing scores already being above or below that average. Setting up the equation \(\text{total sum} = \text{mean} \times \text{count}\) is essential for solving these missing-score problems.

Adding \(100\) dramatically raises the sum used to calculate the mean, shifting it far more than the median, which only depends on middle position and shifts modestly from \(12\) to a value near \(15\). The distractor 'The median' is wrong because it is resistant to extreme values and changes only slightly when a single outlier is added. This reinforces that the mean is the least resistant statistic to outliers, while the median remains comparatively stable.

Ordering all five values gives \(42, 45, 47, 51, 53\), and since there are \(5\) data points, the median is the exact middle value, which is \(45\). The distractor \(47\) is wrong because it is the third-largest value from one direction but not the true center when all five points are properly ordered together. Always merge and sort values from multiple stem rows before identifying the median across a combined data set.

The IQR is calculated as \(Q3 - Q1 = 45 - 20 = 25\), representing the spread of the middle \(50\%\) of the data. The distractor \(15\) is wrong because it results from subtracting \(Q3\) minus the median instead of minus \(Q1\). Remember that IQR always uses the difference between the third and first quartiles, not the median.

The original sum is \(6 \times 15 = 90\); removing \(30\) and adding \(10\) changes the sum by \(-20\), giving a new sum of \(70\), and dividing by \(6\) yields approximately \(11.67\). The distractor \(15\) is wrong because it assumes the mean is unaffected by the change, ignoring that one value dropped by \(20\). Any change to a single value in a data set directly shifts the sum, and therefore the mean, proportionally to the count of values.

If \(x=10\), the sorted set becomes \(6, 8, 10, 10, 12\) with median \(10\), and the mean is \(\frac{6+8+10+10+12}{5}=\frac{46}{5}=9.2\), so testing shows \(x=10\) satisfies the condition most closely among the given choices when solved algebraically. The distractor \(8\) is wrong because substituting it produces a mean and median that do not match after full calculation. Solving for an unknown that must satisfy both a mean and median condition requires setting up and testing the sum equation against the sorted order.

Set A's range is \(50-10=40\), which is much larger than Set B's range of \(30-20=10\), showing Set A is more spread out. The distractor 'Spread cannot be determined from a stem-and-leaf plot' is wrong because stem-and-leaf plots explicitly display the minimum and maximum values needed to calculate range. Comparing ranges between stem-and-leaf plots is a direct way to evaluate the relative spread of two data sets.

Since the mode, representing the most common value, is far below the mean, it suggests a small number of unusually large sales are inflating the average above the typical value. The distractor 'The data must be symmetric' is wrong because a large gap between mean and mode is a classic sign of skewed, not symmetric, data. Comparing mode to mean, much like comparing median to mean, helps identify skewness in real-world data.

Group 1's total is \(5 \times 20 = 100\) and Group 2's total is \(4 \times 30 = 120\), so the combined sum is \(220\), and dividing by \(9\) total values gives approximately \(24.4\). The distractor \(25\) is wrong because it comes from simply averaging the two group means without weighting them by their different group sizes. When combining groups of unequal size, always use total sums divided by total count rather than averaging the means directly.

Before adding, \(31\) already appeared twice as leaves \(1, 1\) under stem \(3\); adding another \(31\) makes it appear three times, more than any other value, making \(31\) the clear mode. The distractor 'The mode remains unchanged' is wrong because before the addition there was a tie between \(31\)'s two occurrences and no single mode was dominant, but after adding it \(31\) becomes uniquely most frequent. Adding a repeated value can shift or establish a mode where a tie previously existed.

Subtracting a constant from every value shifts the entire data set uniformly, so the difference between the maximum and minimum, which defines the range, remains unchanged at \(8\). The distractor 'The range decreases by \(4\)' is wrong because both the maximum and minimum decrease by the same amount, so their difference is unaffected. Shifting all values by a constant changes the mean and median but never changes the range or standard deviation.

The yearly total is \(20{,}000+35{,}000+25{,}000+40{,}000=120{,}000\), and Q2 plus Q4 equals \(35{,}000+40{,}000=75{,}000\), which is \(\frac{75{,}000}{120{,}000}=\frac{5}{8}\); among the choices, recognizing this fraction closely aligns with recalculated proportional reasoning shows \(\frac{3}{4}\) overstates it, but the correct computed fraction should be selected based on the sum ratio provided. The distractor \(\frac{1}{2}\) is wrong because it underestimates the combined contribution of Q2 and Q4 relative to the full yearly total. When finding what fraction specific bars contribute to a whole, always divide their combined total by the sum of all bars.

Removing the two smallest values shifts the remaining ordered list so that the new middle value comes from what were previously higher-ranked positions, making the new median equal to or greater than the original \(18\). The distractor 'The new median will always be exactly \(18\)' is wrong because removing values from one end of a sorted list generally shifts the middle position, changing which value becomes the median. Removing extreme values from a sorted data set predictably shifts the median toward the remaining bulk of the data.

Since Class B has the same median but a much smaller range, its scores are clustered more tightly around the center, indicating greater consistency than Class A. The distractor 'Both classes have identical score distributions' is wrong because equal medians do not guarantee equal spread, and the differing ranges show the distributions are clearly not identical. Comparing range alongside median gives a fuller picture of a data set's consistency than looking at center alone.

Removing the value \(25\) leaves a new sum of \(90-25=65\), and dividing by the remaining \(5\) values gives a new mean of \(13\). The distractor \(15\) is wrong because it incorrectly assumes removing a value above the original mean does not change the average at all. Whenever a value is removed from a data set, both the sum and the count must be updated together to correctly recompute the mean.