AP Statistics Unit 1: Exploring One-Variable Data — Free Review Games.

Histograms display the distribution of quantitative data by showing frequencies in intervals.

The middle value in this ordered set of 5 numbers is 7.

The median is not affected much by extreme values, unlike the mean.

The interquartile range is the difference between the third and first quartiles.

A long tail to the right indicates right (positive) skew.

Standard deviation quantifies typical spread around the mean.

The 1.5*IQR rule defines outliers as values beyond Q1-1.5*IQR or Q3+1.5*IQR.

In right-skewed distributions, the mean is pulled toward the tail, making it larger.

z = (x - mean)/SD, so z=2 means 2 SDs above the mean.

The empirical rule applies specifically to normal (bell-shaped) distributions.

40 and 60 are 2 SDs from the mean. About 95% falls within 2 SDs.

IQR=15, Q3+1.5*IQR=35+22.5=57.5. 60>57.5, so it is an outlier.

Adding a constant shifts the mean by that constant but does not change the spread.

Multiplying by a constant scales the SD by the absolute value of that constant.

The 85th percentile means 85% of the data falls at or below that value.

In a frequency histogram, bar height corresponds to the count (frequency) of data values that fall within each class interval. Choice D describes a relative frequency histogram, where bar heights show proportions or percentages rather than counts. The shape of both histograms is identical; only the y-axis scale differs.

The mode is the value that appears most frequently. In this data set, \(11\) appears three times, \(7\) appears twice, and all other values appear once. Therefore the mode is \(11\). Choice A (\(7\)) is a common distractor because \(7\) also repeats, but it appears only twice — less than \(11\).

Range \(=\) Maximum \(-\) Minimum \(= 19 - 1 = 18\). Choice A (\(16\)) results from subtracting \(3\) (the second-smallest value) instead of the true minimum \(1\). Choice C (\(19\)) confuses the maximum value itself with the range.

Mean, median, and mode all describe the center (typical value) of a distribution. The range — defined as Maximum \(-\) Minimum — describes how spread out the data values are, making it a measure of variability, not center. Other measures of spread include standard deviation, variance, and IQR.

Dotplots place individual data values as dots along a number line, making them ideal for small to moderate-sized quantitative data sets where individual values matter. They clearly reveal shape, center, and spread. Scatterplots (choice A) handle two-variable relationships, bar charts (choice C) display categorical distributions, and line graphs (choice D) track change over time.

Standard deviation is the square root of an average of squared deviations, so it is always \(\geq 0\). It equals zero only when all values are identical. The other choices are all achievable: \(s = 0\) when every value is the same; \(s\) can exceed the mean (e.g., \(\{0, 0, 10\}\)); and \(s\) can be any positive decimal less than one.

In a perfectly symmetric, unimodal distribution (such as the normal distribution), the mean, median, and mode all coincide at the single central peak. Skewness is what causes these three measures to diverge. Choice A describes a right-skewed distribution (mean pulled toward the high tail), and choice B describes a left-skewed distribution.

The mean uses every value in its calculation, so the extremely low score of 36 pulls it down significantly: new mean \(= \frac{8(72) + 36}{9} = \frac{612}{9} = 68\), a drop of 4 points. The median depends only on the middle-ranked value and is resistant to extreme scores, so it will shift by at most one rank position. This illustrates why the mean is non-resistant to outliers while the median is resistant.

The z-score formula is \(z = \frac{x - \mu}{\sigma} = \frac{65 - 80}{10} = \frac{-15}{10} = -1.5\). The negative sign indicates that \(65\) is below the mean. Choice A drops the negative sign, losing the directional information. Choices C and D appear to divide \(65\) directly by \(100\), which is not the z-score formula.

\(\text{IQR} = Q_3 - Q_1 = 50 - 30 = 20\). Upper fence \(= Q_3 + 1.5 \times \text{IQR} = 50 + 1.5(20) = 50 + 30 = 80\). Choice B (\(60\)) results from adding \(1.5 \times \text{IQR}\) to \(Q_1\) instead of \(Q_3\): \(30 + 30 = 60\). Choice C (\(70\)) uses a coefficient of \(1\) rather than \(1.5\): \(Q_3 + \text{IQR} = 50 + 20 = 70\). Choice D uses a coefficient of \(2\): \(50 + 40 = 90\).

Standard deviation measures spread (variability), not center. A larger standard deviation means scores deviate more from the mean on average. Since Class A has \(s = 8\) and Class B has \(s = 3\), Class A's scores are more spread out. Choices A and B are incorrect because standard deviation gives no information about which class scored higher — comparing averages requires the means, not the standard deviations.

A percentile rank of 72 means the student's score is at or above approximately 72% of all scores in the reference distribution. Percentile is a relative ranking measure, not a measure of raw performance. Choice A confuses percentile rank with percent correct. Choice C has the direction reversed. Choice D confuses percentile with a z-score or deviation from the mean.

New mean \(= \frac{5 + 8 + 10 + 12 + 25}{5} = \frac{60}{5} = 12\), up from \(10\). The standard deviation also increases: the new maximum of 25 is \(25 - 12 = 13\) units from the new mean, far larger than the original gap of \(15 - 10 = 5\) units, creating greater spread. The median of \(\{5, 8, 10, 12, 25\}\) is still \(10\) — the middle value is unchanged when only the maximum is replaced — so choice A is wrong.

In a left-skewed (negatively skewed) distribution, the tail extends toward lower values on the left. The mean is pulled in the direction of the tail — toward smaller values — so \(\bar{x} < M\). The median, being resistant to extreme values, stays closer to the bulk of the data. Choice A describes a right-skewed distribution, where the tail extends right and pulls the mean upward.

In a relative frequency histogram, each bar's height represents that class interval's proportion (or percentage) of the entire data set rather than the raw count. This makes it easier to compare distributions from samples of different sizes. The shape of the histogram is identical to the corresponding frequency histogram — only the y-axis scale changes.

The five-number summary consists of: Minimum, \(Q_1\), Median, \(Q_3\), and Maximum. The mean is not part of it. The five-number summary relies entirely on order statistics (ranked positions), making it resistant to outliers. The median — not the mean — serves as the measure of center. A boxplot is constructed directly from the five-number summary.

Standard deviation directly quantifies variability: a larger \(s\) means values are, on average, farther from the mean. Data Set B (\(s_B = 10\)) has substantially greater spread than Data Set A (\(s_A = 2\)), regardless of their equal means. Choice C is a frequent misconception — the mean describes center; knowing two distributions share a mean says nothing about their spreads.

A tail extending toward lower (left) values indicates left-skew (negative skew). The mean is pulled toward the tail — toward lower values — so \(\bar{x} < M\). The median remains closer to the main cluster of high scores. Choice A describes a right-skewed distribution, which has its tail extending toward higher values and \(\bar{x} > M\).

The combined mean requires a weighted average: $\bar{x}_{\text{combined}} = \frac{n_A \bar{x}_A + n_B \bar{x}_B}{n_A + n_B} = \frac{20(75) + 30(85)}{50} = \frac{1500 + 2550}{50} = \frac{4050}{50} = 81$. The simple unweighted average \(\frac{75 + 85}{2} = 80\) (choice A) incorrectly treats both classes as equal in size. Because Class B has more students, the combined mean is pulled closer to 85 than to 75.

Solving \(z = \frac{x - \mu}{\sigma}\) for \(x\) gives \(x = \mu + z\sigma = 70 + (-2.4)(5) = 70 - 12 = 58\). Choice B (\(60\)) results from using \(z = -2.0\) instead of \(-2.4\). Choice C (\(55\)) results from using \(z = -3.0\), computing \(70 - 15 = 55\). These errors come from rounding or misreading the z-score before substituting.

Unusualness is measured by the z-score, not raw distance. In Distribution X: \(z = \frac{115 - 100}{15} = 1.0\). In Distribution Y: \(z = \frac{115 - 100}{5} = 3.0\). A z-score of \(3.0\) is far more extreme than \(1.0\). In Distribution Y, where data is tightly clustered, a value 15 units above the mean is very rare. In Distribution X, with its wider spread, the same gap is relatively unremarkable. Choice B is wrong because absolute distance is meaningless without accounting for the scale of variation.

For a linear transformation \(y = ax + b\), the new mean \(= a\bar{x} + b\) and the new standard deviation \(= |a| \cdot s\). Here \(a = 2\) and \(b = -3\). New mean \(= 2(40) - 3 = 80 - 3 = 77\). New standard deviation \(= 2(8) = 16\). Subtracting a constant shifts all values equally, so it changes the mean but does not affect spread. Choice B adds \(3\) instead of subtracting it. Choices C and D incorrectly apply the subtraction to the standard deviation as well.

Sample variance \(= s^2 = \frac{\sum(x_i - \bar{x})^2}{n - 1} = \frac{150}{7 - 1} = \frac{150}{6} = 25\). Sample standard deviation \(= s = \sqrt{25} = 5\). Choice B (\(\approx 4.63\)) uses \(n\) instead of \(n - 1\) in the denominator: \(\sqrt{150/7} \approx 4.63\) — this computes the population standard deviation, not the sample standard deviation. Choice C takes \(\sqrt{150} \approx 12.25\), forgetting to divide by \(n - 1\) first. Choice D reports the variance (\(25\)) rather than the standard deviation.

The word 'average' typically refers to the mean, and 'typical' suggests the resistant median. Since mean (\(\\)450{,}000$) \(>\) median (\(\\)320{,}000$), the mean is being pulled upward by a small number of very high-priced homes — the hallmark of a right-skewed distribution with its tail toward higher values. In right-skewed data, \(\bar{x} > M\). Choice A incorrectly labels this as left-skewed; a left-skewed distribution would have \(\bar{x} < M\). Choice D is wrong because a symmetric distribution would have mean \(\approx\) median.

Quality control depends on both accuracy (mean near target) and precision (low standard deviation). Machine A's mean is only \(0.1\) mm off target — a minor, correctable bias — while \(s = 0.2\) mm means nearly all bolts fall within a very tight, predictable range. Machine B has a perfect mean of \(10.0\) mm, but \(s = 0.8\) mm creates highly inconsistent output: many individual bolts will deviate significantly from target. A small consistent bias is generally preferable to high variability. This question illustrates that center and spread are separate, equally important properties of a distribution.

The mode is the value that appears most frequently in a data set. A data set can have one mode (unimodal), two modes (bimodal), or no mode if all values are distinct. The middle value describes the median, the arithmetic average describes the mean, and the difference between max and min describes the range.

In a frequency histogram, the height of each bar equals the number of data values (frequency) that fall within that interval. A relative frequency histogram uses proportions or percentages instead of counts. A cumulative frequency graph (ogive) shows the running total, which is different from a standard histogram.

The mean is most affected by outliers because it incorporates every value in its calculation: \(\bar{x} = \frac{\sum x_i}{n}\). A single extreme value can pull the mean significantly toward it. The median depends only on the middle value(s) and is resistant to outliers. The mode is unaffected unless the outlier itself happens to be the most frequent value.

Standard deviation measures how far data values typically fall from the mean. A small standard deviation indicates values are clustered near the mean; a large one indicates values are widely spread. The middle value is the median, the most common value is the mode, and the difference between max and min is the range.

In a perfectly symmetric, unimodal distribution (such as the normal distribution), the mean, median, and mode all coincide at the center. Symmetry prevents either tail from pulling the mean away from the median, and with a single peak, the mode also aligns at that central value.

An outlier is a value that is significantly separated from the bulk of the data. A formal rule uses the $1.5 \times IQR$ criterion: values below $Q_1 - 1.5(IQR)$ or above $Q_3 + 1.5(IQR)$ are flagged as outliers. Not every maximum is an outlier, and outliers can appear on either end of the distribution, so choices A and C are too restrictive.

A dotplot places one dot above a number line for each data value, stacking dots when values repeat. This shows every individual observation and its frequency clearly. Grouped intervals describe a histogram, scatterplots show relationships between two variables, and dotplots become impractical with very large data sets due to clutter.

The range equals the maximum value minus the minimum value: \(22 - 4 = 18\). The range measures total spread but is sensitive to outliers because it depends only on the two most extreme values. Here, \(13\) is the median and \(22\) is the maximum, not the range.

The data set is already ordered and contains \(7\) values. The median is the middle (4th) value: \(3, 7, 7, \underline{9}, 12, 15, 20\). The median is \(9\). The value \(10.43\) is approximately the mean (\(73 \div 7 \approx 10.43\)), which differs from the median. The mode is \(7\) because it appears twice, but it is not the median.

In a right-skewed distribution, the long tail extends toward higher values. These large values pull the mean upward while the median — being resistant to extreme values — stays closer to the bulk of the data. Therefore, the mean exceeds the median. This is the opposite of a left-skewed distribution, where the mean is pulled below the median.

Adding a constant \(k\) to every value shifts the entire distribution by \(k\), so the mean increases by \(k\): \(\bar{x}_{\text{new}} = \bar{x} + k\). However, the spread is unchanged because differences between values remain the same. Therefore, the standard deviation and variance are unaffected: \(s_{\text{new}} = s\). Only the center shifts, not the spread.

Original mean: \(\frac{86}{7} \approx 12.3\). After removing \(52\): \(\frac{34}{6} \approx 5.7\) — a large drop. Original median (4th of 7 values): \(6\). New median (average of 3rd and 4th of 6 values): \(\frac{4+6}{2} = 5\) — only a small change. This illustrates that the mean is sensitive to outliers while the median is resistant. The mean is pulled much more strongly by the extreme value \(52\).

A bimodal distribution has two distinct peaks (local modes), which appear as two clusters of tall bars separated by a lower region in a histogram. A single tall bar in the center describes a symmetric, unimodal distribution. Bars that increase from left to right suggest right-skewness. Equal-height bars suggest a uniform distribution.

Standard deviation measures spread around the mean. Since every value equals \(10\), the mean is \(\bar{x} = 10\), and every deviation from the mean is \(10 - 10 = 0\). Therefore \(s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}} = \sqrt{\frac{0}{4}} = 0\). A standard deviation of \(0\) means there is no variability — all values are identical.

The incorrect entry of \(91\) inflated the total sum by \(91 - 64 = 27\) points above the true value. Correcting it reduces the sum by \(27\), and since \(\bar{x} = \frac{\sum x_i}{n}\), the mean decreases. The class size is not needed to determine the direction of change: any correction that lowers one value while keeping everything else the same must lower the mean.

In a dotplot, each dot represents one observation. When dots are stacked vertically above a value, each dot represents one occurrence of that value. Five dots above \(18\) means \(18\) was recorded five times. The stacking tells us only about frequency at that value — it does not indicate whether \(18\) is the mean, median, or an outlier.

The $IQR = Q_3 - Q_1$ captures the spread of the middle 50 percent of the data, from the 25th percentile to the 75th percentile. Because it ignores the lowest and highest 25 percent, the $IQR$ is resistant to outliers. The total spread from min to max is the range, and average distance from the mean relates to standard deviation.

For a linear transformation \(y = ax + b\), the new mean is \(\bar{y} = a\bar{x} + b\) and the new standard deviation is \(s_y = |a| \cdot s_x\). Here \(a = 3\) and \(b = -5\), so \(\bar{y} = 3(20) - 5 = 55\) and \(s_y = 3(4) = 12\). The constant \(-5\) shifts the distribution but does not change its spread, so it affects the mean but not the standard deviation.

First, compute $IQR = Q_3 - Q_1 = 30 - 15 = 15$. The fences are: lower fence $= Q_1 - 1.5(IQR) = 15 - 22.5 = -7.5$ and upper fence $= Q_3 + 1.5(IQR) = 30 + 22.5 = 52.5$. Values outside \((-7.5,\, 52.5)\) are outliers. Checking: \(0 > -7.5\) (not an outlier), \(10 > -7.5\) (not an outlier), \(50 < 52.5\) (not an outlier), \(55 > 52.5\) (outlier). Only \(55\) exceeds the upper fence.

The sample standard deviation formula is \(s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}\). With \(n = 10\): \(s = \sqrt{\frac{360}{9}} = \sqrt{40} \approx 6.32\). A common error is dividing by \(n = 10\) instead of \(n - 1 = 9\), which gives \(\sqrt{36} = 6\) (choice A). The denominator \(n - 1\) is used because we are estimating a population parameter from a sample, and dividing by \(n\) would underestimate the true variability.

Compute the \(z\)-score for each: \(z_A = \frac{70 - 60}{5} = 2.0\) and \(z_B = \frac{70 - 80}{10} = -1.0\). Since \(|z_A| = 2.0 > |z_B| = 1.0\), the score of \(70\) is more extreme relative to Distribution A. Choice C is a tempting shortcut but is incorrect — a smaller standard deviation does not automatically make a value extreme; it depends on how far the score falls from that distribution's mean.

The original total sum is \(n \cdot \bar{x} = 9 \times 24 = 216\). The error caused the sum to be inflated by \(30 - 12 = 18\). The corrected sum is \(216 - 18 = 198\), giving a corrected mean of \(\frac{198}{9} = 22\). This method uses the relationship \(\sum x_i = n\bar{x}\) to adjust without needing every individual value.

Since \(120 > 100\), the new value becomes the maximum, so the maximum is guaranteed to change from \(100\) to \(120\). The $IQR = Q_3 - Q_1 = 30$ may be unaffected because \(Q_1\) and \(Q_3\) depend on the middle portion of the data and are resistant to extreme values. Whether the median changes depends on the sample size, which is not given. The maximum, however, must update whenever a new value exceeds it.

Both samples have mean \(\bar{x} = \frac{150}{5} = 30\). Their spreads differ greatly. For Sample 1: \(\sum(x_i - 30)^2 = 400 + 100 + 0 + 100 + 400 = 1000\), so \(s_1 = \sqrt{\frac{1000}{4}} = \sqrt{250} \approx 15.8\). For Sample 2: \(\sum(x_i - 30)^2 = 4+1+0+1+4 = 10\), so \(s_2 = \sqrt{\frac{10}{4}} \approx 1.6\). Same center, very different spread.

The frequencies \(1, 4, 10, 18, 22, 18, 10, 4, 1\) rise to a single peak at the middle interval and decrease symmetrically on both sides — a classic bell shape. This indicates an approximately symmetric, unimodal distribution. Right-skewness would show a slow taper on the right; left-skewness would show a slow taper on the left. A bimodal distribution would have two separate peaks rather than one central peak.

A histogram groups data into consecutive intervals (bins) and displays how many observations (frequency) or what proportion (relative frequency) fall within each bin. It does NOT show individual data values — that is what a dotplot or stemplot does. Scatterplots, not histograms, display relationships between two quantitative variables.

The mode is the value that appears most frequently. In \(\{3, 5, 7, 5, 9, 3, 5, 7\}\): the value \(3\) appears twice, \(5\) appears three times, \(7\) appears twice, and \(9\) appears once. Since \(5\) has the highest frequency, it is the mode.

A dotplot places one dot above the number line for each data value, so all individual values remain visible. A histogram groups values into intervals (bins), which causes individual values to be lost. For large datasets, histograms are preferred because dotplots become cluttered, making choice A incorrect. Grouping into intervals (choice B) is what histograms do, not dotplots.

With \(n = 5\) values (an odd number), the median is the middle value at position \(\frac{n+1}{2} = 3\). The 3rd value in \(\{12, 15, 18, 21, 27\}\) is \(18\). The value \(18.6\) is the mean (\(\frac{12+15+18+21+27}{5} = \frac{93}{5} = 18.6\)), which is a common distractor but is not the median.

Standard deviation quantifies how much data values typically deviate from the mean. A larger standard deviation means greater spread; a smaller standard deviation means values cluster close to the mean. Choice D describes the range, which is a different (simpler) measure of spread. Choice A describes measures of center such as the mean or median, not spread.

In a perfectly symmetric, unimodal distribution — such as the normal distribution — the mean, median, and mode all coincide at the center. Skewness is what causes these three measures to differ: in a right-skewed distribution the mean exceeds the median, and in a left-skewed distribution the mean falls below the median. The mode is always at the peak, not the tail.

Range \(=\) maximum \(-\) minimum. It is the simplest measure of spread but is sensitive to extreme values because it uses only the two most extreme data points. The difference \(Q_3 - Q_1\) (choice B) is the interquartile range (IQR), which measures the spread of the middle 50% of data. The average distance from the mean (choice A) is related to mean absolute deviation, not the range.

In a right-skewed distribution, extreme high values pull the mean in the direction of the tail (upward), while the median — which depends only on rank, not magnitude — is far less affected. Therefore \(\bar{x} > M\) in a right-skewed distribution. This is why the median is preferred over the mean when summarizing skewed income data.

In a left-skewed distribution, the tail extends to the left. Extreme low values pull the mean downward, so \(\bar{x}\) falls below the median. The mode is typically near the peak on the right side of the distribution. The typical order is \(\bar{x} < M < \text{mode}\). This is the mirror image of a right-skewed distribution, where \(\text{mode} < M < \bar{x}\).

With \(n = 7\), the median is the 4th value, which is \(40\). After adding \(200\), the ordered dataset becomes \(\{10, 20, 30, 40, 50, 60, 70, 200\}\) with \(n = 8\). For an even number of values, the median is the average of the 4th and 5th values: \(\frac{40 + 50}{2} = 45\). Choice A (\(40\)) is wrong because the median does shift when \(n\) changes and the middle positions move. Choice D (\(60\)) is the new mean, illustrating why the median is preferred over the mean when outliers are present.

In a relative frequency histogram, bar heights represent the proportion (or percentage) of observations in each interval, not the raw count. This allows meaningful comparison between datasets of different sizes. All bar heights sum to \(1\) (or \(100\%\)). Choice A describes a standard frequency histogram. Choice D describes a cumulative frequency distribution, which is a different display entirely.

A percentile rank of 80 means the student's score is at or above approximately 80% of the scores in the reference distribution. It says nothing about the percent of questions answered correctly (choice A), the number of points above the mean (choice C), or a specific rank — a rank of 80th from the top (choice D) depends on the total number of test takers, which is unrelated to the percentile.

The \(z\)-score formula is \(z = \dfrac{x - \bar{x}}{s} = \dfrac{66 - 50}{8} = \dfrac{16}{8} = 2.0\). A \(z\)-score of \(2.0\) means the value \(66\) is \(2\) standard deviations above the mean. Choice A (\(z = 1.5\)) would correspond to \(x = 50 + 1.5(8) = 62\), not \(66\). Choice C (\(z = 16\)) is the raw difference \(x - \bar{x}\), before dividing by \(s\).

The median is resistant to outliers because it depends only on the rank (order) of values, not their magnitudes. A single extreme value can substantially shift the mean, since the mean uses every value in its calculation. The midrange (average of the minimum and maximum) is the least resistant measure of center — it is entirely determined by the two most extreme values in the dataset.

Standard deviation measures how far data values typically fall from their mean. Class B has scores ranging from \(50\) to \(100\) — a spread of \(50\) points — while Class A's scores span only \(10\) points (\(75\) to \(85\)). Greater spread from the mean means a larger standard deviation. Choice C is incorrect: two datasets can share the same mean but have very different standard deviations.

When every data value is multiplied by a constant \(c\), the mean scales as \(\bar{x}_{\text{new}} = c\bar{x}\) and the standard deviation scales as \(s_{\text{new}} = |c| \cdot s\), since all deviations from the mean are stretched by the same factor. Multiplying by \(3\) gives \(s_{\text{new}} = 3s\). Choice D (\(\times 9\)) would be correct for the variance (\(s^2_{\text{new}} = 9s^2\)), since variance is the square of the standard deviation.

The total number of observations equals the sum of all bar frequencies: \(4 + 9 + 15 + 8 + 4 = 40\). The number of bars (\(5\)) is the number of intervals, not the sample size. The tallest bar frequency (\(15\)) represents the modal class, not the total count. Choice D (\(8\)) is the frequency of the 4th bar, not the total.

A histogram with one central peak and bars tapering symmetrically toward both tails is described as unimodal and symmetric — often called approximately bell-shaped or normal. A uniform distribution (choice A) would have bars of roughly equal height with no visible peak. A bimodal distribution (choice B) would have two distinct peaks. A right-skewed distribution (choice D) would have a longer tail extending to the right with the peak shifted toward the left.

Adding a constant to every value shifts the distribution's center but does not change its spread. Therefore the variance and standard deviation remain unchanged after adding \(10\). The original standard deviation is \(s = \sqrt{36} = 6\), and it stays \(6\) after the shift. Choice A (\(46\)) incorrectly adds the constant to the variance. Choice D (\(16\)) incorrectly adds the constant directly to the standard deviation.

The combined mean is a weighted average: \(\bar{x} = \dfrac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \dfrac{20(75) + 30(85)}{50} = \dfrac{1500 + 2550}{50} = \dfrac{4050}{50} = 81\). A simple unweighted average of \(75\) and \(85\) gives \(80\) (choice A), but this ignores the fact that Group 2 has more students and should carry more weight in the combined mean.

The sample variance formula is \(s^2 = \dfrac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n - 1}\). With \(s = 4\), we have \(s^2 = 16\). Solving for the sum of squared deviations: \(\sum(x_i - \bar{x})^2 = s^2(n - 1) = 16 \times 4 = 64\). Choice B (\(80\)) results from incorrectly dividing by \(n = 5\) instead of \(n - 1 = 4\). Choice D (\(16\)) is just the sample variance \(s^2\), not the sum of squared deviations.

With \(n = 8\), \(Q_1 = \dfrac{8 + 12}{2} = 10\) and \(Q_3 = \dfrac{22 + 25}{2} = 23.5\), so $IQR = 23.5 - 10 = 13.5$. The upper fence is $Q_3 + 1.5 \times IQR = 23.5 + 1.5(13.5) = 23.5 + 20.25 = 43.75$. Since \(60 > 43.75\), the value \(60\) is classified as an outlier. Being the maximum value (choice A) does not exempt a point from being an outlier. The $1.5 \times IQR$ rule uses quartiles, not standard deviations (choice D).

Changing a recorded score from \(74\) to \(44\) decreases the total sum by \(30\), lowering the mean by \(\frac{30}{30} = 1\) point to \(\bar{x} = 73\). The corrected score (\(44\)) is now far from the new mean (\(73\)), introducing a large squared deviation and inflating the standard deviation. So the mean decreases while the standard deviation increases — a single data entry error can simultaneously affect both center and spread.

Using \(z = \dfrac{x - \bar{x}}{s}\): \(-2.5 = \dfrac{x - 100}{15}\), so \(x = 100 + (-2.5)(15) = 62.5\). In a bell-shaped distribution, values with \(|z| > 2\) are considered unusual (the Empirical Rule places about 95% of data within \(2\) standard deviations of the mean). Choice C incorrectly applies a positive deviation: \(100 + 2.5(15) = 137.5\). Choice D is wrong — a negative \(z\)-score simply means below average, not automatically extreme.

The combined mean is \(\bar{x} = \dfrac{10(50) + 10(70)}{20} = 60\). Even though each group has \(s = 5\) internally, the combined dataset has values scattered around both \(50\) and \(70\) — far from the combined mean of \(60\). This between-group spread adds substantial variation beyond the within-group standard deviation of \(5\). Simply averaging the two standard deviations (choices A and D) ignores the fact that the group means differ by \(20\) points. Adding standard deviations using the Pythagorean theorem (choice B) is also not a valid technique here.

The mode is the value that occurs most often in the dataset. The middle value of an ordered dataset is the median (choice B), the arithmetic average is the mean (choice C), and the difference between the maximum and minimum is the range (choice D). A dataset may have no mode, one mode, or multiple modes.

In a right-skewed distribution, the long tail extends toward the right (toward higher values), while most data are clustered on the left side near the peak. Choice A describes left skew (negative skew). Choice C describes a symmetric distribution, and choice D describes a bimodal distribution.

In a dotplot, every dot corresponds to exactly one data value. If three dots appear above the number \(7\), it means the value \(7\) occurs three times in the dataset. This makes dotplots ideal for displaying small datasets and preserving every individual observation, unlike histograms which group values into intervals.

With \(n = 5\) values (an odd count), the median is the value in position \(\frac{n+1}{2} = \frac{6}{2} = 3\) in the ordered list. The third value is \(9\), so the median is \(9\). Choice B (\(8.8\)) is the mean, not the median: \(\frac{3+7+9+11+14}{5} = \frac{44}{5} = 8.8\). The median requires only identifying the center position, not averaging.

Standard deviation is computed as \(s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}\). This equals zero only when every squared deviation \((x_i - \bar{x})^2 = 0\), which requires every \(x_i = \bar{x}\) — meaning all values are identical. A mean of zero (choice C) does not eliminate variability; for example, \(\{-3, 0, 3\}\) has mean \(0\) but \(s > 0\). A two-value dataset (choice D) generally has \(s > 0\) unless both values happen to be equal.

The $IQR = Q_3 - Q_1$ is based only on the middle 50% of the data, so extreme values in the tails have no effect on it. The range depends entirely on the two most extreme values, making it the most sensitive measure of spread. Standard deviation and variance involve every value's squared deviation from the mean, so a single extreme outlier can substantially inflate them.

In a frequency histogram, the height of each bar equals the count (frequency) of data values in that class interval. A bar height of \(15\) means exactly \(15\) values fall between \(30\) and \(40\). This differs from a relative frequency histogram, where bar heights represent proportions (such as \(0.15\) or \(15\%\)) rather than raw counts.

In a perfectly symmetric unimodal distribution — such as a normal (bell-shaped) distribution — the single peak (mode), the balance point (mean), and the center position (median) all coincide exactly at the axis of symmetry. Choices A and B describe skewed distributions: right skew pulls the mean above the median, while left skew pulls the mean below. Choice D would imply asymmetry between the mean and the mode/median.

Replacing \(15\) with \(150\) changes the sum from \(50\) to \(185\), so the new mean is \(\frac{185}{5} = 37\) — a large increase. The new ordered set is \(\{5, 8, 10, 12, 150\}\), and the middle value (third of five) is still \(10\), so the median is unchanged. This illustrates the median's resistance: only the count and position of values matters for the median, not how extreme the outlier is.

Two distinct clusters produce two peaks in the distribution, making it bimodal. A unimodal distribution (choice A) has a single peak. A right-skewed distribution (choice C) has one peak with a long tail extending to the right — not two separate clusters. A uniform distribution (choice D) has roughly equal frequency across all values with no peaks.

The frequency pattern \(2, 5, 11, 5, 2\) is symmetric: the first bar equals the last, the second equals the fourth, and the tallest bar sits in the center. This mirror-image arrangement produces a bell-shaped, symmetric distribution. Skewed distributions have more area in one tail than the other, and a uniform distribution would show bars of approximately equal height throughout.

Adding a constant to every value shifts the entire distribution to the right by \(c\), increasing the mean (and median and mode) by \(10\). However, measures of spread — standard deviation, variance, and IQR — depend on the distances between values, not their absolute positions. Since all values shift by the same amount, the gaps between them are unchanged. Only center changes; spread does not.

First compute $IQR = Q_3 - Q_1 = 42 - 18 = 24$, then $1.5 \times IQR = 36$. The lower fence is \(Q_1 - 36 = 18 - 36 = -18\) and the upper fence is \(Q_3 + 36 = 42 + 36 = 78\). Any value below \(-18\) or above \(78\) is flagged as an outlier. Among the choices: \(-10 > -18\) (not an outlier), \(5 > -18\) (not an outlier), \(60 < 78\) (not an outlier), but \(80 > 78\), so \(80\) is the outlier.

All relative frequencies in a distribution must sum to \(1.00\) because they account for the entire dataset. The sum of the first five bars is \(0.05 + 0.15 + 0.30 + 0.25 + 0.10 = 0.85\). Therefore the sixth bar must have relative frequency \(1.00 - 0.85 = 0.15\). This constraint — that all relative frequencies sum to \(1\) — is a useful check when constructing or interpreting relative frequency histograms.

The original sum is \(10 \times 45 = 450\). Adding \(x = 90\) gives a new sum of \(450 + 90 = 540\). The new mean is \(\frac{540}{11} \approx 49.09 \approx 49.1\). Choice C (\(52.5\)) is the simple average \(\frac{45 + 90}{2}\), which incorrectly treats both as having equal weight. Because \(x = 90\) is only one of eleven values, it raises the mean modestly rather than dramatically.

All four datasets have mean \(10\). Choice A has \(s = 0\) (no variation). For the others, compare how far values deviate from the mean. Choice C: deviations are \(-1, 0, 0, 0, 1\), giving \(s \approx 0.71\). Choice B: deviations are \(-2, -1, 0, 1, 2\), giving \(s \approx 1.58\). Choice D: deviations are \(-6, -3, 0, 3, 6\), giving \(s = \sqrt{\frac{90}{4}} \approx 4.74\). Dataset D has values most spread from the mean and thus the largest standard deviation.

In a right-skewed distribution, the mean is pulled upward toward the long upper tail by the extreme high-income values, making it unrepresentatively high for a 'typical' household. The median, as the middle value of the ordered dataset, is determined only by position — not by how extreme the outliers are — making it resistant to skew. Note that range (choice D) is a measure of spread, not center, so it cannot represent a typical value.

The original sum is \(8 \times 25 = 200\). The correction removes the erroneous \(13\) and adds the correct \(37\): new sum \(= 200 - 13 + 37 = 224\). The corrected mean is \(\frac{224}{8} = 28.00\). This method is efficient because knowing the sum and sample size is sufficient — you do not need to know the other seven values individually. A common error is computing \(\frac{25 + (37-13)}{1}\) without adjusting the total sum first.

The original sum is \(n \times 40\). Removing one value equal to \(40\) gives a new sum of \(n \times 40 - 40 = 40(n-1)\). Dividing by the new count: \(\frac{40(n-1)}{n-1} = 40\). The mean is unchanged. Intuitively, removing a value exactly at the mean does not tip the balance in either direction. Choice D is incorrect because the algebraic result holds for any dataset satisfying the given conditions — no additional information is needed.

The transformation \(z = \frac{x - \mu}{\sigma}\) is called standardization. Subtracting \(\mu = 70\) shifts every value so the new mean is \(\frac{\mu - 70}{5} = 0\). Dividing by \(\sigma = 5\) rescales the spread so the new standard deviation is \(\frac{\sigma}{5} = \frac{5}{5} = 1\). This \(z\)-score transformation always produces mean \(0\) and standard deviation \(1\) regardless of the original distribution's shape or parameters.

For a linear transformation \(y = ax + b\), the standard deviation scales by \(|a|\) and is unaffected by the additive constant \(b\). Here \(a = 3\) and \(b = -7\), so \(s_Q = |3| \times s_P = 3 \times 15.81 \approx 47.43\). Choice A ignores the multiplication entirely. Choice B subtracts \(7\) from the standard deviation, which is incorrect — adding or subtracting a constant never changes standard deviation because it shifts all values equally without altering any gaps between them.

Using the computational formula: $s^2 = \frac{\sum x_i^2 - \frac{(\sum x_i)^2}{n}}{n-1} = \frac{2500 - \frac{120^2}{6}}{6-1} = \frac{2500 - \frac{14400}{6}}{5} = \frac{2500 - 2400}{5} = \frac{100}{5} = 20$. Choice B (\(16.67\)) results from dividing the numerator by \(n = 6\) instead of \(n - 1 = 5\), which would give the population variance rather than the sample variance. Choice D (\(100\)) is the uncorrected numerator before dividing by \(n - 1\).

The upper fence is $Q_3 + 1.5 \times IQR = 30 + 1.5(16) = 30 + 24 = 54$. Since \(95 > 54\), the value \(95\) is an outlier. Since \(34 < 54\), the value \(34\) is not flagged as an outlier. Choice A uses an incorrect fence of \(60\), which would result from \(1.5 \times 20 = 30\), as though $IQR = 20$. Always recompute $1.5 \times IQR$ from the actual $IQR$ before drawing conclusions about outlier status.

The combined mean requires a weighted average: $\bar{x}_{combined} = \frac{n_A \bar{x}_A + n_B \bar{x}_B}{n_A + n_B} = \frac{15(78) + 25(86)}{40} = \frac{1170 + 2150}{40} = \frac{3320}{40} = 83$. The simple average \(\frac{78 + 86}{2} = 82\) (choice B) is incorrect because it treats both classes as equal in size. Since Class B is larger, its higher mean of \(86\) receives more weight, pulling the combined mean above the midpoint of \(82\).

The new mean is \(\frac{20(50) + 95}{21} = \frac{1095}{21} \approx 52.1\), which exceeds \(50\), so the mean increases. The \(z\)-score of the new value relative to the original distribution is \(\frac{95 - 50}{8} = 5.625\) — nearly six standard deviations above the mean. Adding a value this extreme dramatically increases the spread, so the standard deviation also increases. Choice A would only apply if the new value were closer to the mean than the typical spread, reducing variability — clearly not the case here.

The median is the middle value of an ordered dataset with an odd number of values. With \(n = 7\) values, the median is the \(\frac{7+1}{2} = 4\)th value, which is \(9\). The value \(7\) is the 3rd value, not the middle. The mean is \(\frac{3+5+7+9+12+14+18}{7} = \frac{68}{7} \approx 9.7\), which differs from the median.

Favorite school subject is categorical (nominal) data, so arithmetic operations like addition or ordering are not meaningful. The mode — the most frequently chosen category — is the only appropriate measure of center. Mean and median require numerical values that can be ordered or averaged, making them inapplicable here. Standard deviation is a measure of spread, not center.

The mode is the value that appears most frequently. In this dataset, \(4\) appears three times, which is more than any other value (\(2\) appears twice; \(3\), \(5\), and \(6\) each appear once). Therefore the mode is \(4\). The value \(2\) is a common distractor because it also appears more than once, but it occurs less often than \(4\).

A distribution is right-skewed (positively skewed) when its tail extends to the right. The bulk of the data sits on the left, and a few unusually large values pull the tail rightward. Left-skewed (negatively skewed) is the opposite: the peak is on the right with a tail extending left. A symmetric distribution has tails of equal length on both sides, and a bimodal distribution has two distinct peaks.

Standard deviation measures spread — specifically, how far data values typically deviate from the mean. A small standard deviation means values cluster tightly around the mean; a large standard deviation means they are more dispersed. Choice C describes the range, not standard deviation. Choice A describes the median, and Choice D describes the mode.

The IQR is \(Q_3 - Q_1\), where \(Q_1\) is the 25th percentile and \(Q_3\) is the 75th percentile. It represents the spread of the middle \(50\%\) of the data. Choice C is the range, which measures total spread but is sensitive to outliers. Choices B and D each capture only one half of the middle portion rather than the full interquartile interval.

The modal class is the class interval with the highest frequency. Here, $[5, 10)$ has frequency \(8\), which is greater than \(3\), \(5\), and \(2\) for the other classes. In a histogram, the modal class corresponds to the tallest bar. Note that the modal class identifies the interval containing the most values, not the exact mode of individual data points.

The mean is \(\bar{x} = \frac{\sum x_i}{n} = \frac{10 + 15 + 15 + 20 + 25}{5} = \frac{85}{5} = 17\). The median of this dataset is \(15\) (the 3rd value in order), which is a common distractor when mean and median are confused. The mean \(17\) is pulled above the median because the value \(25\) is larger than the rest, illustrating sensitivity of the mean to larger values.

With \(n = 6\) values (even), the median is the average of the \(\frac{n}{2} = 3\)rd and \(\frac{n}{2} + 1 = 4\)th ordered values. The 3rd value is \(8\) and the 4th value is \(10\), so the median is \(\frac{8 + 10}{2} = 9\). A common error is selecting \(8\) or \(10\) alone as the median. With an even number of values, the median is always the mean of the two middle values.

In a right-skewed distribution, a few unusually large values pull the mean toward higher values, while the median — being resistant to extreme values — stays closer to the bulk of the data. Therefore the mean is typically greater than the median. In a left-skewed distribution the opposite holds: extreme low values pull the mean below the median. In a symmetric distribution the mean and median are approximately equal.

The original sum is \(9 \times 30 = 270\). After removing the outlier \(90\), the new sum is \(270 - 90 = 180\), and the new mean is \(\frac{180}{8} = 22.5\). Because the removed value (\(90\)) was larger than the original mean (\(30\)), removing it lowers the mean. The general rule: removing a value above the mean decreases the mean; removing a value below the mean increases it.

Standard deviation measures how spread out values are from the mean. Dataset A has values within \(2\) units of \(50\), so its deviations are \(\{-2, -1, 0, 1, 2\}\). Dataset B has values as far as \(20\) units from \(50\), with deviations \(\{-20, -10, 0, 10, 20\}\). Since B's deviations are far larger, Dataset B has the larger standard deviation. No calculation is required — visual inspection of spread around the common mean suffices.

When every value is multiplied by a constant \(k\), the standard deviation is also multiplied by \(|k|\). Multiplying by \(k = 3\) scales all deviations from the mean by a factor of \(3\), so the new standard deviation is \(3s\). In contrast, adding a constant shifts all values without changing spread, leaving \(s\) unchanged. Choice B confuses multiplication with addition of the constant.

The total number of data values is \(5 + 10 + 8 + 7 = 30\). The relative frequency of the class with frequency \(10\) is \(\frac{10}{30} = \frac{1}{3} \approx 0.333\). The value \(0.25\) would result if the total were \(40\), and \(0.40\) would result if the total were \(25\). Relative frequency is always a proportion between \(0\) and \(1\) expressing what fraction of the whole a given class represents.

When the mean exceeds the median (\(62 > 55\)), the distribution is likely right-skewed. High outliers or a long right tail pull the mean above the median while the median remains close to the center of the bulk of data. In a left-skewed distribution the mean would fall below the median. In a symmetric distribution the mean and median are approximately equal. This relationship is a useful diagnostic for skewness.

The IQR is \(Q_3 - Q_1 = 40 - 15 = 25\). It measures the spread of the middle \(50\%\) of the data. Choice A (\(65\)) is the range: \(\text{Max} - \text{Min} = 70 - 5 = 65\). Choice C (\(15\)) is \(Q_1\) itself, not a spread measure. Choice D (\(20\)) is \(\text{Median} - Q_1 = 25 - 15\), which is only one side of the interquartile interval.

With \(n = 10\) values (even), the median is the average of the 5th and 6th values. Counting through the ordered list: the 5th value is \(7\) and the 6th value is also \(7\), so the median is \(\frac{7 + 7}{2} = 7\). A common error is choosing \(7.5\), as if the two middle values were \(7\) and \(8\), but both positions 5 and 6 hold the value \(7\). Careful counting in dotplots prevents this mistake.

The upper fence is \(Q_3 + 1.5 \times \text{IQR} = 50 + 1.5(30) = 50 + 45 = 95\). Any value above \(95\) is a potential high outlier. The lower fence is \(Q_1 - 1.5 \times \text{IQR} = 20 - 45 = -25\). Choice A (\(80\)) results from adding \(Q_3 + \text{IQR}\) without the factor of \(1.5\). Choice B (\(75\)) is \(Q_3 + Q_1 = 50 + 25\), which has no statistical meaning here.

The squared deviations from \(\bar{x} = 5\) are \((1-5)^2 = 16\), \((3-5)^2 = 4\), \((5-5)^2 = 0\), \((7-5)^2 = 4\), \((9-5)^2 = 16\). Their sum is \(16 + 4 + 0 + 4 + 16 = 40\). The sample variance uses \(n - 1 = 4\) in the denominator: \(s^2 = \frac{40}{4} = 10\), so \(s = \sqrt{10}\). Using \(n = 5\) instead of \(n - 1 = 4\) gives \(s = \sqrt{8}\), which is the population standard deviation — a common error on AP Statistics.

The combined mean must be a weighted average accounting for different group sizes: $\bar{x}_{\text{combined}} = \frac{n_A \bar{x}_A + n_B \bar{x}_B}{n_A + n_B} = \frac{20(75) + 10(90)}{30} = \frac{1500 + 900}{30} = \frac{2400}{30} = 80$. Choice A (\(82.5\)) is the simple average \(\frac{75 + 90}{2}\), which ignores the fact that Class A has twice as many students as Class B. Combined means must always be weighted by group size.

After inserting \(6\), the new ordered dataset is \(\{3, 5, 6, 7, 9, 11\}\) with \(n = 6\) values. The median is the average of the 3rd and 4th values: \(\frac{6 + 7}{2} = 6.5\). The median shifted from \(7\) to \(6.5\) because inserting a value below the old median changes which positions form the middle pair. Choice B (\(7\)) incorrectly assumes the median is unchanged; adding any new value changes \(n\) and thus which positions are central.

For a linear transformation \(y = ax + b\), the new mean is \(\mu_y = a\mu_x + b = 2(40) - 5 = 75\), and the new standard deviation is \(\sigma_y = |a| \cdot \sigma_x = 2(8) = 16\). Adding or subtracting a constant shifts the distribution but does not change spread, so only the multiplier \(2\) affects the standard deviation. Choice C gives mean \(80\) by applying only the multiplication and forgetting to subtract \(5\); Choice B incorrectly applies \(-5\) to the standard deviation as well.

First find the mean: \(\bar{x} = \frac{25}{5} = 5\). Then apply the computational formula for sample variance: \(s^2 = \frac{\sum x_i^2 - n\bar{x}^2}{n - 1} = \frac{145 - 5(25)}{4} = \frac{145 - 125}{4} = \frac{20}{4} = 5\). Therefore \(s = \sqrt{5}\). A common error is dividing by \(n = 5\) instead of \(n - 1 = 4\), which gives \(s^2 = 4\) and \(s = 2\) (Choice B) — this is the population standard deviation, not the sample standard deviation.

The original sum is \(7 \times 12 = 84\). Replacing the erroneous value \(5\) with the correct value \(19\) changes the total sum by \(19 - 5 = 14\). The new sum is \(84 + 14 = 98\), and the new mean is \(\frac{98}{7} = 14\). The key insight is that correcting one value changes only the total sum while \(n\) stays at \(7\). Choice A (\(13\)) results from an arithmetic error, and Choice C (\(14.5\)) incorrectly uses \(n = 8\) as if a new value were added rather than corrected.

With \(n = 12\) and \(p = 75\): \(L = \frac{75}{100} \times 12 = 9\). Since \(L = 9\) is a whole number, the 75th percentile is the average of the 9th and 10th values. Counting the ordered list: the 9th value is \(11\) and the 10th value is \(12\), so \(P_{75} = \frac{11 + 12}{2} = 11.5\). Choosing \(11\) or \(12\) alone corresponds to reading only one value rather than averaging the two. When \(L\) is a whole number, the AP convention requires averaging the \(L\)th and \((L+1)\)th values.

In a dotplot, each dot is placed above the number line at the position corresponding to its value. The number of dots above a given position equals the frequency of that value in the dataset. Dotplots display individual data points, making it easy to see shape, center, and spread for small datasets.

The median is resistant to outliers because it depends only on the middle value(s) of an ordered dataset, regardless of how extreme other values are. The mean incorporates every value in its calculation, so one very large or very small value can pull it substantially away from the center of the bulk of the data.

The mode is the value that appears most frequently. In this dataset, \(5\) appears twice, \(7\) appears three times, and all other values appear once. Because \(7\) appears more often than any other value, there is exactly one mode: \(7\). Choice B is incorrect because a dataset has two modes only when two values are tied for the highest frequency, which is not the case here.

The range is calculated as \(\text{Max} - \text{Min} = 19 - 3 = 16\). The maximum value is \(19\) and the minimum is \(3\). Choice A (\(13\)) results from subtracting \(6\) from \(19\) instead of using the true minimum of \(3\).

In a right-skewed distribution, the tail extends toward larger values on the right. Because the mean is sensitive to extreme values, those large values in the right tail pull the mean above the median. This gives the characteristic relationship \(\text{Mean} > \text{Median}\). The opposite holds for left-skewed distributions.

The \(\text{IQR} = Q_3 - Q_1\) measures the spread of the middle \(50\%\) of the data. It is a resistant measure of spread because it is unaffected by values outside the first and third quartiles. Choice C describes the range, which uses the most extreme values and is not resistant to outliers.

Standard deviation measures spread around the mean. A value of \(s = 0\) means every data point has zero deviation from the mean, so every value equals the mean and all values are identical. For example, \(\{5, 5, 5, 5\}\) has \(s = 0\) and \(\bar{x} = 5 \neq 0\), which disproves choice A. A dataset with \(s = 0\) can contain multiple identical values, so choice C is not necessarily true.

A histogram groups continuous quantitative data into intervals (bins) and uses bar height to show frequency. Unlike a bar chart, which displays categorical data with gaps between bars, histogram bars are adjacent to reflect continuous data. A dotplot plots individual values rather than grouped intervals.

The original sum is \(n \cdot \bar{x} = 9 \times 14 = 126\). Adding \(x = 23\) gives a new sum of \(126 + 23 = 149\). The new mean is \(\frac{149}{10} = 14.9\). Choice D (\(18.5\)) results from incorrectly averaging just the old mean and the new value: \(\frac{14 + 23}{2} = 18.5\), which ignores that the dataset already contains \(9\) values.

Adding a constant to every value shifts the entire distribution by \(7\), so the mean increases by \(7\). However, standard deviation measures spread — the distances between data values. Since every value shifts by the same amount, those distances do not change and the standard deviation remains the same. Choice A incorrectly assumes that shifting a distribution also stretches it.

For an even number of values, the median is the average of the two middle values. With \(n = 6\), the two middle values are the 3rd and 4th: \(11\) and \(14\). The median is \(\frac{11 + 14}{2} = 12.5\). Choices A and C report the individual middle values rather than their average.

First, \(\text{IQR} = Q_3 - Q_1 = 28 - 12 = 16\). The upper fence is \(Q_3 + 1.5 \times \text{IQR} = 28 + 1.5(16) = 28 + 24 = 52\). Any value above \(52\) is a potential outlier. Choice B (\(44\)) results from using \(Q_3 + \text{IQR}\) without the \(1.5\) multiplier.

When every value is multiplied by a constant \(k\), the standard deviation is multiplied by \(|k|\). So the new standard deviation is \(4 \times 5 = 20\). Choice D (\(100\)) confuses variance with standard deviation: the new variance is \(k^2 s^2 = 16 \times 25 = 400\), giving $s_{new} = \sqrt{400} = 20$, not \(100\).

A distribution that rises to a single central peak and falls off equally on both sides is approximately symmetric and unimodal, often described as bell-shaped. Right-skewed distributions have a longer tail extending to the right, left-skewed distributions have a longer tail to the left, and bimodal distributions have two distinct peaks.

The mean is \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{3(2) + 5(4) + 2(6)}{3 + 5 + 2} = \frac{6 + 20 + 12}{10} = \frac{38}{10} = 3.8\). Choice A (\(4.0\)) results from averaging the three distinct values without weighting by frequency: \(\frac{2 + 4 + 6}{3} = 4.0\).

Standard deviation measures how far values typically deviate from the mean. Dataset B spans from \(2\) to \(18\) (range \(= 16\)), indicating far greater spread than Dataset A, which spans from \(8\) to \(12\) (range \(= 4\)). Choice C is a common misconception: equal means do not imply equal standard deviations.

When the mean exceeds the median (\(72 > 65\)), large values in the right tail are pulling the mean upward relative to the median. This pattern is characteristic of a right-skewed distribution. A left-skewed distribution would produce \(\text{Mean} < \text{Median}\), and a symmetric distribution would have \(\text{Mean} \approx \text{Median}\).

The population mean is \(\mu = \frac{2+4+4+6+9}{5} = \frac{25}{5} = 5\). The squared deviations are \((2-5)^2 = 9\), \((4-5)^2 = 1\), \((4-5)^2 = 1\), \((6-5)^2 = 1\), \((9-5)^2 = 16\), summing to \(28\). The population variance is \(\sigma^2 = \frac{28}{5} = 5.6\), so \(\sigma = \sqrt{5.6}\). Choice A uses the denominator \(n - 1 = 4\) (sample variance formula), giving \(\frac{28}{4} = 7\) and \(\sqrt{7}\).

The new sum is \(6(20) + 20 = 140\), so $\bar{x}_{new} = \frac{140}{7} = 20$. For the standard deviation: the original sum of squared deviations is \(s^2(n-1) = 25(5) = 125\). The new value contributes \((20-20)^2 = 0\) to the numerator. The new sample variance is \(\frac{125 + 0}{7 - 1} = \frac{125}{6} \approx 20.83\), giving $s_{new} \approx 4.56 < 5$. Adding a value at the mean increases the denominator \(n-1\) while the numerator stays the same, decreasing the standard deviation.

The sample variance formula is \(s^2 = \frac{\sum(x_i - \bar{x})^2}{n - 1}\). Substituting: \(12 = \frac{60}{n-1}\), so \(n - 1 = \frac{60}{12} = 5\), giving \(n = 6\). Choice B (\(n = 5\)) would yield \(s^2 = \frac{60}{4} = 15 \neq 12\).

With \(n = 19\) values, the median is the \(\frac{19+1}{2} = 10\)th ordered value. Cumulative frequencies: $[0, 5)$ contains values 1 through 3; $[5, 10)$ contains values 4 through 10. The 10th value is the last value in $[5, 10)$, so the median lies in $[5, 10)$. Choice C ($[10, 15)$) results from counting only the frequency of $[5, 10)$ as \(7\) without adding the prior cumulative count of \(3\).

The \(z\)-score formula is \(z = \frac{x - \mu}{\sigma}\). Solving for \(x\): \(x = \mu + z\sigma = 50 + (-2.5)(8) = 50 - 20 = 30\). The negative \(z\)-score confirms that \(x\) lies below the mean. Choice D (\(35\)) results from an arithmetic error such as computing \(50 - 2.5(6) = 35\).

Step 1: \(\sum x_i = 8 \times 10 = 80\). The known values sum to \(4+6+7+8+9+11+13 = 58\), so \(x = 80 - 58 = 22\). Step 2: \(Q_1 = \frac{6+7}{2} = 6.5\) (median of the lower half \(\{4,6,7,8\}\)) and \(Q_3 = \frac{11+13}{2} = 12\) (median of the upper half \(\{9,11,13,22\}\)), giving \(\text{IQR} = 12 - 6.5 = 5.5\). Step 3: Upper fence \(= Q_3 + 1.5(5.5) = 12 + 8.25 = 20.25\). Since \(x = 22 > 20.25\), \(x\) is an outlier.

When every value is divided by a constant \(k = 2\), the standard deviation is divided by \(|k|\) and the variance is divided by \(k^2 = 4\). New variance \(= \frac{36}{4} = 9\). This can also be verified: the new standard deviation is \(\frac{\sqrt{36}}{2} = \frac{6}{2} = 3\), and \((3)^2 = 9\). Choice A (\(18\)) incorrectly divides the variance by \(k\) rather than \(k^2\).

The original sum of squared deviations is \(s^2(n-1) = 4 \times 4 = 16\). For the original \(5\) values relative to the new mean of \(11\): \(\sum(x_i - 11)^2 = \sum[(x_i - 10) - 1]^2 = \sum(x_i-10)^2 - 2\sum(x_i-10) + 5 = 16 - 0 + 5 = 21\), since \(\sum(x_i - 10) = 0\) when the original mean is \(10\). The new value contributes \((16-11)^2 = 25\). Total new sum of squared deviations \(= 21 + 25 = 46\). The new sample variance is $s_{new}^2 = \frac{46}{6-1} = \frac{46}{5} = 9.2$, so $s_{new} = \sqrt{9.2}$.

The height of each bar in a histogram represents the frequency (count) or relative frequency (proportion) of data values that fall within that class interval. The width of the bar represents the span of the class interval. Choice D describes a cumulative frequency histogram, which is a different type of display.

The mode is the value that appears most frequently in a dataset. In \(\{3, 5, 5, 7, 9\}\), the value \(5\) appears twice while all other values appear once, so the mode is \(5\). The value \(6\) is the mean of this dataset but does not appear in the data and therefore cannot be the mode.

The mean is calculated using every value in the dataset, so an extreme outlier directly pulls the mean toward itself. The median depends only on the middle value(s) after sorting, making it resistant to outliers. The mode depends only on frequency, not magnitude. Because of this, the mean is described as the least resistant measure of center.

In a dotplot, each dot represents one data value. A tall stack of dots above a number means that value appears many times — it occurs frequently. This is not an indicator of an outlier (outliers appear isolated, far from other stacks) nor does it confirm the value equals the mean (the mean can fall anywhere on the number line).

Standard deviation measures the typical (average) distance of data values from the mean. A large standard deviation indicates data is spread far from the mean; a small one indicates data clusters close to the mean. Choice A describes the mode, choice B describes the median, and choice D describes the range.

An outlier is a data value that falls far from the rest of the data. The values \(10\) through \(15\) form a tight cluster spanning only \(5\) units. The value \(45\) is \(30\) units above the next highest value and is clearly separated from the cluster. The other choices are all within the main group of data and are not candidates for outliers.

In a perfectly symmetric, unimodal distribution (such as an ideal bell curve), the center of symmetry is simultaneously the balance point (mean), middle value (median), and most frequent value (mode) — all three coincide. In contrast, right-skewed distributions produce mean \(>\) median \(>\) mode, and left-skewed distributions produce mean \(<\) median \(<\) mode.

In a right-skewed distribution, the long right tail contains large values that pull the mean upward. The mode sits at the peak of the histogram (the most frequent value, toward the left). The median is pulled slightly right of the mode but remains resistant to extreme values. The mean is pulled furthest right. This produces the ordering \(\text{Mode} < \text{Median} < \text{Mean}\). Choice A describes a left-skewed distribution.

The sum of all five values must equal \(n \cdot \bar{x} = 5 \times 14 = 70\). The sum of the four known values is \(8 + 12 + 16 + 20 = 56\). Therefore the fifth value is \(70 - 56 = 14\). A common error is guessing the median; however, the correct approach uses the relationship \(\sum x_i = n\bar{x}\) to work backwards.

Standard deviation measures the typical distance of values from the mean. Dataset A has values clustered near the center, meaning most values are close to the mean, resulting in a small standard deviation. Dataset B has values spread across the full range, placing values farther from the mean on average and producing a larger standard deviation. Having the same range does not guarantee equal standard deviations — range only captures the gap between the minimum and maximum.

When every value is multiplied by a constant \(c\), both the mean and standard deviation are multiplied by \(|c|\). New mean \(= 2 \times 10 = 20\) and new standard deviation \(= 2 \times 3 = 6\). This contrasts with adding a constant: adding \(c\) to every value shifts the mean by \(c\) but leaves the standard deviation unchanged. Multiplying rescales both the center and the spread.

$IQR = Q_3 - Q_1 = 40 - 20 = 20$. The lower fence is $Q_1 - 1.5 \times IQR = 20 - 30 = -10$ and the upper fence is $Q_3 + 1.5 \times IQR = 40 + 30 = 70$. Any value below \(-10\) or above \(70\) is classified as an outlier. Only \(-15 < -10\) falls outside the fences. The values \(55\), \(0\), and \(65\) all fall within \([-10,\; 70]\) and are not outliers.

The sum of all relative frequencies in a relative frequency histogram must equal \(1.00\), because relative frequencies are proportions of the total dataset and every data value must fall in exactly one interval. The first three intervals account for \(0.15 + 0.35 + 0.30 = 0.80\). Therefore the fourth interval must have relative frequency \(1.00 - 0.80 = 0.20\).

For \(n = 7\) ordered values, the median is the 4th value. The original median is \(9\). After replacing \(15\) with \(45\), the new ordered dataset is \(\{3, 5, 7, 9, 11, 13, 45\}\) and the 4th value is still \(9\) — the median is unchanged. However, the sum increases by \(45 - 15 = 30\), so the mean increases by \(\frac{30}{7} \approx 4.29\). The median is resistant to this extreme change; the mean is not.

All four datasets have mean \(\bar{x} = 5\). Standard deviation depends on how far values deviate from the mean. Dataset A has all deviations equal to \(0\), giving \(s = 0\). Dataset B has deviations \(-1, 0, 0, 0, 1\). Dataset D has deviations \(-2, -1, 0, 1, 2\). Dataset C has deviations \(-4, -2, 0, 2, 4\) — the largest of any option. Since \(s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}}\), dataset C produces the largest standard deviation.

Reading the dotplot: \(2\) appears once, \(4\) appears three times, \(6\) appears twice, and \(10\) appears once. Total \(n = 1 + 3 + 2 + 1 = 7\). Sum \(= 1(2) + 3(4) + 2(6) + 1(10) = 2 + 12 + 12 + 10 = 36\). Mean \(= \frac{36}{7} \approx 5.14\). The value \(5\) is incorrect — it would require a sum of \(35\). The value \(6\) is the median of this dataset, not the mean.

The outlier score \(98\) is well above all other data values. It pulls the mean upward, so removing it causes the mean to decrease. It also contributes a large squared deviation \((98 - \bar{x})^2\) to the standard deviation calculation. Removing this value eliminates that large contribution, reducing the overall spread. Both the mean and standard deviation decrease when a high outlier is removed.

For Dataset A (symmetric, bell-shaped), the mean and median are approximately equal, so median \(\approx 50\). For Dataset B (right-skewed), the long right tail pulls the mean above the median, meaning median $< $ mean \(= 50\). Since Dataset A's median \(\approx 50\) and Dataset B's median \(< 50\), Dataset A has the larger median. Equal means do not imply equal medians — the shape of the distribution governs the relationship between mean and median.

First find the mean: \(\bar{x} = \frac{2+4+6+8+10}{5} = \frac{30}{5} = 6\). Deviations from the mean: \(-4, -2, 0, 2, 4\). Squared deviations: \(16, 4, 0, 4, 16\). Sum of squared deviations \(= 40\). Sample variance: \(s^2 = \frac{40}{n-1} = \frac{40}{4} = 10\). Therefore \(s = \sqrt{10} \approx 3.16\). The divisor is \(n - 1 = 4\) (not \(n = 5\)) for the sample standard deviation.

The original sum is \(n \cdot \bar{x} = 10 \times 30 = 300\). Correcting the error means removing the wrong value and inserting the correct one: new sum \(= 300 - 80 + 8 = 228\). Corrected mean \(= \frac{228}{10} = 22.8\). Equivalently, the recording error inflated the mean by \(\frac{80 - 8}{10} = \frac{72}{10} = 7.2\), so the corrected mean is \(30 - 7.2 = 22.8\).

The transformation \(y = \frac{x - 100}{15}\) operates in two steps. Step 1 — subtract \(100\) from each value: new mean \(= 100 - 100 = 0\), standard deviation unchanged \(= 15\). Step 2 — divide each value by \(15\): new mean \(= \frac{0}{15} = 0\), new standard deviation \(= \frac{15}{15} = 1\). This process is standardization, and the resulting values are \(z\)-scores. The transformed dataset always has mean \(0\) and standard deviation \(1\).

The combined mean must be computed as a weighted average using total sums. Total sum \(= (20 \times 75) + (30 \times 85) = 1500 + 2550 = 4050\). Combined mean \(= \frac{4050}{50} = 81\). Simply averaging \(75\) and \(85\) gives \(80\), which is incorrect — it ignores that Class B is larger and contributes more to the total. The combined mean is pulled closer to \(85\) because that group has more students.

Adding a value equal to \(\bar{x}\) does not change the mean, since the new point has zero deviation from the current mean and the balance point is preserved. For the standard deviation: the numerator \(\sum(x_i - \bar{x})^2\) is unchanged (the new point contributes \(0^2 = 0\)), but the denominator changes from \(n - 1\) to \(n\). So \(s_{\text{new}}^2 = \frac{\sum(x_i - \bar{x})^2}{n} < \frac{\sum(x_i - \bar{x})^2}{n-1} = s^2\). The standard deviation decreases because the new value is less dispersed than the average.

The midpoints of the three class intervals are \(5\), \(15\), and \(25\). The estimated mean uses frequency-weighted midpoints: \(\frac{6(5) + 12(15) + 6(25)}{6 + 12 + 6} = \frac{30 + 180 + 150}{24} = \frac{360}{24} = 15\). Because the histogram is symmetric (frequencies \(6, 12, 6\)), the mean equals the middle midpoint. This midpoint method is the standard technique for estimating the mean when only a histogram is available rather than individual data values.

For \(n = 10\), split into two halves of \(5\): lower half \(\{12, 15, 16, 18, 20\}\) gives \(Q_1 = 16\), and upper half \(\{22, 24, 27, 30, 85\}\) gives \(Q_3 = 27\). Then $IQR = 27 - 16 = 11$. Upper fence $= Q_3 + 1.5 \times IQR = 27 + 1.5(11) = 27 + 16.5 = 43.5$. Since \(85 > 43.5\), the value \(85\) is classified as an outlier. Choice B is incorrect because the $1.5 \times IQR$ rule does not use standard deviations. Being the maximum value does not prevent a value from being an outlier.