Explore data, compare distributions, calculate normal probabilities, and fit regression lines, with worked examples and practice.
How long does a trip to school usually take? How much do travel times vary? Does leaving earlier help? Statistics gives us tools to organize observations, describe patterns, and assess claims.
Statistics is the field concerned with collecting, classifying, organizing, summarizing, analyzing, and interpreting data. Descriptive statistics organizes, summarizes, and describes the observed data. Inferential statistics uses a sample to draw conclusions about characteristics of its population, with uncertainty.
A variable is a characteristic that can vary across individuals or objects, or over time. An experimental unit is the individual or object on which a variable is measured. A data value is the result of measuring or recording that variable on one unit.
A population is the collection of all measurements of interest; a sample is a selected subcollection of those measurements.
Suppose we want to study the heights of all Grade 11 students in a school at the start of the school year. The population consists of all their height measurements. If we select 30 students and measure their heights, those 30 measurements form a sample.
If two students both measure 160 cm, keep both values: each measurement belongs to a different student.
A parameter is a numerical characteristic of the population; a statistic is a numerical summary calculated from the sample.
Summarizing the measurements in a sample is descriptive. Using that summary to estimate a population characteristic is inferential. The commute study below shows the distinction.
To describe today’s journeys among students who travelled to school, you ask 40 of those students to record their travel times in minutes.
The sample mean describes those 40 trips. Generalizing to the entire school depends on how the sample was selected.
How many characteristics do we record for each unit? This determines whether we study one variable or relationships among several.
Univariate data record one variable per unit; bivariate data record two. Here, multivariate data record more than two variables per unit.
Five students with one height each give univariate data. One hundred students with height and travel time each give bivariate data.
| How many variables? | Example | Useful question |
|---|---|---|
| Univariate | Travel time for each student | What is a typical travel time? |
| Bivariate | Departure time and travel time | Are earlier departures associated with shorter trips? |
| Multivariate | Departure time, distance, transport mode, travel time | How do several characteristics relate to travel time? |
To explore students’ academic profiles, a researcher records five characteristics for each of five selected undergraduates. Each row belongs to one student; the first column identifies that student rather than adding a measured characteristic. Read across a row to connect one unit to its data values, then down a column to follow one variable.
| Student ID | GPA | Gender (recorded code) | Year | Major | Enrolled units |
|---|---|---|---|---|---|
| 1 | 2.0 | F | First | Psychology | 16 |
| 2 | 2.3 | F | Second | Mathematics | 15 |
| 3 | 2.9 | M | Second | English | 17 |
| 4 | 2.7 | M | First | English | 15 |
| 5 | 2.6 | F | Third | Business | 14 |
For student 3, the student is the experimental unit, GPA is the variable, and 2.9 is its observed data value. The full five-variable record is multivariate. Studying GPA alone gives univariate data: the five GPAs form the sample, while all undergraduates’ GPAs at the specified university and time form the population of measurements.
Using the student-record table, identify the sample for enrolled units, count its observations, and classify a study using only GPA and enrolled units. Does the mean GPA of these five selected students necessarily equal the university’s mean GPA?
The sample is 16, 15, 17, 15, 14 enrolled-unit measurements. Its size is five; the two 15s belong to different students and both count. GPA together with enrolled units gives two variables per student, so the data are bivariate. The sample mean GPA is (2.0 + 2.3 + 2.9 + 2.7 + 2.6)/5 = 2.5, a statistic. It need not equal the population mean, a parameter. The selection method is unspecified, so representativeness cannot be assumed.
A large sample can still be biased. Surveying only students who drive leaves out other transport modes. Check who was included, who was missed, the units, and how the question was asked.
A survey of 60 volunteers reports a median study time of 90 minutes. A student concludes that every student in the school studies for 90 minutes. What is wrong?
The median summarizes the volunteers; it does not describe every individual. Volunteers may not represent the school. Even a representative sample would support an estimate with uncertainty, not an exact claim about everyone.
A qualitative (categorical) variable records a quality or category on each unit. A quantitative (numerical) variable records an amount.
A discrete quantitative variable has a finite or countably infinite set of possible values. A continuous quantitative variable is modeled as capable of taking any real value in an interval.
Classify a variable by its possible values, not just the finite list collected in one study.
A number used as an identifier, such as a student number, is still a label. Number of siblings is discrete; elapsed travel time is modeled as continuous. Rounding a time to the nearest minute makes its recorded values discrete without changing the underlying continuous quantity. Discrete values need not be integers: a score restricted to half-points is discrete.
A recovery rate must be defined carefully before classifying it. For a fixed cohort of N patients, the observed proportion recovered is R/N with R = 0, 1, …, N, so it has only N + 1 possible values and is discrete. A continuous rate model requires an additional modeling assumption. Likewise, QPI/GPA is quantitative, but its discrete or continuous treatment depends on the permitted scores and recording precision.
| Level | What comparisons mean | Example |
|---|---|---|
| Nominal | Same or different category; no natural ordering | Transport mode: walk, bus, train |
| Ordinal | An order exists; gaps are not necessarily equal | Satisfaction: low, medium, high |
| Interval | Equal differences are meaningful; zero is conventional | Temperature in °C |
| Ratio | Equal differences and ratios are meaningful; zero represents none | Elapsed time, distance, number of books |
Twice a Celsius temperature does not mean twice as hot. Twice an elapsed time does mean twice the duration. Numeric codes for categories do not make their averages meaningful.
Classification depends on what the recorded value represents, rather than its appearance.
Before choosing an average or a graph, we need to know what the recorded values mean. A transport label, a finishing rank, and a journey time allow different comparisons. We will classify the variable and its measurement level so that our later calculations answer a meaningful question.
Mean: add all observations and divide by the number of observations, x̄=Σxᵢn.
Median: sort the observations. For odd n, choose the middle value. For even n, average the two middle values.
Mode: the most frequent value or category.
Several values can tie for the highest frequency. Here, a dataset with no repeated values is described as having no mode.
Seven students score 60, 60, 70, 80, 60, 70, 50 on a 100-point assessment. A teacher wants a representative score for this group. Compare the mean, median, and mode.
Sorted: 50, 60, 60, 60, 70, 70, 80.
Mean: 450/7 ≈ 64.29. Median: the 4th value, 60. Mode: 60, appearing three times.
The mean balances all seven scores, the median locates the middle observation, and the mode identifies the most common score. They answer different questions, so they need not agree.
It depends on the question. Are we describing the middle observation, finding the most common value, or using a total to plan resources? The mean uses every observation, but unusually large or small values can pull it away from most of the data. Compare these two situations before choosing a summary.
Five families have 2,4,3,5,7 children. We want to describe how many children a family has, on average. Which average should we use?
The mean is (2+4+3+5+7)/5=21/5=4.2 children per family. It uses all five counts and expresses the total of 21 children per five families.
In order, the counts are 2,3,4,5,7. The median is 4 children: this is the middle count, with two counts below it and two above it. There is no mode, since every count occurs once.
Both the mean and median are reasonable summaries here. The values have no strongly isolated extreme, and the two summaries are close. Use 4.2 to express the total number of children per family, or 4 to describe the middle of the ordered counts. The general question alone does not make one uniquely best.
No individual family can have 4.2 children, but a mean need not be a possible individual observation. A decimal result alone does not make the mean inappropriate. In the salary example below, the two highest values create a much larger difference between the mean and median.
Ten staff members earn monthly salaries of ₱15,000, ₱18,000, ₱16,000, ₱14,000, ₱15,000, ₱15,000, ₱12,000, ₱17,000, ₱90,000, and ₱95,000. A job applicant wants to understand what staff usually earn.
| Measure | Monthly salary | What it describes |
|---|---|---|
| Mean | ₱30,700 | Total payroll divided by 10 staff members |
| Median | ₱15,500 | The midpoint of the two middle salaries |
| Mode | ₱15,000 | The most common salary |
Sorted in thousands of pesos, the salaries are 12,14,15,15,15,16,17,18,90,95. The median is (15+16)/2=15.5, and 15 occurs three times. The total is ₱307,000, giving a mean of ₱30,700.
Eight of the ten salaries are ₱18,000 or less. The two highest salaries pull the mean upward. The median is suitable for describing the middle salary; the mode answers which salary is most common. The mean remains useful for payroll: 10 × 30,700 = 307,000 pesos.
To make the large gap visible, we can report the eight lower salaries and the two highest salaries separately, while also giving the overall summary:
Eight staff members earn ₱12,000–₱18,000, with a mean of ₱15,250. The other two earn ₱90,000 and ₱95,000. Across all ten staff members, the mean is ₱30,700.
The eight lower salaries total ₱122,000, so their mean is 122,000/8=15,250. This is the mean of that subgroup, not the mean of all staff.
Do not silently discard unusually high or low observations, or assume they are errors. State which values were separated and why. If there are genuinely different job groups, reporting those groups separately may be more informative; salaries alone do not establish those roles. A formal rule for flagging potential outliers appears with box plots later in this lesson.
Edit the dataset, compare presets, or add an extreme value. Predict which summaries will change the most.
Changes stay in this page until you leave or reload it. Sample calculations need at least two values.
Find the mean, median, and mode(s) of 101,101,106,108,108,106.
Sorted: 101,101,106,106,108,108. Mean: 630/6=105. Median: (106+106)/2=106. All three values 101, 106, and 108 share the highest frequency of 2, so they are tied modes.
We have decided what kind of data we have. Now imagine explaining a whole list of observations to someone who has not seen it: which value would you call typical? A very large observation pulls the mean toward it, while the median depends on position. Keep that contrast in mind as we compare the three summaries.
A pooled mean summarizes all observations from disjoint groups. If their sizes are \(n_1,n_2\) and means are \(\bar x_1,\bar x_2\), then \(\bar x=(n_1\bar x_1+n_2\bar x_2)/(n_1+n_2)\). The group sizes are weights because each individual observation must count once.
Ten walkers have a mean travel time of \(12\) minutes; thirty bus riders have a mean of \(28\) minutes. Find the overall mean and explain why \((12+28)/2\) is misleading.
The group totals are \(120\) and \(840\) minutes. Thus \(\bar x=960/40=24\) minutes. The unweighted result \(20\) gives ten walkers as much influence as thirty riders. The correct mean lies between the two means and closer to the larger group's mean, which is a useful reasonableness check.
Combining means needs counts. Combining variances needs still more information: groups may also differ in their centers. Do not average summary statistics without checking what they summarize.
Five students have a mean score of 12. Every score increases by 3. What is the new mean?
Two groups can have the same mean and very different consistency. Compare {35,35,35,35,35} with {10,25,35,45,60}: both have mean 35, but only one has no spread.
Range: maximum − minimum.
Population variance: σ²=Σ(xᵢ−μ)²N. Population SD: σ=√σ².
Sample variance: s²=Σ(xᵢ−x̄)²n−1. Sample SD: s=√s².
The sample formula requires n > 1; the population formula requires N > 0.
Positive and negative deviations from the mean cancel when added. Squaring prevents this cancellation. Taking the square root returns the result to the original units; standard deviation is not the average absolute distance.
A survey of ten fast-food restaurants recorded the calorie content of a midsized hamburger at each restaurant. Treat the ten calorie measurements as a sample and calculate the mean and sample standard deviation, rounding to two decimal places.
514, 507, 502, 498, 496, 506, 458, 478, 463, 514
The range is 514−458=56 calories. Use the Burger calories preset above to compare the sample and population formulas.
Variance has squared units. Standard deviation has the original units. Use n−1 for a sample variance used to estimate population variation, and N when describing the complete population. Do not round intermediate deviations.
Find the sample standard deviation of {25,35,35,35,45}. Compare it with {35,35,35,35,35}.
The first mean is 35. Squared deviations total 100+0+0+0+100=200. Thus s²=200/4=50 and s=√50≈7.07. The second dataset has all deviations equal to 0, so its standard deviation is 0.
A typical value is a useful beginning, but it does not tell the whole story. Two classes can have the same average and very different results. We will now measure how far the observations tend to lie from their center, so that our summary describes consistency as well as level.
A z-score expresses a value in standard deviation units:
z=x−μσ for population summaries, or z=x−x̄s for sample summaries.
The standard deviation must be positive.
For this lesson, percentile rank uses the strictly-below convention: number of observations below xtotal number of observations × 100%.
A positive z-score is above the mean and a negative one is below. Percentile rank instead counts observations below a value. Other percentile conventions handle ties differently; use the stated method.
An employer summarizes staff ages with a mean of 36 years and standard deviation of 8.3 years. Compare Tammy (23) and Joy (44): who is farther from the mean in standard deviation units?
Tammy, age 23: z=(23−36)/8.3≈−1.57. Joy, age 44: z=(44−36)/8.3≈0.96.
Tammy is farther from the mean in standard deviation units. These two numbers alone do not tell us how common either age is; that requires the distribution’s shape.
A group receives a mean of 43 pieces of candy, with SD 2. Find the z-score for 20 pieces. If Zia’s z-score is 6, how many pieces did she receive?
z=(20−43)/2=−11.5. To recover an observation, use x=μ+zσ. Zia received 43+6(2)=55 pieces. These are algebraic results; no normal distribution assumption is needed to compute them.
A percentile rank of 77 does not mean a test score of 77%.
We can describe a center and a spread, but how should we compare observations from different distributions? A standardized score measures position relative to a mean in units of standard deviation. A percentile answers a different question about relative rank. Keep the measurement of distance separate from the proportion of observations below a value.
This is the median-of-halves convention used throughout this lesson. Other software may interpolate differently.
The interquartile range is IQR=Q₃−Q₁.
This measures the distance between the lower and upper quartiles. Unlike the range, it uses the middle portion of the ordered data rather than the two extremes.
Lower fence: Q₁−1.5(IQR). Upper fence: Q₃+1.5(IQR). Values strictly outside these fences are potential outliers.
Draw a box from Q₁ to Q₃, a line at the median, and whiskers to the most extreme observed values at or within the fences. Plot potential outliers as separate points.
Travel times (minutes): 10,12,14,16,18,20,22,24,60.
Q₂=18; lower half 10,12,14,16 gives Q₁=13; upper half 20,22,24,60 gives Q₃=23.
IQR=10. Fences: −2 and 38. The value 60 is an outlier; whiskers end at 10 and 24.
Check for a recording error, a different process, or an unusual but valid observation. Explain any exclusions. A fence is a calculated cutoff; a whisker ends at an actual observed value.
Eight students record their waiting times for a school shuttle, in minutes: 2,4,5,7,8,10,12,14. Summarize the middle half of the waits and check for unusually long waits. Using the median-of-halves convention, find the minimum, Q₁, median, Q₃, maximum, and any potential outliers.
Five-number summary: (2,4.5,7.5,11,14). IQR=6.5 minutes. Fences: −5.25 and 20.75 minutes. There are no potential outliers under this rule, so the whiskers reach 2 and 14 minutes. The box spans 4.5 to 11 minutes; it describes the middle half of these recorded waits, not a guarantee for future journeys.
Quartiles are cut points in ordered numerical data: the first quartile marks approximately the lowest quarter, the second is the median, and the third marks approximately the lowest three quarters. For finite data, their numerical values depend on the stated quartile convention.
A box plot displays these quartiles and two whisker endpoints on a common numerical scale. The box spans the first to third quartile, representing the middle half of the ordered data approximately; a line marks the median.
A longer box means greater spread in that middle portion, not more observations. Comparing plots on the same scale helps compare typical values, variability, and unusual observations between groups.
This lesson uses a modified box plot: whiskers end at the most extreme observed values within the outlier fences, and values beyond them appear separately.
Some box plots instead extend whiskers to the minimum and maximum, so check the convention. A box plot does not show every observation or reveal multiple clusters as clearly as a dot plot or histogram.
The mean and standard deviation describe center and spread, but an unusually long commute can pull them away from most students’ experience. We now want a summary based on positions in the ordered data: where is the middle, and where does the middle half lie? Quartiles mark those positions; the interquartile range measures the width of the middle half, and a box plot displays them alongside the tails and possible outliers. We will sort the observations, find the quartiles using one stated convention, and then turn that summary into a picture.
Build a histogram in four stages: raw data → ordered array → frequency distribution table → histogram.
These are the ten-year population growth rates (%) of 50 Philippine cities and municipalities from 2000 to 2010. They are historical teaching data.
16.9, 4.3, 7.4, 12.2, 19.2, 29.4, 23.8, 13.2, 23.3, 24.6, 20.5, 24.2, 44, 10.7, 14.9, 19, 4.5, 32.6, 9.4, 37.1, 9.7, 5.3, 21.3, 18.5, 15.9, 15.6, 16.5, 30.4, 25.6, 34, 12.4, 22.2, 38.5, 18.2, 8.4, 13.9, 54.3, 26.3, 14.6, 11.8, 2.2, 19.5, 11.8, 8.3, 20.1, -3.2, 21.9, 18, 53.4, 28.1
Arrange the observations from lowest to highest. Here, \(L=-3.2\), \(H=54.3\), and \(n=50\).
-3.2, 2.2, 4.3, 4.5, 5.3, 7.4, 8.3, 8.4, 9.4, 9.7, 10.7, 11.8, 11.8, 12.2, 12.4, 13.2, 13.9, 14.6, 14.9, 15.6, 15.9, 16.5, 16.9, 18, 18.2, 18.5, 19, 19.2, 19.5, 20.1, 20.5, 21.3, 21.9, 22.2, 23.3, 23.8, 24.2, 24.6, 25.6, 26.3, 28.1, 29.4, 30.4, 32.6, 34, 37.1, 38.5, 44, 53.4, 54.3
Class labels such as “−4 to 5” name the class limits; the boundaries determine membership. For example, 5.3 belongs in the first class because \(-4.5\le5.3<5.5\). Do not round the observations.
| Class (%) | LCB | UCB | Frequency |
|---|---|---|---|
| -4.5 | 5.5 | 5 | |
| 5.5 | 15.5 | 14 | |
| 15.5 | 25.5 | 19 | |
| 25.5 | 35.5 | 7 | |
| 35.5 | 45.5 | 3 | |
| 45.5 | 55.5 | 2 | |
| Total | 50 | ||
Select a class to inspect the values counted in it.
The horizontal axis shows classes, with bar edges at their boundaries. The vertical axis shows frequencies. Each rectangle has width 10 and a height equal to its class frequency. Adjacent rectangles touch.
Point to, tap, or focus a bar or class button to connect the histogram with its frequency table. The numbers remain unchanged.
Which class would contain 15.5? How many of the actual observations are in the class 16 to 25, and what percentage of the data is that?
15.5 belongs to the class 16 to 25 because it is that class’s lower boundary. There are 19 actual observations in this class: \(19/50=38\%\).
The frequency of a category or class interval is the number of observations assigned to it. A frequency distribution records those frequencies across the categories or intervals.
A histogram represents a numerical frequency distribution by adjacent rectangles over class intervals.
A histogram places numerical intervals along a number line and draws adjoining bars to reveal where values cluster, how widely they spread, and whether there are gaps or long tails. Here the intervals have equal widths, so their bar heights can directly represent counts.
A long list makes it difficult to see where observations cluster. Grouping values into intervals gives us a frequency table; drawing those counts produces a histogram. We will keep the same dataset throughout, so you can trace how each observation contributes to a row and then to a bar.
A frequency density is a class frequency divided by its width.
In a frequency-density histogram, rectangle area represents frequency. The equal-width histogram above can use frequency as height because every bar has the same width. Unequal widths require this extra care.
A survey has \(8\) travel times in \([0,10)\) minutes and \(12\) in \([10,30)\). Which interval has more observations, and which has the higher concentration per minute?
| Interval (minutes) | Frequency | Width | Frequency density |
|---|---|---|---|
| [0,10) | 8 | 10 | 0.8 |
| [10,30) | 12 | 20 | 0.6 |
The second interval has more observations, but they are spread across twice the width. The first has higher density: \(8/10=0.8\) versus \(12/20=0.6\) observations per minute. Bar areas are \(10(0.8)=8\) and \(20(0.6)=12\). A taller bar does not necessarily contain more observations when widths differ.
When reading a histogram, check the vertical-axis label and the class widths together. Equal widths let you compare counts by height; unequal widths require comparing areas.
Two classes each contain 12 observations. The second class is twice as wide. In a frequency-density histogram, how does its height compare?
Look at the graph before revealing the explanation. Check which impression comes from the data and which comes from the way it is drawn.
How much larger does Group A look than the other groups?
The bars begin at 50%, so their visible lengths are 12, 4, and 4. Group A looks three times as large, even though 62% is only about 1.15 times 54%. The gap is 8 percentage points.
For a magnitude comparison using bar lengths, a zero baseline preserves the ratio. Changing the axis does not change 62%, 54%, or 54%.
Does this series look almost flat? Read the values as well as the slope.
A 0–40 scale compresses the differences. A 0–15 scale uses the plotting area more effectively. Both show the same five observations; a flatter line does not mean there was no change.
From 2019 to 2020 the value doubles from 6 to 12 in both versions. These values are a teaching reconstruction of the plotted points. There is no single mandatory range for a line chart: show clear ticks and enough context for the question.
If you saw only the first two years, would you expect the growth to continue?
A selected interval can emphasize an increase while hiding a later reversal. Here the first two points are taken from the same complete five-year series.
The wider view shows a peak followed by a decline. The data are an illustrative reconstruction of the five-year pattern; the selected view is a true subset, so no values change between the two graphs.
Can these three percentages be slices of one whole?
The labels total 152%, not 100%. A pie chart falsely suggests mutually exclusive parts of a single whole. The original labels cannot be valid pie-slice percentages.
A bar chart compares the three percentages without claiming they partition one whole. Keep the original percentages; silently normalizing them to 100% would change their meaning.
Which region attracts your attention first? Does darker shading mean a higher density?
The darkest areas represent the lowest densities in the first view. Readers often expect darker shading to mean more. A correct legend helps, but a reversed convention can still invite a quick misreading.
Both schematic maps use the same six values. Only the shading direction changes. These simplified regions demonstrate the density-map concept; they are not geographic or current population estimates. Always read the legend and units.
A graph should help us understand the data, but its design can also create a misleading impression. We will compare displays of the same information and ask what the axes, scale, time window, and shading encourage us to believe. Read the actual values before accepting that first impression.
A normal distribution is a continuous probability distribution with density
\[f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}},\qquad x\in\mathbb R,\quad \sigma>0.\]
The parameters μ and σ are its mean and standard deviation.
This formula produces a symmetric bell-shaped curve centered at μ; its mean, median, and mode coincide. Larger σ spreads the curve more widely. You do not need to integrate the formula by hand to use the probability explorer below.
The total area under the density curve is 1. Area over an interval represents probability; height at one point does not. For a continuous normal variable, the probability of any one exact value is zero.
| Interval | Approximate proportion |
|---|---|
| μ−σ to μ+σ | 68% |
| μ−2σ to μ+2σ | 95% |
| μ−3σ to μ+3σ | 99.7% |
Assume tomato weights are normally distributed with μ=0.61 lb and σ=0.15 lb.
These are empirical-rule approximations. A normal CDF gives slightly different, more precise values.
The 68–95–99.7 rule is for approximately normal data. Strong skew, multiple peaks, or extreme outliers can make it unsuitable. A symmetric appearance alone does not prove that a distribution is normal.
Assume grades are normal with mean 81 and SD 5. Approximately what proportion lies between 86 and 91? In a class of 40, what is the expected number in that interval?
The bounds are μ+σ and μ+2σ. Using the empirical rule, the interval contains (95%−68%)/2=13.5%. Expected count: 40(0.135)=5.4, or about 5 students. An expected count is an average prediction, not a guaranteed whole-number outcome.
So far we have described the data in front of us. A probability model lets us ask what values might be expected in similar situations. The normal curve is one such model; we should check whether its shape is a sensible description before using its areas as probabilities.
If X is normal with mean μ and SD σ, then Z=(X−μ)/σ has mean 0 and SD 1. Let Φ(z)=P(Z≤z).
Change the mean, standard deviation, or bounds. The area updates after you choose Calculate.
Numerical CDF approximation; calculations use unrounded z-scores. The plot shows ±4 standard deviations, while probabilities include the full tails.
For a left-tail table, Φ(1.50)≈0.9332. The area above 1.50 is 1−0.9332=0.0668. By symmetry, Φ(−1.50)≈0.0668.
Some tables report the area between 0 and z instead. Read the table heading before using its numbers.
The curve is a model; the area beneath a specified part of it gives the probability we want. Sketch the region before consulting a table or calculator. Decide whether you need a left tail, a right tail, or an interval, so that the number you obtain answers the original question.
Linear regression models the relationship between an explanatory variable x and a response y using ŷ=ax+b. The hat marks a predicted value.
A residual is e=y−ŷ: observed minus predicted. Least squares chooses a and b to minimize Σe².
A positive residual means the observed point lies above the fitted line; a negative residual means it lies below. Squaring these vertical errors prevents positive and negative errors from cancelling.
a=Σ(xᵢ−x̄)(yᵢ−ȳ)Σ(xᵢ−x̄)² and b=ȳ−ax̄.
At least two distinct x-values are needed. Always look at the scatter plot before treating a line as a useful model.
The first dataset records machine hours (x) and sweets produced (y). Each line below contains one observed pair.
x: machine hours; y: sweets produced.
Interpolation predicts within the observed x-range. Extrapolation predicts outside it, where the relationship may change. An intercept at x=0 is not automatically meaningful if zero is far outside the observed range.
A line predicts 284 sweets for a particular run. The observed output is 281. Find and interpret the residual.
e=281−284=−3. The model overpredicted by 3 sweets. The point lies below the fitted line. Its contribution to the sum of squared errors is 9 sweets².
A regression prediction is the output of a model for a specified input; it is not a replacement for the measured observation. A residual is the observed value minus that prediction.
Its sign identifies which side of the fitted line the observation lies on, and its size measures the error in the response variable’s units.
We have mainly studied one variable at a time. Now suppose we want to understand how two measurements move together. A fitted line summarizes that relationship, while the residuals show what the line has failed to explain. Both deserve our attention.
For paired observations with neither variable constant, the Pearson correlation coefficient is r=Σ(xᵢ−x̄)(yᵢ−ȳ)√[Σ(xᵢ−x̄)² · Σ(yᵢ−ȳ)²].
Try the Curved relationship preset in the regression explorer. It produces y=x² at symmetric inputs: the relationship is exact, yet the linear correlation is zero.
Ice cream sales and swimming incidents may both rise during hotter months. Their association alone does not show that buying ice cream causes an incident. Temperature, season, and the number of people swimming are possible common influences.
Likewise, a fitted line for family size and aggregate poverty incidence describes an association in that dataset. It cannot establish what changing one family’s size would cause.
For ordinary least-squares regression with an intercept, r² is the proportion of observed variation in y accounted for by the fitted linear relationship. It is not prediction accuracy, a causal percentage, or proof that the model will generalize.
The fitted line and correlation describe an association, but they do not tell us what caused it. Suppose two measurements rise together: could a third factor influence both? Before turning a numerical pattern into an explanation, ask how the data were collected and what other explanations remain possible.
An aggregate rate combines subgroup rates using their sizes as weights. Changing the mixture can reverse a comparison, even when each subgroup comparison points the same way. This is called Simpson’s paradox. It is a reason to investigate how data were collected, not permission to select whichever comparison supports a preferred claim.
Two programs report passes in a foundation test. Among beginners, A has 8 passes out of 10 and B has 63 out of 90. Among experienced students, A has 81 out of 90 and B has 10 out of 10. Compare overall rates and each starting level. Does either overall rate establish which program is better?
Overall A has \(89/100=89\%\) and B has \(73/100=73\%\). Within beginners, A is \(80\%\) and B is \(70\%\); within experienced students, A is \(90\%\) and B is \(100\%\). Here the subgroup rankings conflict, so neither program dominates both groups. The overall figure mainly mixes different starting populations and cannot establish a causal effect.
Now A has 7/10 beginner passes and 81/90 experienced passes. B has 72/90 beginner passes and 10/10 experienced passes. Prove that B has higher rates at both levels but A has a higher overall rate. Explain the source of the reversal.
For beginners, \(80\%>70\%\); for experienced students, \(100\%>90\%\): B wins each comparison. Yet A has \(88/100=88\%\) overall and B \(82/100=82\%\). A serves mostly experienced students, while B serves mostly beginners. The overall percentages use different weights; they are not a like-for-like comparison. Without a defensible study design, neither comparison by itself proves causation.
Enter measurements from a small study—such as travel times in minutes. You will calculate a frequency distribution table (FDT), then a histogram, and finally a modified box plot. Each correct answer unlocks the next step. The graphs appear only after you have checked their required calculations.
Write down your reasoning before opening each solution. Choose a graph as well as a numerical summary when the question calls for a description of data.
Daily reading times are 10,15,15,20,25,35. Find the mean, median, mode, range, sample variance, and sample SD.
Mean=20; median=17.5; mode=15; range=25. Squared deviations total 400. Sample variance=400/5=80; sample SD=√80≈8.94 minutes.
Replace 35 by 95 in the previous dataset. Find the new mean and median. Which is more stable?
New data: 10,15,15,20,25,95. Mean=180/6=30; median=(15+20)/2=17.5. The mean increased by 10 minutes while the median stayed unchanged.
A student scores 84 on a test with mean 72 and SD 8. Find the z-score. Can you determine an exact percentile from these facts alone?
z=(84−72)/8=1.5. The score is 1.5 SD above the mean. An exact percentile requires the distribution. Under a normal model, the percentile would be about 93.32.
Use the median-of-halves convention for 1,2,3,4,5,6,7,20. Find Q₁, median, Q₃, IQR, and the whisker endpoints.
Q₁=2.5; median=4.5; Q₃=6.5; IQR=4. Fences: −3.5 and 12.5. The value 20 is a potential outlier. Whiskers: 1 and 7.
Assume delivery times are normal with mean 30 minutes and SD 4. Find the proportion taking 26 to 34 minutes, and the expected count among 200 deliveries.
The interval is within one SD of the mean. Empirical rule: about 68%, or 136 deliveries. The normal CDF gives approximately 68.27%, or 136.54 expected deliveries. Different precision explains the small difference.
Fit a least-squares line to (1,2),(2,3),(3,5). Predict y at x=2.5. Find the residual for (2,3). Is predicting at x=10 interpolation?
x̄=2, ȳ=10/3; centered cross-products total 3 and squared x-deviations total 2. Thus slope=1.5 and intercept=1/3.
ŷ=1.5x+1/3. At x=2.5, ŷ≈4.0833. At x=2, ŷ=3.3333…, so the residual is −1/3. Predicting at 10 is extrapolation beyond the observed range [1,3].
State the question and units. Describe how the data were collected. Choose summaries and displays that fit the variable. Interpret the result in context, and explain what the evidence cannot establish.
Use the mean equation and the positions in the sorted list.
State the variance convention. For quartiles, use medians of halves, excluding the overall median when the size is odd.
Show the standardization step. Use the empirical rule where requested.
Find the least-squares line and correlation coefficient before making predictions.
A display compresses data. Before using it to make a decision, ask which information survives the compression and which has been lost. The following problems connect the frequency table, box plot and numerical summaries already developed in this lesson.
Two groups report waiting times in minutes. Group A: \(0,2,2,4,6,8,8,10\). Group B: \(0,1,3,4,6,7,9,10\). Use the median-of-halves convention. Determine each five-number summary, decide whether their modified box plots coincide, and compare population variances. What cannot be concluded from equal box plots?
Both lists are sorted. Compute the medians of the four-value halves, then compare the squared deviations from the common mean.
For both groups, the median is \((4+6)/2=5\), \(Q_1=2\), \(Q_3=8\), minimum \(0\), and maximum \(10\). The IQR is \(6\); fences are \(2-1.5(6)=-7\) and \(8+1.5(6)=17\). Every value lies inside the fences, so the whiskers are \(0\) and \(10\): the box plots coincide.
Both means are \(5\). Group A has squared-deviation sum \(25+9+9+1+1+9+9+25=88\), giving population variance \(88/8=11\). Group B has sum \(25+16+4+1+1+4+16+25=92\), giving \(92/8=11.5\). A box plot does not reveal every observation, the exact frequencies inside a quartile, or the variance. Equal box plots do not imply identical distributions.
A class records \(32\) whole-number travel times, with minimum \(6\) and maximum \(36\). Choose the smallest \(k\) with \(2^k>32\), compute the initial width \(w=\lceil(H-L)/k\rceil\), then check whether classes starting at \(6\) cover the maximum. Give corrected limits and boundaries if needed. Can frequencies be recovered from this information alone?
The rule is strict. The last upper boundary must be above the maximum, not below it. The range does not tell you how observations are distributed.
Because \(2^5=32\) and \(2^6=64>32\), take \(k=6\). The initial width is \(\lceil30/6\rceil=5\). Six classes with whole-number limits start at \(6,11,16,21,26,31\); the final upper limit is \(35\), with upper boundary \(35.5\). This excludes \(36\), so increase the width to \(6\).
The corrected limits are \(6\text{–}11,12\text{–}17,18\text{–}23,24\text{–}29,30\text{–}35,36\text{–}41\). Boundaries are \([5.5,11.5),[11.5,17.5),[17.5,23.5),[23.5,29.5),[29.5,35.5),[35.5,41.5)\). Now the maximum is included. Frequencies cannot be calculated without the actual observations; many datasets share the same size and extrema.
A histogram uses classes with whole-number limits \(0\text{–}4,5\text{–}9,10\text{–}14\), each with frequency \(2\). All observations are integers. Determine the smallest and largest possible exact means. Find the midpoint estimate, and explain why it need not equal the exact mean.
To minimize a total, place each observation at its permitted lower limit. To maximize it, use the upper limits.
The smallest possible total is \(2(0)+2(5)+2(10)=30\), giving mean \(5\); it is attained by \(0,0,5,5,10,10\). The largest is \(2(4)+2(9)+2(14)=54\), giving mean \(9\), attained by \(4,4,9,9,14,14\). The midpoint estimate is \([2(2)+2(7)+2(12)]/6=7\). It replaces unknown observations by class midpoints, so it is an estimate, not a uniquely determined exact mean. Both extreme datasets produce the same grouped frequencies.
Opening your learning space…