| Mode | Frequency |
|---|---|
| Walk | 96 |
| Bike | 38 |
| Bus | 51 |
| Car | 55 |
| Total | 240 |
2. Summarizing Data
Math 115 - Examples for Topic 2
These worked examples cover the summary statistics and graphical reads from topic 2: proportions from a frequency table, mean vs. median and skew, the five-number summary with the 1.5×IQR outlier fences, and comparing groups with boxplots.
Example 1: Proportions from a Frequency Table
Scenario: A campus sustainability office surveys 240 students about how they usually get to campus.
(a) What proportion of the students bike to campus?
(b) What percentage walk?
(c) What proportion do not drive?
TipSolution
(a) Proportion who bike:
A proportion is a count divided by the total:
\[\text{proportion} = \frac{\text{count}}{\text{total}} = \frac{38}{240} = 0.158\]
(b) Percentage who walk:
First the proportion, then multiply by 100 to convert to a percentage:
\[\frac{96}{240} = 0.40 \quad\longrightarrow\quad 0.40 \times 100 = 40\%\]
(c) Proportion who do not drive:
A not-category goes through the complement: everyone except the Car riders.
\[240 - 55 = 185 \text{ students} \quad\longrightarrow\quad \frac{185}{240} = 0.771\]
Common error: using the largest remaining category (Walk, 96/240) instead of the complement. “Not driving” includes walkers, bikers, and bus riders — all 185 of them.
Example 2: Mean vs. Median and Skew
Scenario: Ten homes sold in a neighborhood last quarter. The sale prices, in thousands of dollars, were:
\[182, \quad 195, \quad 201, \quad 208, \quad 212, \quad 219, \quad 224, \quad 230, \quad 238, \quad 610\]
(a) Compute the mean sale price.
(b) Compute the median sale price.
(c) Why are the two so different, and what does that suggest about the shape of the distribution?
(d) Which measure of center is the better one to report here?
TipSolution
(a) Mean:
Add all ten values and divide by 10:
\[\bar{x} = \frac{182 + 195 + 201 + 208 + 212 + 219 + 224 + 230 + 238 + 610}{10} = \frac{2519}{10} = 251.9\]
The mean sale price is $251,900.
(b) Median:
The values are already in order. With \(n = 10\) (even), the median is the average of the 5th and 6th ordered values:
\[\text{median} = \frac{212 + 219}{2} = 215.5\]
The median sale price is $215,500.
(c) Why they differ, and the shape:
The mean sits far above the median because of the $610k sale — one giant value moves the mean a lot and the median not at all. When the mean is well above the median, the distribution is right-skewed: a tail of high values pulls the mean toward it.
(d) Which to report:
The median. It is the resistant measure — a single extreme sale barely affects it. For skewed data like home prices, the median describes the typical value better than the mean, which is why housing reports quote median sale prices.
Example 3: Five-Number Summary, IQR, and the 1.5×IQR Fences
Scenario: A coffee shop tracks its daily customer counts for several months. The five-number summary:
| Min | Q1 | Median | Q3 | Max |
|---|---|---|---|---|
| 42 | 118 | 145 | 176 | 297 |
(a) Compute the IQR and the range.
(b) Compute the 1.5×IQR outlier fences.
(c) Does this dataset have any outliers on either end?
TipSolution
(a) IQR and range:
\[\text{IQR} = Q_3 - Q_1 = 176 - 118 = 58\]
\[\text{range} = \text{Max} - \text{Min} = 297 - 42 = 255\]
(b) The fences:
First compute \(1.5 \times \text{IQR} = 1.5 \times 58 = 87\). Then:
Lower fence — subtract from Q1, watching the signs:
\[Q_1 - 1.5 \times \text{IQR} = 118 - 87 = 31\]
Upper fence — add to Q3:
\[Q_3 + 1.5 \times \text{IQR} = 176 + 87 = 263\]
(c) Outliers:
- Low end: the minimum is 42, and \(42 > 31\), so no value falls below the lower fence — no low outliers.
- High end: the maximum is 297, and \(297 > 263\), so at least the maximum falls above the upper fence — at least one high outlier.
Note what the summary can and can’t tell us: we know the max is an outlier, but we can’t tell from five numbers alone how many values exceed 263.
Example 4: Comparing Two Groups with Boxplots
Scenario: A commuter survey records door-to-door commute times for random samples of 110 bus riders and 110 train riders.
(a) Which mode has the longer typical commute?
(b) Approximate each group’s IQR. Which group’s commute times are more variable?
(c) Describe the shape of the Bus distribution.
(d) How do the Bus group’s mean and median compare?
TipSolution
(a) Typical commute:
Read the medians from the center lines of the boxes — Train ≈ 44 minutes vs. Bus ≈ 31 minutes. Train runs longer for the typical rider.
(b) IQR and variability:
The IQR is the width of the box: Bus ≈ 13 minutes, Train ≈ 10 minutes. Bus commutes are more variable, even though they are typically shorter.
(c) Shape of the Bus distribution:
The Bus box has a long right whisker and high outliers: the distribution is right-skewed. Most bus commutes are moderate, but a few run very long.
(d) Mean vs. median for the Bus group:
The boxplot does not draw the mean — but the shape licenses an inference: a long right tail pulls the mean up, so the Bus group’s mean sits above its median.
“Can’t be determined” is the tempting answer precisely because no mean marker exists on the plot. But shape → mean-vs-median is exactly the topic 2 skill that answers it anyway: right skew means the mean exceeds the median.