Classifying Statistical Variables (Data Taxonomy)
Statistical inquiry begins by identifying the nature of the data collected, which determines appropriate presentation techniques and calculations.
- Qualitative (Categorical) Data: Non-numerical descriptors or labels.
- Nominal: Categories with no natural order (e.g., eye color, nationality, blood type).
- Ordinal: Categories with a meaningful sequence or ranking (e.g., satisfaction level: low/medium/high, shirt sizes S/M/L).
- Quantitative (Numerical) Data: Quantities measured or counted numerically.
- Discrete: Countable quantities taking only specific isolated values, usually whole integers (e.g., number of siblings, shoe sizes, goals scored).
- Continuous: Measurable quantities that can take any real value along a continuum, limited only by instrument precision (e.g., height, temperature, time, mass).
Data Sources & Collection Methodologies
Evaluating how data is gathered is vital for Criterion D (Applying mathematics in real-world contexts) and recognizing potential biases.
- Primary Data: Data collected first-hand by the researcher specifically for the current investigation (e.g., conducting surveys, lab experiments, direct counting).
Pros: Highly relevant, fresh, researcher controls quality. Cons: Costly, time-consuming. - Secondary Data: Pre-existing data gathered by an external organization or previous study (e.g., census databases, published scientific journals, meteorological archives).
Pros: Inexpensive, instant access, massive sample sizes. Cons: May contain hidden bias, outdated definitions, or irrelevant variables.
Frequency Distribution Tables & Grouped Data
Raw ungrouped data is unwieldy. Organizing observations into frequency tables reveals underlying distributions.
- Absolute Frequency ($f$): The count of occurrences in a category or class interval. Total observations $n = \sum f$.
- Relative Frequency: The proportion of the total represented by a class: $$\text{Relative Frequency} = \frac{f}{\sum f} = \frac{f}{n}$$
- Cumulative Frequency ($cf$): Running sum of frequencies up to and including the current class interval.
When continuous data is grouped into class intervals such as $20 \le x < 30$:
- Class Width ($w$): Upper boundary minus lower boundary ($30 - 20 = 10$).
- Mid-Interval Value ($x$): The average of the class boundaries: $$x_{\text{mid}} = \frac{\text{Lower Boundary} + \text{Upper Boundary}}{2} = \frac{20 + 30}{2} = 25$$
- Mid-interval values act as the representative score when estimating summary statistics (mean, variance).
Two-Way Contingency Tables
A two-way table categorizes bivariate categorical data across two different variables (one displayed in rows, the other in columns), providing the foundation for conditional analysis and probability.
| Category | Condition $B$ | Condition $B'$ | Row Total |
|---|---|---|---|
| Condition $A$ | $n(A \cap B)$ | $n(A \cap B')$ | $n(A)$ |
| Condition $A'$ | $n(A' \cap B)$ | $n(A' \cap B')$ | $n(A')$ |
| Column Total | $n(B)$ | $n(B')$ | Grand Total ($N$) |
Worked Example 1: Grouped Frequency Table Construction
The completion times (in seconds) of 20 students in a science lab puzzle are recorded below:
$$34, 42, 28, 55, 39, 47, 51, 33, 44, 29, 38, 46, 58, 31, 49, 43, 36, 52, 41, 48$$
a) Construct a grouped frequency table using intervals of width $10$: $20 \le t < 30$, $30 \le t < 40$, etc.
b) Calculate the mid-interval values and cumulative frequencies for each class.
c) State the modal class.
Step-by-step Solution:
Part a & b: Table Construction
- $20 \le t < 30$: Values are $28, 29 \implies \text{Frequency } f = 2$. Midpoint: $\frac{20+30}{2} = 25$. Cumulative $cf = 2$. [M1]
- $30 \le t < 40$: Values are $34, 39, 33, 38, 31, 36 \implies f = 6$. Midpoint: $35$. Cumulative $cf = 2 + 6 = 8$. [A1]
- $40 \le t < 50$: Values are $42, 47, 44, 46, 49, 43, 41, 48 \implies f = 8$. Midpoint: $45$. Cumulative $cf = 8 + 8 = 16$. [A1]
- $50 \le t < 60$: Values are $55, 51, 58, 52 \implies f = 4$. Midpoint: $55$. Cumulative $cf = 16 + 4 = 20$. [A1]
Part c: Modal class
The class interval with the highest frequency ($f = 8$) is $40 \le t < 50$. [Criterion A: Knowledge & Understanding]
Worked Example 2: Two-Way Contingency Table Analysis
A survey of $120$ high school students investigated preferred learning mode (Online vs. In-Person) across Grade 9 and Grade 10 students. $50$ of the students were in Grade 9. Of the Grade 10 students, $45$ preferred In-Person learning. In total, $80$ students preferred In-Person learning.
a) Complete the two-way table for this data.
b) What percentage of Grade 9 students prefer Online learning?
c) What proportion of all students who prefer Online learning are in Grade 10?
Step-by-step Solution:
Part a: Table Completion
Total students $= 120$. Grade 9 students $= 50 \implies$ Grade 10 students $= 120 - 50 = 70$. [M1]
Total In-Person $= 80 \implies$ Total Online $= 120 - 80 = 40$.
For Grade 10: In-Person $= 45 \implies$ Online $= 70 - 45 = 25$. [M1]
For Grade 9: Total Online $= 40 \implies$ Grade 9 Online $= 40 - 25 = 15$.
Grade 9 In-Person $= 50 - 15 = 35$ (Check: $35 + 45 = 80$). [A1]
Part b: Percentage of Grade 9 preferring Online:
$\frac{\text{Grade 9 Online}}{\text{Total Grade 9}} = \frac{15}{50} = 0.30 = 30\%$. [A1]
Part c: Proportion of Online students who are in Grade 10:
$\frac{\text{Grade 10 Online}}{\text{Total Online}} = \frac{25}{40} = \frac{5}{8} = 0.625 = 62.5\%$. [Criterion C: Communication & Interpretation]
Tiered Practice Exercises
10 problems advancing from variable taxonomy and table calculations to grouped data boundaries, two-way tables, and real-world bias investigations.
Classify each of the following variables as qualitative (nominal/ordinal) or quantitative (discrete/continuous):
a) Number of cars passing a toll booth per hour.
b) Marathon finish times recorded in minutes and seconds.
c) Student ranking in a spelling bee (1st, 2nd, 3rd).
d) Make of mobile phones owned by students (Apple, Samsung, Google).
View Solution & Markscheme
Method & Marks:
a) Quantitative Discrete (countable count of cars). [A1]
b) Quantitative Continuous (measurable time continuum). [A1]
c) Qualitative Ordinal (categorical ranking order). [A1]
d) Qualitative Nominal (unordered categories). [A1]
A researcher wants to study student study habits. State whether collecting data from the national ministry of education website is primary or secondary data, and give one advantage and one disadvantage.
View Solution & Markscheme
Method & Marks:
It is secondary data because the data was originally collected by a third party. [A1]
Advantage: Readily accessible, inexpensive, and typically encompasses a large, representative national sample. [A1]
Disadvantage: May not address the researcher's exact study habit questions or may be slightly outdated. [R1]
For the grouped class interval $45 \le m < 60$, state:
a) The lower class boundary
b) The upper class boundary
c) The class width
d) The mid-interval value
View Solution & Markscheme
Method & Marks:
a) Lower boundary $= 45$. [A1]
b) Upper boundary $= 60$. [A1]
c) Class width $= 60 - 45 = 15$. [A1]
d) Mid-interval value $= \frac{45 + 60}{2} = 52.5$. [A1]
The test marks of 50 students are grouped into four categories: $0-24$ ($f = 8$), $25-49$ ($f = 14$), $50-74$ ($f = 18$), $75-100$ ($f = 10$).
a) Construct the relative frequency for each category.
b) Calculate the cumulative frequency for each category.
View Solution & Markscheme
Method & Marks:
Total $n = 50$.
a) Relative Frequencies: [M1]
$0-24$: $\frac{8}{50} = 0.16$ ($16\%$); $25-49$: $\frac{14}{50} = 0.28$ ($28\%$);
$50-74$: $\frac{18}{50} = 0.36$ ($36\%$); $75-100$: $\frac{10}{50} = 0.20$ ($20\%$). [A1]
b) Cumulative Frequencies:
$cf_1 = 8$; $cf_2 = 8 + 14 = 22$; $cf_3 = 22 + 18 = 40$; $cf_4 = 40 + 10 = 50$. [A1]
In a school of 200 staff members, 120 drink coffee, 110 drink tea, and 45 drink neither.
a) Complete a two-way frequency table representing this data.
b) How many staff members drink both coffee and tea?
View Solution & Markscheme
Method & Marks:
Total staff $= 200$. Those drinking at least one $= 200 - 45 = 155$. [M1]
By the principle of inclusion-exclusion: $n(\text{Coffee} \cup \text{Tea}) = n(\text{Coffee}) + n(\text{Tea}) - n(\text{Both})$ [M1]
$155 = 120 + 110 - n(\text{Both}) \implies 155 = 230 - n(\text{Both}) \implies n(\text{Both}) = 75$. [A1]
Table values:
Coffee & Tea: $75$; Coffee only: $120 - 75 = 45$;
Tea only: $110 - 75 = 35$; Neither: $45$. Check total: $75 + 45 + 35 + 45 = 200$. [A1]
A biologist measures the lengths of 80 leaves (in cm). The grouped frequency table has intervals $4 \le L < 6$ ($f = 12$), $6 \le L < 8$ ($f = 26$), $8 \le L < 10$ ($f = 30$), $10 \le L < 12$ ($f = 12$).
a) Find the mid-interval value $x$ for each class.
b) Estimate the total sum of all leaf lengths $\sum fx$.
c) Estimate the mean leaf length.
View Solution & Markscheme
Method & Marks:
a) Midpoints: $x_1 = 5$, $x_2 = 7$, $x_3 = 9$, $x_4 = 11$. [A1]
b) $\sum fx = (12 \times 5) + (26 \times 7) + (30 \times 9) + (12 \times 11) = 60 + 182 + 270 + 132 = 644\text{ cm}$. [M1][A1]
c) Estimated mean $= \frac{\sum fx}{\sum f} = \frac{644}{80} = 8.05\text{ cm}$. [A1]
A clinical trial classifies 150 patients by treatment (Drug vs. Placebo) and symptom outcome (Recovered vs. Not Recovered). 90 patients received the Drug. 75 patients recovered in total, of which 55 were in the Drug group.
a) Construct the full two-way table.
b) Compare the recovery rate of the Drug group to the Placebo group.
View Solution & Markscheme
Method & Marks:
a) Placebo group size $= 150 - 90 = 60$.
Drug group: Recovered $= 55 \implies$ Not Recovered $= 90 - 55 = 35$. [M1]
Placebo group: Recovered $= 75 - 55 = 20 \implies$ Not Recovered $= 60 - 20 = 40$. [A1]
b) Recovery rates:
Drug recovery rate $= \frac{55}{90} \approx 61.1\%$.
Placebo recovery rate $= \frac{20}{60} \approx 33.3\%$.
The Drug group shows an approximately $27.8\%$ higher recovery rate than the placebo. [A1][R1]
A dataset of continuous measurements has unequal class intervals: $0 \le x < 10$ ($f = 15$), $10 \le x < 20$ ($f = 25$), $20 \le x < 40$ ($f = 30$), $40 \le x < 70$ ($f = 18$).
a) Determine the class width for each interval.
b) Calculate the frequency density for each interval (where Frequency Density $= \frac{\text{Frequency}}{\text{Class Width}}$).
c) Explain why frequency density must be used instead of raw frequency when constructing a histogram for this data.
View Solution & Markscheme
Method & Marks:
a) Widths: $w_1 = 10, w_2 = 10, w_3 = 20, w_4 = 30$. [A1]
b) Frequency Densities: [M1]
$FD_1 = \frac{15}{10} = 1.5$; $FD_2 = \frac{25}{10} = 2.5$;
$FD_3 = \frac{30}{20} = 1.5$; $FD_4 = \frac{18}{30} = 0.6$. [A1]
c) In a histogram, area represents frequency ($\text{Area} = \text{width} \times \text{height}$). If raw frequencies are used as heights for wider intervals, their areas will be disproportionately inflated, giving a falsely distorted visual impression of the data distribution. [R1]
The ages of participants in an athletic league are recorded as discrete integer values: $16-19$, $20-24$, $25-29$, $30-39$.
a) Explain the difference between recorded class limits and true continuous class boundaries for this discrete integer age data.
b) State the exact mathematical class boundaries and resulting class widths for each group.
View Solution & Markscheme
Method & Marks:
a) Because age advances continuously but is reported rounded to the nearest integer, there is a gap of $1$ year between classes (e.g. between 19 and 20). To eliminate gaps, true class boundaries extend $0.5$ below the lower limit and $0.5$ above the upper limit. [M1][R1]
b) Boundaries and Widths:
$16-19 \implies [15.5, 19.5]$, width $= 19.5 - 15.5 = 4$. [A1]
$20-24 \implies [19.5, 24.5]$, width $= 24.5 - 19.5 = 5$. [A1]
$25-29 \implies [24.5, 29.5]$, width $= 29.5 - 24.5 = 5$.
$30-39 \implies [29.5, 39.5]$, width $= 39.5 - 29.5 = 10$. [A1]
Criterion C & D Investigation on Questionnaire Bias: An eco-club wants to survey community attitudes towards banning single-use plastic bottles.
a) Critique this survey question: "Given the catastrophic pollution of our oceans, don't you agree that single-use plastic bottles should be banned immediately?" Identify two sources of bias.
b) Rewrite the question to be neutral, objective, and statistically valid.
c) The club collects 50 responses by handing questionnaires to patrons outside an organic vegan café on Sunday morning. Evaluate why this data collection method produces sampling bias.
View Solution & Markscheme
Method & Marks:
Part a:
1. Leading language: The phrase "catastrophic pollution" emotionally primes the respondent.
2. Loaded question structure: The phrase "don't you agree" pushes the respondent toward agreement rather than offering a balanced choice. [R1: Criterion C Communication]
Part b: Neutral rewrite:
"To what extent do you support or oppose a ban on single-use plastic bottles?" with balanced Likert options: Strongly Support / Support / Neutral / Oppose / Strongly Oppose. [A1]
Part c: Sampling bias evaluation:
Sampling patrons outside an organic vegan café is convenience sampling and suffers from selection bias. Patrons of this specific location are likely to be significantly more environmentally conscious than the general community, overestimating community support for the ban. [R1: Criterion D Contextual Reflection]
Data Presentation & Grouping Reference Guide
Quantitative Data
$$\text{Discrete vs Continuous}$$
Counts (integers) vs measurements (real numbers).Relative Frequency
$$\text{RF} = \frac{f}{\sum f} = \frac{f}{n}$$
Proportion of total observations.Class Width
$$w = \text{Upper} - \text{Lower Boundary}$$
Width of the continuous interval.Mid-Interval Value
$$x_{\text{mid}} = \frac{L + U}{2}$$
Class midpoint for calculations.Frequency Density
$$\text{FD} = \frac{\text{Frequency}}{\text{Class Width}}$$
Used for unequal histogram intervals.Estimated Mean
$$\bar{x} \approx \frac{\sum fx_{\text{mid}}}{\sum f}$$
Using mid-interval values.Top 5 Common MYP Data Collection Pitfalls
- Confusing Discrete and Continuous Data: Money or shoe sizes can have fractions (e.g., $9.5$), but they are still discrete because they jump in fixed increments and cannot take every continuous real value between them. Continuous data is measured on an unbroken scale.
- Overlapping Class Interval Boundaries: Writing intervals as $10 - 20$ and $20 - 30$ creates ambiguity for a score of $20$. Always use rigorous inequality notation such as $10 \le x < 20$ and $20 \le x < 30$.
- Using Raw Frequencies for Unequal Interval Widths: If intervals have different widths, drawing bar heights using raw frequencies distorts the visual area. You must calculate and plot Frequency Density.
- Ignoring Questionnaire & Sampling Bias: Always evaluate how questions are phrased (leading vs neutral) and where participants are selected. A large sample with systematic bias is worse than a smaller random sample.
- Mistaking True Value for Midpoint: In grouped frequency calculations, $\sum fx$ gives an estimate of the mean, because individual data values are lost within each interval.
Self-Grading Mini Test
5 quiz questions on data taxonomy, frequency tables, and statistical collection.
Which of the following variables is quantitative and continuous?
Explanation: Time is measured on an unbroken continuous scale, meaning it can take any real value depending on timer precision.
What is the mid-interval value for the class interval $30 \le x < 50$?
Explanation: $x_{\text{mid}} = \frac{30 + 50}{2} = \frac{80}{2} = 40$.
A student personally records the arrival times of 50 school buses at the front gate. This is an example of:
Explanation: Because the student directly collects the data first-hand specifically for their own study, it is primary data.
A class interval has boundaries $20 \le x < 50$ and a frequency of $18$. What is its frequency density?
Explanation: Class width $w = 50 - 20 = 30$. Frequency Density $= \frac{\text{Frequency}}{\text{Width}} = \frac{18}{30} = 0.6$.
In a survey of 100 people, 60 like coffee, 50 like tea, and 20 like both. How many like neither?
Explanation: People who like at least one $= 60 + 50 - 20 = 90$. Those who like neither $= 100 - 90 = 10$.