Sampling and Data Displays | IB Math AA SL

SL 4.1 · Core Concepts

1. Populations, Samples, and Sampling Methods

A population is the entire group of individuals under consideration; a sample is a subset selected to represent the population. Reliable statistical inferences require minimizing selection bias.

Simple Random Sampling: Every member of the population has an equal and independent chance of selection (e.g. using a random number generator). Free from bias, but can be impractical for large or geographically dispersed populations.
Systematic Sampling: Selecting every $k^{\text{th}}$ individual from an ordered list after a random starting point between $1$ and $k$, where $k = \frac{N}{n}$. Simple to implement, but vulnerable to bias if periodic cycles exist in the population ordering.
Stratified Sampling: The population is divided into mutually exclusive strata (e.g., age groups, grades), and random samples are drawn proportionally from each stratum: $$\text{Sample size from stratum } i = \frac{\text{Size of stratum } i}{N} \times n$$ Guarantees representation of all sub-groups.
Quota Sampling: Stratified sampling without random selection; the researcher fills pre-set quotas using convenience. Prone to interviewer bias.
Convenience Sampling: Selecting individuals who are readily available. Highly prone to severe selection bias and non-representative conclusions.

Worked Example 1.1 · Stratified Sample Sizing

A high school has $1200$ students: $420$ in Grade 9, $330$ in Grade 10, $270$ in Grade 11, and $180$ in Grade 12. A researcher intends to survey a stratified sample of $80$ students. Determine how many students should be sampled from each grade.

Total population $N = 1200$, sample size $n = 80$. Sampling fraction $= \frac{80}{1200} = \frac{1}{15}$.
Grade 9: $\frac{420}{15} = 28$ students.
Grade 10: $\frac{330}{15} = 22$ students.
Grade 11: $\frac{270}{15} = 18$ students.
Grade 12: $\frac{180}{15} = 12$ students.
Verify total: $28 + 22 + 18 + 12 = 80$.

Answer: 28 Grade 9, 22 Grade 10, 18 Grade 11, and 12 Grade 12 students.

SL 4.2 · Continuous Data

2. Histograms and Frequency Density

For grouped continuous data with unequal class intervals, the vertical axis of a histogram must represent frequency density, not raw frequency. The area of each bar is proportional to its frequency.

Frequency Density Formula: $$\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}$$
Frequency as Area: $$\text{Frequency} = \text{Frequency Density} \times \text{Class Width} = \text{Area of bar}$$
If class intervals are equal, the height can be directly proportional to raw frequency, but labeling the vertical axis as frequency density is always mathematically rigorous.

Worked Example 1.2 · Frequency from Unequal Intervals

In a histogram displaying commute times, a bar spans the interval $10 \le t < 25$ minutes with a frequency density of $1.6\text{ min}^{-1}$. An adjacent bar spans $25 \le t < 30$ minutes with a frequency of $15$. Calculate the frequency of the first class and the height of the second bar.

For the first class $10 \le t < 25$: Class width $= 25 - 10 = 15$.
$\text{Frequency} = \text{Density} \times \text{Width} = 1.6 \times 15 = 24$.
For the second class $25 \le t < 30$: Class width $= 30 - 25 = 5$.
$\text{Frequency Density} = \frac{\text{Frequency}}{\text{Width}} = \frac{15}{5} = 3.0\text{ min}^{-1}$.

Answer: Frequency $= 24$; second bar height $= 3.0\text{ min}^{-1}$.

SL 4.2 · Visual Analytics

3. Cumulative Frequency Curves and Box Plots

A cumulative frequency curve (ogive) plots running totals against upper class boundaries. Projecting horizontal quartiles directly generates the 5-number summary and box-and-whisker plot.

Plotting Rule: Points are plotted at the upper class boundary of each interval with cumulative frequency on the vertical axis. The curve begins at the lower boundary of the first interval with a cumulative frequency of $0$.
Key Quartiles from $n$ Observations:
- Lower Quartile ($Q_1$): Cumulative frequency $= 0.25n$
- Median ($Q_2$): Cumulative frequency $= 0.50n$
- Upper Quartile ($Q_3$): Cumulative frequency $= 0.75n$
- Interquartile Range ($\text{IQR}$): $\text{IQR} = Q_3 - Q_1$
Outlier Identification: An observation $x$ is an outlier if: $$x < Q_1 - 1.5 \times \text{IQR} \quad \text{or} \quad x > Q_3 + 1.5 \times \text{IQR}$$

Figure 4.1: Horizontal quartile projections from cumulative frequency heights $0.25n, 0.50n, 0.75n$ determine the five-number summary and box-and-whisker plot.

Worked Example 1.3 · Stratified Sampling Allocation

A multinational company employs $1200$ workers across three divisions: $650$ in Operations, $350$ in Marketing, and $200$ in Research. A representative stratified sample of $180$ employees is to be surveyed.
(a) Calculate the number of employees that must be selected from each division. [3 marks]
(b) State one major advantage of using stratified sampling over simple random sampling in this scenario. [1 mark]

Part (a): Proportional allocation formula $n_i = \frac{N_i}{N} \times n$: $$\text{Operations: } \frac{650}{1200} \times 180 = 97.5 \implies 98 \quad [M1][A1]$$ $$\text{Marketing: } \frac{350}{1200} \times 180 = 52.5 \implies 52 \quad [A1]$$ $$\text{Research: } \frac{200}{1200} \times 180 = 30 \quad [A1]$$ Check: $98 + 52 + 30 = 180$, satisfying the total sample requirement.
Part (b): Advantage of stratified sampling: Stratified sampling guarantees that all subgroups, particularly smaller divisions like Research ($200$ workers), are proportionally represented without risk of omission [R1].

Answers: (a) Operations: $98$, Marketing: $52$, Research: $30$; (b) Guarantees proportional representation of all subgroups.

Worked Example 1.4 · Cumulative Frequency & Percentile Analysis

An examination is taken by $240$ students. From the cumulative frequency graph, the 60th percentile score is $68$ marks and the 85th percentile is $82$ marks.
(a) Find the number of students who scored between $68$ and $82$ marks. [3 marks]
(b) The top $10\%$ of candidates are awarded a Grade 7. Find the minimum mark required for a Grade 7. [2 marks]

Part (a): Number of students between 60th and 85th percentiles: $$\text{Cumulative count at 60th percentile} = 0.60 \times 240 = 144 \quad [M1]$$ $$\text{Cumulative count at 85th percentile} = 0.85 \times 240 = 204 \quad [A1]$$ $$\text{Students in range} = 204 - 144 = 60 \text{ students} \quad [A1]$$
Part (b): Grade 7 threshold (top 10% corresponds to 90th percentile): $$\text{Rank} = 0.90 \times 240 = 216\text{th student} \quad [M1]$$ Read the score corresponding to cumulative frequency $216$ from the graph [A1].

Answers: (a) $60$ students; (b) Score at cumulative frequency $216$.

Worked Example 1.5 · Boxplot Outlier Verification

A dataset has lower quartile $Q_1 = 42$ and upper quartile $Q_3 = 66$. The minimum value is $12$ and the maximum value is $105$.
(a) Determine whether $12$ is an outlier. [3 marks]
(b) Determine whether $105$ is an outlier. [2 marks]

Part (a): Compute Interquartile Range and lower outlier boundary: $\text{IQR} = Q_3 - Q_1 = 66 - 42 = 24 \quad [M1]$ $\text{Lower fence} = Q_1 - 1.5 \times \text{IQR} = 42 - 1.5(24) = 42 - 36 = 6 \quad [A1]$ Since $12 > 6$, the value $12$ lies above the lower fence, so $12$ is not an outlier [R1][A1].
Part (b): Compute upper outlier boundary: $\text{Upper fence} = Q_3 + 1.5 \times \text{IQR} = 66 + 1.5(24) = 66 + 36 = 102 \quad [M1]$ Since $105 > 102$, the data value $105$ lies beyond the upper fence and is confirmed as an outlier [R1][A1].

Answers: (a) $12 > 6$, not an outlier; (b) $105 > 102$, confirmed outlier.

Topic 4.1 & 4.2 Practice Exercises

10 tiered problems covering sampling methods, frequency density, cumulative frequency ogives, box plots, and outlier boundaries.

Question 1

A sports club has $500$ members: $250$ adults, $150$ juniors, and $100$ seniors. An analyst takes a stratified sample of $60$ members. Calculate the number of seniors in the sample.

Show Step-by-Step Solution

Step 1: Identify the proportion of seniors in the entire club:

$$\text{Proportion} = \frac{100}{500} = \frac{1}{5} = 0.20$$

Step 2: Apply the proportion to the desired sample size $n = 60$:

$$\text{Seniors sampled} = 60 \times \frac{1}{5} = 12$$

Final Answer: 12 seniors

Question 2

In a histogram with unequal class widths, an interval $20 \le x < 35$ has a frequency density of $2.4$. Calculate the frequency of this interval.

Show Step-by-Step Solution

Step 1: Calculate class width:

$$\text{Class Width} = 35 - 20 = 15$$

Step 2: Apply the frequency formula:

$$\text{Frequency} = \text{Frequency Density} \times \text{Class Width} = 2.4 \times 15 = 36$$

Final Answer: 36

Question 3

A dataset has lower quartile $Q_1 = 34$ and upper quartile $Q_3 = 58$. Determine whether an observed score of $96$ is an outlier according to the standard $1.5 \times \text{IQR}$ rule.

Show Step-by-Step Solution

Step 1: Calculate the Interquartile Range ($\text{IQR}$):

$$\text{IQR} = Q_3 - Q_1 = 58 - 34 = 24$$

Step 2: Calculate the upper outlier boundary:

$$\text{Upper Fence} = Q_3 + 1.5 \times \text{IQR} = 58 + 1.5(24) = 58 + 36 = 94$$

Step 3: Compare the observation:

Since $96 > 94$, the observation lies strictly above the upper fence.

Final Answer: Yes, 96 is an outlier because $96 > Q_3 + 1.5 \times \text{IQR} = 94$.

Question 4

State which sampling method is used in each scenario:
(a) A quality control officer tests every $20^{\text{th}}$ smartphone off the production line.
(b) A student surveys the first 15 classmates who enter the cafeteria at lunchtime.

Show Step-by-Step Solution

Part (a): Selecting every $k^{\text{th}}$ item from an ordered list/stream is Systematic Sampling.

Part (b): Selecting subjects purely based on accessibility without random selection is Convenience Sampling.

Final Answer: (a) Systematic Sampling; (b) Convenience Sampling

Question 5

A cumulative frequency graph representing the test scores of $160$ students shows the following coordinates $(x, \text{cf})$:
$(40, 24)$, $(52, 40)$, $(65, 80)$, $(78, 120)$, $(85, 144)$, $(100, 160)$.
(a) Estimate the median score.
(b) Estimate the Interquartile Range ($\text{IQR}$).
(c) The passing mark was $85$. Estimate the percentage of students who passed the test.

Show Step-by-Step Solution

Part (a): Total students $n = 160$. Median position is $0.5n = 80$.
From the data, $\text{cf} = 80$ corresponds to $x = 65$. So Median $= 65$.

Part (b): Lower quartile $Q_1$ position is $0.25(160) = 40 \implies Q_1 = 52$.
Upper quartile $Q_3$ position is $0.75(160) = 120 \implies Q_3 = 78$.
$$\text{IQR} = Q_3 - Q_1 = 78 - 52 = 26$$

Part (c): At mark $x = 85$, the cumulative frequency is $144$. This means $144$ students scored $85$ or less.
Number of students who passed ($> 85$) $= 160 - 144 = 16$.
$$\text{Percentage passing} = \frac{16}{160} \times 100\% = 10\%$$

Final Answer: (a) Median $= 65$; (b) $\text{IQR} = 26$; (c) $10\%$ passed

Question 6

The heights of trees in two forests, Forest A and Forest B, are summarized below in five-number summaries $[\text{Min}, Q_1, Q_2, Q_3, \text{Max}]$:
Forest A: $[4, 12, 18, 24, 38]$
Forest B: $[8, 17, 21, 26, 32]$
(a) Compare the medians and spreads ($\text{IQR}$) of the two forests.
(b) Check Forest A for potential outliers.

Show Step-by-Step Solution

Part (a):
• Medians: Forest B has a higher median ($21\text{ m}$) compared to Forest A ($18\text{ m}$), indicating that typical tree height is greater in Forest B.
• Spread: $\text{IQR}_A = 24 - 12 = 12\text{ m}$; $\text{IQR}_B = 26 - 17 = 9\text{ m}$. Forest A exhibits greater dispersion in its central $50\%$ than Forest B.

Part (b): Outlier Check for Forest A:
$\text{IQR}_A = 12$.
Lower fence: $Q_1 - 1.5(\text{IQR}) = 12 - 1.5(12) = 12 - 18 = -6$. Since $\text{Min} = 4 > -6$, there are no lower outliers.
Upper fence: $Q_3 + 1.5(\text{IQR}) = 24 + 1.5(12) = 24 + 18 = 42$. Since $\text{Max} = 38 < 42$, there are no upper outliers.

Final Answer: (a) Forest B has higher median ($21 > 18$), Forest A has greater spread ($\text{IQR } 12 > 9$); (b) No outliers in Forest A (fences are $-6$ and $42$).

Question 7

A market research company wishes to survey opinions on electric vehicles in a city. They send surveyors to stand outside a luxury shopping mall from 14:00 to 16:00 on a Tuesday afternoon.
(a) Identify the sampling technique used.
(b) State two distinct sources of bias inherent to this sampling design and describe how each affects the validity of the conclusions.

Show Step-by-Step Solution

Part (a): The technique is Convenience Sampling (or opportunity sampling).

Part (b) Sources of Bias:
1. Location bias (Socio-economic bias): Patrons of a luxury shopping mall tend to have significantly higher household incomes than the general city population, leading to an overestimation of EV affordability and willingness to purchase.
2. Timing bias: Surveying between 14:00 and 16:00 on a weekday excludes regular full-time working individuals and students, over-representing retirees or non-working individuals.

Final Answer: (a) Convenience sampling; (b) Location bias (higher income) and Timing bias (excludes full-time workers).

Question 8

The table below displays the waiting times $w$ (in minutes) of 80 hospital patients:

Time interval ($w$)	$0 \le w < 10$	$10 \le w < 20$	$20 \le w < 40$	$40 \le w < 70$
Frequency	18	26	24	12

(a) Calculate the frequency density for each interval.
(b) Estimate the mean waiting time using interval midpoints.

Show Step-by-Step Solution

Part (a): Frequency Densities:
• $0 \le w < 10$: $\text{width} = 10 \implies \text{density} = \frac{18}{10} = 1.8$
• $10 \le w < 20$: $\text{width} = 10 \implies \text{density} = \frac{26}{10} = 2.6$
• $20 \le w < 40$: $\text{width} = 20 \implies \text{density} = \frac{24}{20} = 1.2$
• $40 \le w < 70$: $\text{width} = 30 \implies \text{density} = \frac{12}{30} = 0.4$

Part (b): Mean Estimate:
Midpoints $m_i$: $5, 15, 30, 55$.
$$\sum f_i m_i = 18(5) + 26(15) + 24(30) + 12(55) = 90 + 390 + 720 + 660 = 1860$$ $$\bar{w} = \frac{\sum f_i m_i}{n} = \frac{1860}{80} = 23.25\text{ minutes}$$

Final Answer: (a) Densities: $1.8, 2.6, 1.2, 0.4$; (b) Estimated mean $= 23.25\text{ minutes}$.

Question 9

A set of raw continuous exam scores $X$ has median $54$, lower quartile $40$, and upper quartile $66$. The examining board adjusts every student's score using the linear transformation $Y = 1.2X + 8$.
(a) Determine the median and interquartile range of the adjusted scores $Y$.
(b) A student had a raw score of $x = 2$. Determine whether this raw score is an outlier in the original distribution, and prove whether their transformed score $y$ is an outlier in the adjusted distribution.

Show Step-by-Step Solution

Part (a):
Under linear transformation $Y = aX + b$ ($a = 1.2, b = 8$):
$$\text{Median}_Y = 1.2(\text{Median}_X) + 8 = 1.2(54) + 8 = 64.8 + 8 = 72.8$$ $$Q_{1, Y} = 1.2(40) + 8 = 48 + 8 = 56$$ $$Q_{3, Y} = 1.2(66) + 8 = 79.2 + 8 = 87.2$$ $$\text{IQR}_Y = Q_{3, Y} - Q_{1, Y} = 87.2 - 56 = 31.2$$ Notice that $\text{IQR}_Y = 1.2 \times \text{IQR}_X = 1.2(66 - 40) = 1.2(26) = 31.2$ (the constant shift $+8$ cancels out in the spread!).

Part (b): Outlier Fences:
Original lower fence: $$Q_{1, X} - 1.5(\text{IQR}_X) = 40 - 1.5(26) = 40 - 39 = 1$$ Since raw score $x = 2 > 1$, it is not an outlier in the original distribution.
Now test transformed score: $y = 1.2(2) + 8 = 10.4$.
Adjusted lower fence: $$Q_{1, Y} - 1.5(\text{IQR}_Y) = 56 - 1.5(31.2) = 56 - 46.8 = 9.2$$ Since $y = 10.4 > 9.2$, the transformed score is also not an outlier.
Proof principle: Linear transformations preserve relative outlier boundaries because $y - \text{Fence}_Y = (1.2x + 8) - (1.2 \cdot \text{Fence}_X + 8) = 1.2(x - \text{Fence}_X)$.

Final Answer: (a) Median $= 72.8$, $\text{IQR} = 31.2$; (b) Not an outlier in either distribution ($2 > 1$ and $10.4 > 9.2$).

Question 10

A study measures the reaction times $t$ (in milliseconds) of $200$ airline pilots. The cumulative frequency distribution is tabulated below:

Reaction time ($t \le \dots$)	200	250	300	350	400	450	500
Cumulative Frequency	0	18	62	140	182	196	200

(a) Use linear interpolation to estimate the median reaction time.
(b) The fastest $15\%$ of pilots receive an elite accreditation. Estimate the threshold reaction time for this award.
(c) An external observer claims that the distribution of reaction times is positively skewed. Justify whether the quartiles support this claim.

Show Step-by-Step Solution

Part (a): Median Estimation:
Total pilots $n = 200$. Median position is at cumulative frequency $= 100$.
This falls within the interval $300 < t \le 350$, where cumulative frequency increases from $62$ to $140$ (frequency of interval $= 78$).
Using linear interpolation: $$t_{\text{med}} = 300 + \left(\frac{100 - 62}{140 - 62}\right) \times (350 - 300) = 300 + \left(\frac{38}{78}\right) \times 50 \approx 300 + 24.36 = 324.4\text{ ms}$$

Part (b): Fastest 15%:
"Fastest" reaction time means lowest time values. The lowest $15\%$ corresponds to: $$0.15 \times 200 = 30^{\text{th}}\text{ pilot}$$ This falls in $250 < t \le 300$, where $\text{cf}$ goes from $18$ to $62$ (frequency $= 44$).
$$t_{15\%} = 250 + \left(\frac{30 - 18}{62 - 18}\right) \times (300 - 250) = 250 + \left(\frac{12}{44}\right) \times 50 \approx 250 + 13.64 = 263.6\text{ ms}$$

Part (c): Skewness Assessment:
$Q_1$ position is at $\text{cf} = 50$: $Q_1 = 250 + \frac{50 - 18}{44} \times 50 = 250 + 36.36 = 286.4\text{ ms}$.
$Q_3$ position is at $\text{cf} = 150$: falls in $350 < t \le 400$ ($\text{cf}$ goes $140 \to 182$, freq $= 42$).
$Q_3 = 350 + \frac{150 - 140}{42} \times 50 = 350 + 11.9 = 361.9\text{ ms}$.
Compare distances from median ($M = 324.4$):
$M - Q_1 = 324.4 - 286.4 = 38.0\text{ ms}$
$Q_3 - M = 361.9 - 324.4 = 37.5\text{ ms}$
Since $(M - Q_1) \approx (Q_3 - M)$ (with $M - Q_1$ slightly larger), the central $50\%$ is virtually symmetric or very slightly negatively skewed, not positively skewed. The observer's claim is refuted.

Final Answer: (a) $324.4\text{ ms}$; (b) $263.6\text{ ms}$; (c) Refuted: $M - Q_1 \approx 38.0$ and $Q_3 - M \approx 37.5$, showing approximately symmetric distribution.

IB Formula Booklet References (Topic 4)

Interquartile Range:

$$\text{IQR} = Q_3 - Q_1$$

Mean of Grouped Data:

$$\bar{x} = \frac{\sum_{i=1}^k f_i x_i}{n}, \quad \text{where } n = \sum_{i=1}^k f_i$$

Must-Memorize Formulas & Definitions

Frequency Density:

$$\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}$$

Outlier Fences ($1.5 \times \text{IQR}$ Rule):

$$\text{Lower Fence} = Q_1 - 1.5(\text{IQR}), \quad \text{Upper Fence} = Q_3 + 1.5(\text{IQR})$$

Proportional Stratified Sampling:

$$n_i = \left(\frac{N_i}{N}\right) \times n$$

Critical IB Examination Pitfalls

Plotting at Midpoints on Cumulative Frequency Curves: Always plot cumulative frequency at the upper class boundary, NEVER at the midpoint. (Midpoints are used for histograms and estimated means, not ogives!).
Reading Histograms with Unequal Widths: On histograms with varying class widths, height does NOT represent frequency. You must multiply $\text{height} \times \text{width}$ to get the count.
Confusing Stratified and Quota Sampling: Stratified sampling uses random sampling within each stratum; Quota sampling fills quotas via non-random convenience.
Forgetting to Label Outliers: In box-and-whisker plots, if an observation is beyond the $1.5 \times \text{IQR}$ fence, the whisker ends at the most extreme non-outlier value, and outliers are plotted as separate isolated points (dots or crosses).

Topic 4.1 & 4.2 Diagnostic Mini Test

Section A: 4 Diagnostic MCQs with immediate feedback. Section B: 2 Exam-style Free Response Questions with full markschemes.

Section A: Diagnostic Multiple Choice

Question 1: Which sampling technique divides the population into distinct sub-groups and then selects members at random in proportion to the group sizes?

Question 2: In a histogram, a class interval is $10 \le x < 30$ and has frequency $50$. What is the height (frequency density) of this bar?

Question 3: A dataset of 200 items has $Q_1 = 20$ and $Q_3 = 36$. Which of the following values is an outlier?

Question 4: When plotting a cumulative frequency curve for grouped continuous data, cumulative frequency values should always be plotted against:

Section B: Examination-Style Free Response Questions

Question 5 (6 marks)

A company employs $480$ staff: $120$ in Sales, $240$ in Production, and $120$ in Administration. An HR consultant conducts a survey on job satisfaction.
(a) Calculate the number of staff from each department needed for a proportional stratified sample of size $40$. [3 marks]
(b) The consultant decides instead to wait at the main entrance at 08:45 and survey the first $40$ staff who enter. Identify this sampling method and explain why it may produce a biased sample. [3 marks]

Show Full Markscheme

(a) Stratified sample sizing:
Total staff $N = 480$, sample size $n = 40$. Sampling fraction $= \frac{40}{480} = \frac{1}{12}$. [M1]
Sales: $\frac{120}{12} = 10$ staff [A1]
Production: $\frac{240}{12} = 20$ staff [A1]
Administration: $\frac{120}{12} = 10$ staff

(b) Alternative sampling evaluation:
• Sampling method: Convenience sampling (or opportunity sampling). [A1]
• Bias explanation: Production workers may have shift start times earlier than 08:45 (e.g., 07:00), meaning they are systematically excluded from entering at 08:45. [R1]
• Furthermore, individuals arriving right before 08:45 may experience different commute stress or job roles compared to flexible or remote employees. [R1]

Question 6 (8 marks)

The cumulative frequency curve of examination scores for $240$ candidates is analyzed.
• Median score $= 62$
• Interquartile range $\text{IQR} = 22$
• Lowest score $= 18$, highest score $= 98$
• Lower quartile $Q_1 = 50$

(a) Show that the upper quartile $Q_3$ is $72$. [1 mark]
(b) Determine whether the lowest score ($18$) and highest score ($98$) are statistical outliers. [4 marks]
(c) The top $10\%$ of candidates received a Grade 7. Find the cumulative frequency corresponding to the minimum mark for a Grade 7. [1 mark]
(d) State with a reason whether the distribution of marks shows positive, negative, or zero skewness. [2 marks]

Show Full Markscheme

(a) Upper quartile:
$Q_3 = Q_1 + \text{IQR} = 50 + 22 = 72$. [A1]

(b) Outlier fences:
$\text{Step factor} = 1.5 \times \text{IQR} = 1.5 \times 22 = 33$. [M1]
Lower fence $= Q_1 - 33 = 50 - 33 = 17$. [A1]
Upper fence $= Q_3 + 33 = 72 + 33 = 105$. [A1]
Since $\text{Min} = 18 > 17$ and $\text{Max} = 98 < 105$, neither score is an outlier. [R1]

(c) Cumulative frequency for Grade 7:
Top $10\%$ of candidates means $90\%$ scored below this boundary.
$\text{Cumulative frequency} = 0.90 \times 240 = 216$. [A1]

(d) Skewness:
Compare quartile intervals from median ($M = 62$):
$M - Q_1 = 62 - 50 = 12$
$Q_3 - M = 72 - 62 = 10$ [M1]
Since $M - Q_1 > Q_3 - M$, the lower tail is longer/more stretched than the upper tail, indicating negative skewness (left-skewed). [R1]