IB / IGCSE Statistics

Statistics Mastery Workbook

20 Core Questions · Concept Review · Casio fx Calculator Guide

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📊 Topic 1 — Measures of Central Tendency & Spread

Mean · Median · Mode · Range · Variance · Standard Deviation · IQR

💡 Concept: Mean, Median, Mode
\[ \bar{x} = \frac{\sum x_i}{n} \qquad \text{or for frequency table:} \quad \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \]
🔢 Casio fx — Mean & Standard Deviation
Steps: Press MODE → 2 (STAT) → 1 (1-VAR)
Enter all data values with EXE after each. Then press AC → SHIFT + 1 → 5 (Var) → 2 for \(\bar{x}\) (mean), 3 for \(\sigma x\) (population SD), 4 for \(sx\) (sample SD).
For frequency data: use LIST mode — enter x values in List 1, frequencies in List 2, then SET Freq = List 2.
✏️ Worked Example

Data: 3, 5, 7, 7, 9, 11, 11, 11, 13  (n = 9)

Mean: (3+5+7+7+9+11+11+11+13) ÷ 9 = 77 ÷ 9 ≈ 8.56
Median: Position = (9+1)/2 = 5th value = 9
Mode: 11 appears 3 times → 11
Q 01 Mean from raw data Level 1–2

A student recorded the number of books read each month for 8 months:

2, 5, 3, 8, 5, 4, 5, 8

The mean number of books read per month is equal to 5. If the value 8 appears twice and the student claims the mean is exactly 5, verify whether this is correct and state the mean.

Step-by-step solution:
  1. Sum all values: 2 + 5 + 3 + 8 + 5 + 4 + 5 + 8 = 40
  2. Count: n = 8
  3. Mean = 40 ÷ 8 = 5.0 ✓
  4. The student's claim IS correct.
🔢 Casio fx Check
MODE → 2 → 1, enter: 2 EXE 5 EXE 3 EXE 8 EXE 5 EXE 4 EXE 5 EXE 8 EXE → AC → SHIFT 1 → 5 → 2 → EXE → displays 5.0
Q 02 Finding unknown from mean Level 1–2

The number of siblings of students in a class is:

0, 3, 2, 1, 1, 0, 2, 1, 0, 0, 0, x, 2, 1, 0, 3

The mean number of siblings is equal to 1. Find the value of \(x\).

  1. Sum without x: 0+3+2+1+1+0+2+1+0+0+0+2+1+0+3 = 16
  2. Total n = 16 values. Mean = 1 → total sum must = 16
  3. 16 + x = 16 → \(x = 0\)
Q 03 Median of ordered data Level 1–2

The ages of 11 students in a class are:

14, 16, 15, 13, 17, 15, 14, 16, 15, 14, 13

What is the median age?

  1. Order: 13, 13, 14, 14, 14, 15, 15, 15, 16, 16, 17
  2. n = 11 → median position = (11+1)/2 = 6th value
  3. 6th value = 15
Q 04 Mode & median from frequency table Level 1–2

The number of cups of coffee consumed by 55 teachers is shown below:

Cups (x)012345
Frequency461015128

What are the mode and median for this data?

  1. Mode = value with highest frequency = 3 (f = 15)
  2. n = 55 → median position = (55+1)/2 = 28th value
  3. Cumulative freq: 0→4, 1→10, 2→20, 3→35. The 28th value falls in the group x = 3
  4. Median = 3
🔢 Casio fx — Frequency Table
MODE → 2 → 1. In SETUP (SHIFT + MODE) choose List Freq ON.
Enter x values (0–5) in List 1, frequencies in List 2 → AC → SHIFT 1 → 5 → 2 for mean, 3 for σ.
Q 05 Mean from frequency table Level 3–4

Using the same coffee data from Q4, calculate the mean number of cups consumed.

  1. \(\sum fx\) = 0×4 + 1×6 + 2×10 + 3×15 + 4×12 + 5×8 = 0+6+20+45+48+40 = 159
  2. \(\sum f\) = 4+6+10+15+12+8 = 55
  3. Mean = 159 ÷ 55 = 2.89 (to 3 s.f.)

📐 Topic 2 — Standard Deviation & Consistency

Variance · Population SD · Sample SD · Coefficient of Variation

💡 Concept: Standard Deviation
\[ \sigma = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n}} \qquad \text{CV} = \frac{\sigma}{\bar{x}} \times 100\% \]
🔢 Casio fx — Standard Deviation
After entering data in STAT mode → AC → SHIFT 1 → 5 → select:
3 = \(\sigma x\) (population SD, used when data = entire group)
4 = \(sx\) (sample SD, used when data = sample from larger group)
For school/exam problems, use \(\sigma x\) unless told otherwise.
✏️ Worked Example

Grade 9: mean = 2.5, SD = 1.2  |  Grade 10: mean = 2.1, SD = 1.4
Which grade has more consistent attendance?

Grade 9 has SD = 1.2 which is lower than Grade 10's SD = 1.4.
Lower SD → data closer to mean → Grade 9 is more consistent.
Q 06 Comparing consistency Level 3–4

Two basketball players' points per game over a season are recorded:

Player A: mean = 18.4, standard deviation = 3.2
Player B: mean = 21.0, standard deviation = 2.1

Which player is more consistent and why?

  1. Consistency is measured by standard deviation, not the mean
  2. Player A: σ = 3.2  |  Player B: σ = 2.1
  3. 2.1 < 3.2 → Player B has smaller spread → more consistent
Q 07 Standard deviation calculation Level 3–4

Using the coffee consumption data (Q4), the mean is 2.89 cups. What is the standard deviation (to 2 decimal places)?

  1. \(\sum f(x-\bar{x})^2\): use \(\bar{x} = 2.89\)
  2. = 4(0-2.89)²+6(1-2.89)²+10(2-2.89)²+15(3-2.89)²+12(4-2.89)²+8(5-2.89)²
  3. ≈ 33.37+21.47+7.92+0.18+14.78+35.69 = 113.41
  4. σ = √(113.41/55) = √2.0620 ≈ 1.44
🔢 Casio fx — Quick SD
Enter frequency table data as described in Q4 calc tip → SHIFT 1 → 5 → 3 → EXE → reads σx = 1.44
Q 08 Coefficient of Variation Level 5–6

Two factories produce bolts. Factory X: mean length = 50 mm, SD = 2 mm. Factory Y: mean length = 80 mm, SD = 2.8 mm.

Using the Coefficient of Variation, which factory produces more consistent bolts?

  1. CV = (σ/mean) × 100%
  2. Factory X: CV = (2/50) × 100% = 4%
  3. Factory Y: CV = (2.8/80) × 100% = 3.5%
  4. Lower CV = more consistent → Factory Y (3.5% < 4%)
  5. Note: simply comparing SD is misleading when means differ!

χ² Topic 3 — Chi-Squared Test of Independence

Hypotheses · Expected Frequencies · Degrees of Freedom · Critical Values

💡 Concept: Chi-Squared (χ²) Test
\[ \chi^2_{calc} = \sum \frac{(O - E)^2}{E} \qquad df = (r-1)(c-1) \]
🔢 Casio fx — Chi-Squared Test
MODE → 2 (STAT) → enter observed data in matrix form if supported, OR calculate manually:
For each cell: store O value, calculate E = (row total × col total) / grand total
Then compute (O−E)² / E for each cell and sum them.
Casio fx-991EX/CG series supports χ² directly via STAT → TEST → CHI.
✏️ Worked Example

Fish store data (from exam): 2 rows (Male/Female) × 4 cols (fish types). Grand total = 393.

Expected females buying Goldfish = (row total females × col total goldfish) / grand total
= (170 × 119) / 393 ≈ 51.5
df = (2−1)(4−1) = 3
Q 09 χ² hypotheses Level 3–4

A researcher tests whether gender and preferred music genre are independent. Which pair of hypotheses is correct for a chi-squared test?

  1. The null hypothesis H₀ always states no effect / independence
  2. H₀: Gender and music preference are independent
  3. H₁: Gender and music preference are not independent (there is an association)
Q 10 Expected frequency & degrees of freedom Level 3–4

In a survey of 200 people, data on pet preference (Cat/Dog/Fish) and age group (Under 30 / 30 and over) is collected. What are the degrees of freedom for a chi-squared test of independence?

  1. Rows = 2 (age groups), Columns = 3 (pet types)
  2. df = (r − 1)(c − 1) = (2−1)(3−1) = 1 × 2 = 2
Q 11 Expected frequency calculation Level 5–6

In a chi-squared test, the following data is observed:

BarbsGuppiesGoldfishTetrasTotal
Males57656734223
Females45485225170
Total10211311959393

What is the expected number of females who purchase Goldfish?

  1. E = (row total × column total) / grand total
  2. Row total (Females) = 170
  3. Column total (Goldfish) = 119
  4. Grand total = 393
  5. E = (170 × 119) / 393 = 20230 / 393 ≈ 51.5
Q 12 χ² decision rule Level 5–6

A chi-squared test at the 5% significance level gives \(\chi^2_{calc} = 5.2\) with \(df = 3\). The critical value is \(\chi^2_{crit} = 7.815\).

What is the correct conclusion?

  1. Decision rule: Reject H₀ if \(\chi^2_{calc} > \chi^2_{crit}\)
  2. Here: 5.2 < 7.815 → we fail to reject H₀
  3. Conclusion: Insufficient evidence at 5% level to conclude an association exists

🔢 Topic 4 — Data Types, IQR & Outliers

Discrete · Continuous · Quartiles · Box Plots · Outliers

💡 Concept: Types of Data & IQR
\[ \text{IQR} = Q_3 - Q_1 \qquad \text{Outlier if: } x < Q_1 - 1.5\cdot\text{IQR} \text{ or } x > Q_3 + 1.5\cdot\text{IQR} \]
🔢 Casio fx — Quartiles
After entering data in STAT 1-VAR mode → SHIFT 1 → 5:
6 → minX, 7 → Q1, 8 → Med, 9 → Q3, 0 → maxX
(Menu numbers may vary by model — look for "Min/Q1/Med/Q3/Max")
Q 13 Discrete vs Continuous Level 1–2

Which of the following is an example of continuous data?

  1. A, B — countable whole numbers → discrete
  2. D — shoe sizes come in fixed values (6, 6.5, 7…) → discrete
  3. C — time can be any value (e.g. 12.347 seconds) → continuous ✓
Q 14 IQR calculation Level 3–4

The ordered test scores of 10 students are:

42, 55, 58, 63, 67, 71, 74, 78, 83, 90

Find the interquartile range (IQR).

  1. n = 10. Lower half: 42,55,58,63,67 → Q1 = median = 58
  2. Upper half: 71,74,78,83,90 → Q3 = median = 78
  3. IQR = Q3 − Q1 = 78 − 58 = 20
Q 15 Outlier detection Level 5–6

A dataset has Q1 = 20, Q3 = 35. Using the 1.5 × IQR rule, which value would be classified as an outlier?

  1. IQR = 35 − 20 = 15
  2. Lower fence: Q1 − 1.5×IQR = 20 − 22.5 = −2.5
  3. Upper fence: Q3 + 1.5×IQR = 35 + 22.5 = 57.5
  4. Values outside [−2.5, 57.5] are outliers
  5. A = 5 ✓ (within range), B = 10 ✓, C = 58 → outlier! (58 > 57.5), D = 55 ✓

🎯 Topic 5 — Grouped Data, Histograms & Probability

Grouped frequency · Histograms · Cumulative frequency · Basic probability

💡 Concept: Grouped Data & Estimated Mean
\[ \text{Estimated Mean} = \frac{\sum f_i \cdot m_i}{\sum f_i} \qquad \text{Frequency Density} = \frac{f}{\text{class width}} \]
Q 16 Estimated mean from grouped data Level 3–4

The ages of 40 concert attendees are grouped as follows:

Age group10–2020–3030–4040–50
Frequency814126

What is the estimated mean age?

  1. Midpoints: 15, 25, 35, 45
  2. ∑fm = 8×15 + 14×25 + 12×35 + 6×45 = 120+350+420+270 = 1160
  3. ∑f = 40
  4. Estimated mean = 1160/40 = 29.0
Q 17 Cumulative frequency & percentiles Level 3–4

From the concert data in Q16, what is the cumulative frequency for ages up to 30?

  1. Cumulative frequency up to 30 = freq(10–20) + freq(20–30)
  2. = 8 + 14 = 22
  3. This means 22 out of 40 attendees are under 30 years old (55%)
Q 18 Basic probability from data Level 3–4

From the fish store data in Q11, a customer is chosen at random. What is the probability that the customer is a female who bought Tetras?

  1. P(female AND Tetras) = (number of females buying Tetras) / (total customers)
  2. = 25 / 393
  3. Note: 25/170 would be P(Tetras | Female) — a conditional probability
Q 19 Frequency density Level 5–6

A histogram uses unequal class widths. The class 20 ≤ x < 25 has frequency 30 and class width 5. The class 25 ≤ x < 35 has frequency 40 and class width 10. Which class has the greater frequency density?

  1. Frequency Density = Frequency ÷ Class Width
  2. Class 20–25: FD = 30 ÷ 5 = 6
  3. Class 25–35: FD = 40 ÷ 10 = 4
  4. 6 > 4 → Class 20–25 has greater frequency density
Q 20 Comprehensive: choose the correct analysis Level 5–6

A teacher wants to determine whether there is a statistically significant relationship between students' preferred learning style (Visual / Auditory / Kinaesthetic) and their exam grade (A / B / C / D). The teacher collects data from 120 students.

Which statistical test is most appropriate, and what are the degrees of freedom?

  1. Two categorical variables → use chi-squared test of independence (not t-test)
  2. Rows: 3 learning styles, Columns: 4 grade categories
  3. df = (r−1)(c−1) = (3−1)(4−1) = 2 × 3 = 6
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