Statistics and Probability: Charts, Averages, Spread, Ogives and Chance

Why this topic matters

Expect one statistics question and one probability question in WAEC and NECO Paper 2. The statistics question usually asks for a grouped frequency table, the mean, a cumulative frequency curve (ogive) and the median or quartiles. Paper 1 tests quick averages, pie-chart angles and simple probability.

Presenting data

The number of children in 30 families gives this frequency table:

• 0 children: f = 3
• 1 child: f = 5
• 2 children: f = 9
• 3 children: f = 7
• 4 children: f = 4
• 5 children: f = 2

Totals: Σf = 30 and Σfx = 70. Always check that Σf equals the number of items.

Pie chart angle = (f ÷ Σf) × 360°. For the families data the angles are 36°, 60°, 108°, 84°, 48° and 24°, which add up to 360°.

Histogram vs bar chart:

• In a histogram the bars touch, because the data is continuous.
• The x-axis uses class boundaries: the class 1–10 becomes 0.5–10.5.

Averages

Ungrouped data (the families data):

• Mean = Σfx ÷ Σf = 70 ÷ 30 = 2.33
• Median: there are 30 values, so take the average of the 15th and 16th. The cumulative frequency passes 15 and 16 in the x = 2 row, so the median = 2.
• Mode: the value with the highest frequency (f = 9) is 2.

Mean of grouped data:

• Use the class mark (mid-value), e.g. (1 + 10) ÷ 2 = 5.5.
• For the marks of 40 students: mean = Σfx ÷ Σf = 1290 ÷ 40 = 32.25.
• Assumed-mean shortcut: take A = 35.5 and d = x − A. Then Σfd = −130, so the mean = 35.5 − 130 ÷ 40 = 32.25, the same answer with smaller numbers.

Median (grouped) = L + [(N/2 − F) ÷ f] × c, where:

• L = lower class boundary of the median class;
• F = cumulative frequency before the median class;
• f = frequency of the median class;
• c = class width.

For the 40 marks: N/2 = 20, so the median class is 31–40. Then median = 30.5 + (20 − 17) ÷ 12 × 10 = 33.

Mode (grouped) = L + [D₁ ÷ (D₁ + D₂)] × c, where D₁ = f(modal) − f(class before) and D₂ = f(modal) − f(class after). For the 40 marks: 30.5 + 3 ÷ 8 × 10 = 34.25.

Measures of spread

• Range = highest − lowest. It is quick, but one extreme value affects it.
• Mean deviation = Σ|x − mean| ÷ n. For 2, 4, 6, 8, 10 (mean 6): (4 + 2 + 0 + 2 + 4) ÷ 5 = 2.4.

• Variance = Σ(x − mean)² ÷ n, and standard deviation (SD) = √variance.
• For 2, 4, 6, 8, 10: Σ(x − 6)² = 40, so the variance = 8 and SD = √8 = 2.83.
• From a frequency table: variance = Σf(x − mean)² ÷ Σf. For example, Σfd² = 14 and Σf = 12 give variance 1.17 and SD = 1.08.

A larger SD means the data is more spread out. The deviations always add up to 0, which is why we square them.

Cumulative frequency and the ogive

To draw an ogive:

1. Add each frequency to the running total.
2. Plot the cumulative frequency against the upper class boundary.
3. Start the curve at zero, at the lowest boundary.
4. Join the points with a smooth S-shaped curve.
5. Label both axes and state your scale.

Reading the ogive (N = 40):

• Median: the N/2 = 20th value, about 33.
• Lower quartile Q₁: the N/4 = 10th value, about 22.7.
• Upper quartile Q₃: the 3N/4 = 30th value, about 41.9.
• Interquartile range = Q₃ − Q₁ ≈ 19.2; semi-interquartile range ≈ 9.6.
• Number of students who scored above 40.5 = 40 − 29 = 11.

Probability

P(E) = number of favourable outcomes ÷ total number of possible outcomes, and P(not E) = 1 − P(E). A probability is always between 0 (impossible) and 1 (certain).

Two dice (36 outcomes):

• P(sum = 8) = 5/36
• P(sum ≥ 10) = 6/36 = 1/6
• P(both even) = 9/36 = 1/4
• P(a double) = 6/36 = 1/6

Addition rule ("OR"):

• For events that cannot happen together: P(A or B) = P(A) + P(B). So P(red or green) = 5/10 + 2/10 = 7/10.
• For events that can overlap: P(A or B) = P(A) + P(B) − P(A and B). For a card numbered 1–20 that is even or a multiple of 5: 10/20 + 4/20 − 2/20 = 3/5.

Multiplication rule ("AND") for independent events: P(A and B) = P(A) × P(B).

• Two reds with replacement: 5/10 × 5/10 = 1/4.
• Two reds without replacement: 5/10 × 4/9 = 2/9. Reduce both the top and the bottom on the second draw.

Tree diagrams: multiply along the branches, then add the outcomes you need. With 4 red and 6 blue balls and two draws without replacement:

• P(one of each) = 4/15 + 4/15 = 8/15
• P(at least one red) = 1 − P(both blue) = 1 − 1/3 = 2/3

Top exam traps

• Arrange the data in order before finding the median. With an even number of values, the median is the average of the two middle values.
• Divide Σfx by Σf, not by the number of classes.
• Use class boundaries (30.5), not class limits (31), in the formulas.
• Give the modal value, not the modal frequency.
• A probability greater than 1, or negative, is always wrong.
• "At least one" → work out 1 − P(none).