Statistics and Probability
Statistics and probability are two closely related branches of mathematics that deal with data collection, organization, analysis, and the likelihood of events. For the Post-UTME examination, you need to understand measures of central tendency (mean, median, mode), measures of dispersion (range, variance, standard deviation), probability concepts, and the basic probability rules.
Measures of Central Tendency
These describe the center or typical value of a data set.
Mean (Arithmetic Average)
Mean = (Sum of all values) / (Number of values)
Example: Find the mean of: 12, 15, 18, 22, 25 Sum = 12+15+18+22+25 = 92; Number = 5 Mean = 92/5 = 18.4
Mean for grouped data (discrete): If values x₁, x₂, … with frequencies f₁, f₂, …: Mean = Σ(fx) / Σf
Example: Marks: 3, 4, 5, 6, 7 with frequencies 2, 5, 8, 3, 2 Σ(fx) = 3×2 + 4×5 + 5×8 + 6×3 + 7×2 = 6+20+40+18+14 = 98 Σf = 2+5+8+3+2 = 20 Mean = 98/20 = 4.9
Median
The median is the middle value when data is arranged in order.
Odd number of values: The middle one. Example: 3, 7, 2, 9, 5 → Arrange: 2, 3, 5, 7, 9 → Median = 5
Even number of values: Average of the two middle values. Example: 3, 7, 2, 9, 5, 4 → Arrange: 2, 3, 4, 5, 7, 9 → (4+5)/2 = 4.5
Mode
The mode is the value that appears most frequently. Example: 2, 4, 4, 6, 7, 8, 4, 9 → Mode = 4 (appears 3 times) A data set can have no mode (all values unique), one mode (unimodal), or two modes (bimodal).
Measures of Dispersion
These describe how spread out the data is.
Range
Range = Maximum value − Minimum value Example: Data: 12, 15, 18, 22, 25 → Range = 25 − 12 = 13
Variance and Standard Deviation
Variance (σ²) = Average of squared deviations from the mean Standard Deviation (σ) = √Variance
For ungrouped data:
- Find the mean (x̄)
- Find deviation of each value from the mean (x − x̄)
- Square each deviation: (x − x̄)²
- Find mean of squared deviations = Variance
- Standard deviation = √Variance
Example: Data: 4, 8, 6, 5, 3, 2, 8, 9, 2, 6 Mean = (4+8+6+5+3+2+8+9+2+6)/10 = 53/10 = 5.3 Deviations: −1.3, 2.7, 0.7, −0.3, −2.3, −3.3, 2.7, 3.7, −3.3, 0.7 Squared: 1.69, 7.29, 0.49, 0.09, 5.29, 10.89, 7.29, 13.69, 10.89, 0.49 Sum of squared = 58.1; Variance = 58.1/10 = 5.81 Standard deviation = √5.81 ≈ 2.41
Probability
Probability is a measure of the likelihood that an event will occur. It is expressed as a number between 0 and 1 (or 0% to 100%).
Basic Probability Formula
P(Event) = Number of favorable outcomes / Total number of possible outcomes
Example: What is the probability of getting a prime number when a fair die is thrown? Prime numbers on a die: 2, 3, 5 (3 numbers) Total possible outcomes: 6 P(prime) = 3/6 = 1/2
Complementary Events
P(not A) = 1 − P(A)
Example: P(getting at least one 6 in two throws of a die) = 1 − P(no 6 in either throw) = 1 − (5/6) × (5/6) = 1 − 25/36 = 11/36
Addition Rule (Mutually Exclusive Events)
P(A or B) = P(A) + P(B) Example: P(rolling a 2 or a 3 on a die) = P(2) + P(3) = 1/6 + 1/6 = 1/3
Addition Rule (Non-Mutually Exclusive Events)
P(A or B) = P(A) + P(B) − P(A and B) Example: P(King or Heart from a deck of 52 cards) = P(King) + P(Heart) − P(King of Hearts) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13
Multiplication Rule (Independent Events)
P(A and B) = P(A) × P(B)
Example: Probability of getting two heads in two coin tosses? P(H and H) = (1/2) × (1/2) = 1/4
Conditional Probability
P(A given B) = P(A and B) / P(B)
Key Post-UTME Exam Facts
- Mean = Sum/N; Median = middle value (or avg of 2 middles); Mode = most frequent
- Range = Max − Min
- Standard deviation = √(variance) = √(Σ(x−x̄)²/N) for population
- Probability: Always between 0 and 1
- Complementary: P(not A) = 1 − P(A)
- Addition: P(A or B) = P(A) + P(B) − P(A and B) (for non-mutually exclusive)
- Multiplication: P(A and B) = P(A) × P(B) (for independent events)
- ⚡ Exam tip: When asked for “at least one,” use the complement: P(at least one) = 1 − P(none)
🟢 Lite — Quick Review (1h–1d)
Rapid summary for last-minute revision before your exam.
- Mean: Sum of values divided by count
- Median: Middle value when sorted (or average of 2 middle values for even count)
- Mode: Most frequent value
- Range: Max − Min
- Probability: Favorable outcomes / Total outcomes; always between 0 and 1
- Complement: P(not A) = 1 − P(A)
- Independent events: P(A and B) = P(A) × P(B)
- ⚡ Exam tip: “At least one” problems are easiest by complementary probability
🟡 Standard — Regular Study (2d–2mo)
Standard content for students with a few days to months.
Data Presentation and Interpretation
Frequency Distribution
When data has many values, organize into classes (groups).
Class width = (Max − Min) / Number of classes
Example: Heights of 30 students:
| Class | Tally | Frequency |
|---|---|---|
| 150-154 | ||
| 155-159 | ||
| 160-164 | ||
| 165-169 | ||
| 170-174 |
Histogram
A histogram is a bar graph of a frequency distribution with no gaps between bars.
Finding Median from Grouped Data
Median = L + [(N/2 − cf)/f] × w Where: L = lower boundary of median class, N = total frequency, cf = cumulative frequency before median class, f = frequency of median class, w = class width.
Probability Trees
For multi-step experiments, draw probability trees.
Example: A bag contains 3 red and 2 blue balls. Two balls are drawn without replacement. Find P(one red, one blue).
Tree approach:
- P(R then B) = (3/5) × (2/4) = 6/20
- P(B then R) = (2/5) × (3/4) = 6/20
- P(one of each) = 6/20 + 6/20 = 12/20 = 3/5
🔴 Extended — Deep Study (3mo+)
Comprehensive coverage for students on a longer study timeline.
Measures of Position: Quartiles
Quartiles divide data into four equal parts:
- Q1 (First quartile): 25% of data below
- Q2 (Second quartile): 50% of data below (same as median)
- Q3 (Third quartile): 75% of data below
Interquartile Range (IQR) = Q3 − Q1
Example: Find Q1 and Q3 for: 2, 4, 6, 8, 10, 12, 14, 16 Q2 (median) = (8+10)/2 = 9 Lower half (2,4,6,8): Q1 = (4+6)/2 = 5 Upper half (10,12,14,16): Q3 = (12+14)/2 = 13 IQR = 13 − 5 = 8
Permutations and Combinations
Permutations (Order matters)
nPr = n! / (n−r)! — number of ways to arrange r items from n items.
Example: How many ways to arrange 3 books from 5? 5P3 = 5!/(5-3)! = 120/2 = 60 ways
Combinations (Order doesn’t matter)
nCr = n! / [r!(n−r)!] — number of ways to choose r items from n.
Example: How many ways to choose 3 students from 10 for a committee? 10C3 = 10!/(3!×7!) = (10×9×8)/(3×2×1) = 720/6 = 120 ways
Bayes’ Theorem (Advanced)
P(A|B) = [P(B|A) × P(A)] / P(B)
Used for finding reverse conditional probabilities.
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