Averages and Statistics

Averages and Statistics

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Arithmetic mean of n observations equals the sum of all observations divided by n: Mean = Σxᵢ / n. The median is the middle value once the data is arranged in ascending (or descending) order; for even n, it is the average of the two central values. The mode is the value that occurs most frequently. Range = Max − Min, and standard deviation σ = √[Σ(xᵢ − μ)² / n], which quantifies spread around the mean. For two groups with means x̄₁ and x̄₂ and sizes n₁, n₂, the combined mean = (n₁x̄₁ + n₂x̄₂) / (n₁ + n₂). NAT-I (NTS) tests these as 2–4 MCQs: a single dataset mean, a weighted/cricket-style average, a combined-group mean, and one probability question. Remember: the mean is sensitive to outliers, the median is not — pick the median when a question flags an extreme value.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Measures of Central Tendency

The arithmetic mean summarises a dataset by balancing all values around a single number. For raw data x₁, x₂, …, xₙ, Mean x̄ = (x₁ + x₂ + … + xₙ) / n. For a frequency distribution with class marks mᵢ and frequencies fᵢ, the mean is x̄ = Σ(fᵢmᵢ) / Σfᵢ. The median is the positional average: for odd n it is the ((n+1)/2)-th term after ordering, and for even n it is the mean of the (n/2)-th and (n/2 + 1)-th terms. The mode is simply the value with the highest frequency, and a dataset can be unimodal, bimodal, or multimodal. In a grouped frequency table, the modal class is the class with the largest frequency, and the exact mode is refined using Mode = L + [f₁ − f₀] / [2f₁ − f₀ − f₂] × h, where L is the lower limit, f₁ the modal class frequency, f₀ and f₂ the preceding and succeeding frequencies, and h the class width.

Measures of Dispersion

Range captures only the two extreme values: R = xₘₐₓ − xₘᵢₙ. Variance measures the average of squared deviations from the mean. For a population, σ² = Σ(xᵢ − μ)² / N; for a sample drawn from a larger population, divide by (n − 1) instead. Standard deviation σ = √Variance is the spread measure most often quoted in NAT-I options. A small σ indicates data clustered tightly around the mean; a large σ indicates wide scatter.

Weighted and Combined Means

When groups have unequal sizes, a simple average is wrong. The weighted mean is x̄w = Σ(wᵢxᵢ) / Σwᵢ, where each wᵢ is the weight (count, frequency, distance, time, etc.). Merging two groups with sizes n₁, n₂ and means x̄₁, x̄₂ yields x̄combined = (n₁x̄₁ + n₂x̄₂) / (n₁ + n₂). This is the backbone of “find the missing observation” and “find the new average after one value is added/replaced” problems.

Probability Essentials

P(E) = favourable outcomes / total outcomes, with 0 ≤ P(E) ≤ 1. Complementary events satisfy P(E) + P(E′) = 1. For mutually exclusive events, P(A ∪ B) = P(A) + P(B). For independent events, P(A ∩ B) = P(A) · P(B).

Common NAT-I Patterns

The exam usually gives a small dataset (5–10 numbers) and asks for the mean, asks you to find a missing value given a target mean, presents two groups and asks the combined mean, or pairs a probability question with a percentage/ratio setup. Watch for distractors that swap n and (n − 1), or that use range when standard deviation is requested.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

When to Choose Which Average

The choice between mean, median, and mode is dictated by the shape of the distribution and the presence of outliers. For a symmetric distribution without extreme values, the three measures coincide. For a right-skewed (positively skewed) distribution — e.g. income or property prices — a few large values drag the mean upward, so the median better represents the typical observation. For a left-skewed distribution, the mean understates the centre. Mode is preferred for categorical or discrete data (e.g. “most common shoe size sold”). NAT-I occasionally tests this distinction explicitly: a question may say “which measure best represents the typical salary?” and expects the median when an outlier CEO figure is given.

Outliers and the Mean

A single extreme value shifts the arithmetic mean but leaves the median nearly untouched. Consider {2, 3, 4, 5, 100}: mean = 22.8, median = 4. Whenever a NAT-I dataset contains a value that looks “out of place” or the question mentions a “typical” or “representative” value, default to the median. Conversely, when the question asks for total or aggregate information (“total runs scored”, “combined income”), the mean is the correct tool because back-calculating through the mean yields the correct sum.

Variance, Standard Deviation, and Coefficient of Variation

σ is in the same units as the data, which makes it directly comparable to the mean. The coefficient of variation CV = (σ / μ) × 100% is a unit-free ratio used to compare relative variability across datasets with different means or different units (e.g. comparing marks scored out of 50 versus 100). Two datasets with identical means but different σ have different CVs; the one with the higher CV is more dispersed relative to its mean.

Advanced Combined-Mean Tricks

When a value is replaced, removed, or added to a dataset, use New sum = Old sum ± change, then recompute the mean. If one observation xₖ is replaced by y, the new mean is x̄new = x̄old + (y − xₖ) / n. This shortcut avoids re-summing the entire dataset and is the fastest path to NAT-I answers on “average after replacement” items.

Probability Edge Cases

NAT-I probability questions often involve “at least one” events — apply the complement: P(at least one) = 1 − P(none). “Without replacement” draws change probabilities at each step; “with replacement” keeps them constant. Two events are mutually exclusive only if they cannot occur together; do not confuse mutually exclusive with independent — drawing a king and drawing a queen from a deck are mutually exclusive but not independent.

Practice Prompts

1. The mean of 8 observations is 14. If one observation of value 22 is replaced by 6, what is the new mean? (Answer: 14 + (6 − 22)/8 = 14 − 2 = 12.)
2. Two groups of sizes 40 and 60 have means 52 and 68 respectively. A third group of 20 has mean 45. What is the mean of all 120 observations? (Answer: (40·52 + 60·68 + 20·45)/120 = (2080 + 4080 + 900)/120 = 7060/120 ≈ 58.83.)

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