Tamil Nadu HSE First Year Statistics common mistakes
Tamil Nadu HSE First Year Statistics · Understand the traps and how to avoid them.
When computing the median of a grouped frequency distribution, the candidate uses N/2 as the cumulative frequency target but forgets to check whether the class interval is continuous (e.g., limits 20–30, 30–40 sharing a boundary), so the class width and lower boundary are taken straight from the printed limits and the median comes out shifted by half a unit.
Textbook problems in Tamil Nadu First Year Statistics often present inclusive class limits (like 10–19, 20–29) and the candidate must convert to exclusive limits before applying the formula. Candidates memorise the median formula l + ((N/2 − cf)/f) × h but treat l, cf, f, h as plug-in numbers without understanding that l is the lower boundary of the median class and h is the adjusted class width. The habit of copying limits verbatim from the table is the root cause.
What to do instead
Before applying any grouped-data formula, rewrite the table with exclusive class boundaries: for inclusive limits a–b with gap g, the true boundaries are (a − g/2) to (b + g/2). Then explicitly write l (lower boundary of median class), cf (cumulative frequency of the class just before the median class), f (frequency of median class), and h (upper boundary minus lower boundary of that class). Practise with three inclusive-limit tables and verify the median lies inside the median class as a sanity check.
The candidate computes the arithmetic mean of a grouped frequency distribution using the class midpoints but forgets to divide by the total frequency N = Σf, instead dividing by the number of classes, or uses Σfx without dividing at all when the frequencies are given as relative frequencies or percentages.
In ungrouped data the mean is simply Σx/n, and candidates carry the 'divide by the number of items' reflex into grouped data where n is Σf, not the number of rows. When frequencies are percentages (summing to 100) or relative frequencies (summing to 1), the candidate does not recognise that the same formula still applies because Σf already equals the correct denominator.
What to do instead
Always write the formula as x̄ = Σfx / Σf and compute Σf explicitly in a separate column-sum before dividing. For percentage or relative-frequency tables, state 'Σf = 100' or 'Σf = 1' and divide by that. Cross-check with the assumed-mean or step-deviation method: if both methods give the same answer, the denominator is correct.
When computing the standard deviation, the candidate mixes up the population formula (dividing by N) and the sample formula (dividing by n − 1), or uses the raw-score formula σ = √(Σx²/N − (Σx/N)²) but forgets that Σx² means the sum of squared deviations from the mean, not the sum of the raw values squared.
The two SD formulas look similar and the shorthand notation Σx² is ambiguous. Candidates who memorise 'Σx²/N minus mean squared' often substitute raw x values instead of deviations, producing a wrong (and sometimes negative under the square root) result. The distinction between population and sample SD is also blurred because school problems rarely specify which to use.
What to do instead
Adopt one consistent computational form: σ = √(Σ(x − x̄)²/N) for population and s = √(Σ(x − x̄)²/(n − 1)) for sample, and build a three-column table (x, deviation, deviation squared) so Σ(x − x̄)² is unambiguous. Only use the shortcut Σx²/N − (Σx/N)² after writing out what each Σ represents. If a negative value appears under the root, the shortcut was misapplied.
The candidate computes the coefficient of variation as CV = σ/x̄ without multiplying by 100, or compares two data sets by their standard deviations alone when the means differ substantially, concluding the set with the larger SD is more variable.
The CV formula is often written as (σ/x̄) × 100 in some texts and as σ/x̄ in others, and candidates do not realise the ×100 is just a percentage conversion that does not change the ranking. More fundamentally, candidates confuse absolute dispersion (SD) with relative dispersion (CV) and forget that SD is meaningful only relative to the mean it accompanies.
What to do instead
State clearly: CV = (σ/x̄) × 100% is a unit-free percentage, and comparisons of variability across data sets with different means or units must use CV, not SD. When a question asks 'which is more consistent', compute CV for both and pick the smaller. Write the ×100 explicitly so marks are not lost for the percentage.
The candidate computes the Karl Pearson coefficient of skewness as (Mean − Median)/SD or (Mean − Mode)/SD without checking the sign convention, and reports a positive value for a negatively skewed distribution (or vice versa), or uses the empirical relation Mode = 3 Median − 2 Mean without verifying it is appropriate.
Different textbooks define skewness with Mean − Mode or Mode − Mean in the numerator, and the sign flips accordingly. Candidates memorise one version without understanding that positive skewness means the right tail is longer (Mean > Median > Mode). The empirical relation is only approximate and candidates apply it as if exact.
What to do instead
Fix the convention: Skewness = (Mean − Mode)/SD, so positive means right-skewed (Mean > Median > Mode) and negative means left-skewed. Check the sign against the ordering of Mean, Median, Mode every time. Use Mode = 3 Median − 2 Mean only when the distribution is moderately skewed and the mode is not given directly; if the mode is given, use it directly.
The candidate interprets a high positive correlation coefficient as proof that one variable causes the other, or concludes that r = 0 means no relationship at all, ignoring that r measures only linear association.
Correlation and causation are routinely conflated in everyday reasoning, and the syllabus introduces r as a measure of 'relationship' without always stressing its limits. Candidates also forget that a perfect non-linear relationship (e.g., y = x²) can give r ≈ 0, so zero correlation does not imply independence.
What to do instead
State explicitly in every interpretation: r measures the strength and direction of linear association only, and a high |r| does not establish causation. Give the y = x² counterexample in revision notes: symmetric data around zero yields r = 0 despite a perfect relationship. When a question asks to 'interpret', mention direction (positive/negative), strength (weak/moderate/strong by |r|), and the linear-only caveat.
In regression, the candidate writes the line as y = a + bx but then predicts x from y by substituting into the same equation instead of using the regression of x on y, or swaps the regression coefficients b_xy and b_yx when computing r = ±√(b_xy × b_yx).
There are two regression lines and two coefficients, and they are not reciprocals of each other. Candidates memorise one line and reuse it for both directions, forgetting that the regression of x on y has the form x = c + dy with a different slope. The product formula for r also requires the correct pairing of coefficients.
What to do instead
Label the two lines explicitly: y on x is y = a + b_xy·x with b_xy = r(σ_y/σ_x); x on y is x = c + b_yx·y with b_yx = r(σ_x/σ_y). Then r = ±√(b_xy × b_yx), with the sign taken from the coefficients. When asked to predict x given y, use the x-on-y line. Check that both lines pass through (x̄, ȳ).
The candidate adds probabilities of non-mutually-exclusive events directly (P(A ∪ B) = P(A) + P(B)) without subtracting P(A ∩ B), or multiplies probabilities of dependent events as if they were independent.
The addition and multiplication rules are taught with the 'or' and 'and' keywords, and candidates apply them mechanically without checking whether the events can occur together or whether one event changes the probability of the other. The conditional probability P(B|A) is often ignored when drawing without replacement.
What to do instead
Before computing, write down whether the events are mutually exclusive (cannot co-occur) and whether they are independent (occurrence of one does not change the other). Use P(A ∪ B) = P(A) + P(B) − P(A ∩ B) generally, dropping the last term only when mutually exclusive. Use P(A ∩ B) = P(A)·P(B|A) generally, simplifying to P(A)·P(B) only when independent. For without-replacement problems, always update the denominator.
The candidate computes the expected value of a random variable as E(X) = Σx·P(x) but then computes E(X²) as [E(X)]², or computes Var(X) = E(X²) − [E(X)]² while substituting E(X²) = Σx²·P(x) with the wrong probabilities (using x instead of P(x) as the weight).
The notation E(X²) and [E(X)]² look almost identical, and candidates assume squaring commutes with expectation. The variance shortcut formula is correct, but its validity depends on E(X²) being computed as Σx²·P(x), which candidates often muddle with Σx·P(x)² or similar.
What to do instead
Write the two expectations side by side: E(X) = Σx·P(x) and E(X²) = Σx²·P(x), each weighted by P(x), not by x. Then Var(X) = E(X²) − [E(X)]². Compute E(X²) first, square E(X) separately, and subtract. A quick check: for a fair die, E(X) = 3.5 and E(X²) = 91/6 ≈ 15.17, so Var ≈ 2.92, not 12.25.
The candidate treats a sampling distribution as if it were the population distribution, or applies the central limit theorem to claim the sample mean is normally distributed regardless of sample size, and ignores the finite population correction when sampling without replacement from a small population.
The distinction between the distribution of X in the population and the distribution of X̄ across samples is abstract and is often glossed over in rote learning. The CLT is stated as 'sample means are normal' without the 'for large n' qualifier, and the finite population correction √((N − n)/(N − 1)) is easily forgotten when N is small.
What to do instead
Draw the two distributions separately: the population of individual values, and the sampling distribution of X̄ with mean μ and standard error σ/√n. State the CLT precisely: for large n (rule of thumb n ≥ 30), X̄ is approximately normal regardless of the population shape; for small n, normality of X̄ requires the population itself to be normal. When sampling without replacement and n/N > 0.05, multiply the standard error by √((N − n)/(N − 1)).