Karnataka II PUC Statistics (KSEAB) common mistakes
Karnataka II PUC Statistics (KSEAB) · Understand the traps and how to avoid them.
When computing the median of a grouped frequency distribution, the candidate writes N/2 as the cumulative frequency target but then plugs the frequency of the median class into the formula instead of N/2 minus the preceding cumulative frequency, producing a median that sits at the class boundary.
The formula (l + ((N/2 − cf)/f) × h) is memorised as a string of symbols without attaching meaning to cf; students conflate 'cumulative frequency up to the median class' with 'cumulative frequency before it', and the two differ by exactly f, which is easy to overlook in a hurry.
What to do instead
Before substituting, build a four-column table (class, f, cf, and a column headed 'N/2 − cf_before') and fill the last column only for the median class. Practise five problems where you state in words: 'cf_before is the cf of the class just above the median class.' Cross-check by seeing whether the answer lies inside the median class interval.
The candidate computes standard deviation from a grouped series using the raw class marks x but forgets to multiply by the class width h after using the step-deviation variable d = (x − A)/h, so the SD comes out h times too small.
Step-deviation shortens arithmetic, and the final back-substitution σ = h × √(Σfd²/N − (Σfd/N)²) is treated as an afterthought; students who have practised only with h = 1 (unit-width classes) never internalise that the scale factor must be restored.
What to do instead
Adopt a fixed habit: write 'h = ___' at the top of every grouped problem and circle it. After computing σ_d, write the line 'σ_x = h × σ_d' explicitly before evaluating. Verify with a quick sanity check: the SD of a grouped series should be roughly one-quarter to one-third of the range.
In a question asking which average is most suitable, the candidate answers 'mean' by default for open-ended classes, extreme outliers, or qualitative data such as brand preference rankings.
The arithmetic mean is drilled hardest in practice, and the properties of mean, median and mode are learned as definitions rather than as decision rules tied to data type and class structure.
What to do instead
Memorise a three-line decision table: open-end classes or extreme values → median; rates, ratios and per-unit averages → geometric mean; qualitative or ranked data → mode or median as appropriate. For each past question on suitability, write one sentence justifying the rejection of the other two averages, not just the acceptance of one.
The candidate reports Karl Pearson's coefficient of skewness as negative when the mean exceeds the mode, or reports a skewness coefficient outside the range −1 to +1 without noticing.
The sign convention is tied to the tail direction, which is easy to reverse mentally, and the formula (Mean − Mode)/SD is applied mechanically without checking which of the three averages is largest.
What to do instead
Always compute mean, median and mode first and write the inequality (e.g. mean > median > mode). Then state 'positive skew, tail to the right' before computing. Treat any |Sk| > 1 as a signal to recheck the SD and the mode, since the Pearson coefficient is bounded for unimodal moderately skewed data.
When computing the correlation coefficient, the candidate uses Σxy without converting to deviations from the mean, or mixes the assumed-mean shortcut with raw sums, and ends up with r > 1 or a sign that contradicts the scatter of the data.
The shortcut formula using assumed means and step deviations looks similar to the direct formula, and the correction term (Σdx)(Σdy)/N is dropped under time pressure; the arithmetic is long enough that a single slip propagates silently.
What to do instead
Standardise on one layout: columns for x, y, dx = x − A, dy = y − B, dxdy, dx², dy², with totals row. Compute r = [NΣdxdy − ΣdxΣdy] / √([NΣdx² − (Σdx)²][NΣdy² − (Σdy)²]) and verify that the numerator's sign matches the sign of Σdxdy when the means are near the assumed values. Always check −1 ≤ r ≤ 1.
The candidate writes both regression lines as y on x, or solves for the regression of x on y by simply rearranging the y-on-x equation, producing a regression coefficient that does not satisfy b_yx × b_xy = r².
The two lines are visually similar and the identifying phrase 'of y on x' is read as 'y and x'; students also assume rearrangement is legitimate algebra rather than a different least-squares minimisation.
What to do instead
Anchor on the minimisation: 'y on x' minimises vertical deviations, so it predicts y and has slope b_yx = r(σ_y/σ_x). Write the two equations side by side with their slopes labelled before solving. Finish every regression problem by checking b_yx × b_xy = r² and that both lines pass through (x̄, ȳ).
In index number problems, the candidate computes a weighted index using base-year quantities when the question specifies current-year weights (Paasche), or applies the time-reversal and factor-reversal tests to the wrong formula.
Laspeyres and Paasche formulas differ only in the subscript on q, and under exam pressure the subscripts blur; the tests are memorised as names without a worked example of each.
What to do instead
Write the weight symbol explicitly (p₀q₀ for Laspeyres, p₀q₁ for Paasche) above each term before calculating. Memorise that Laspeyres satisfies the time-reversal test only in the factor-reversal sense and that Fisher's ideal index satisfies both tests; verify by substituting the formula into the test definition once by hand for each index you are likely to be asked about.
When fitting a straight-line trend by least squares, the candidate uses the year numbers (e.g. 2015, 2016, …) directly as the x variable, producing enormous normal equations and arithmetic errors, or assigns the wrong origin when the series has an even number of years.
The shortcut of coding x as deviations from the middle year is taught as a trick rather than as a rule, and students do not realise that for an even number of years the origin falls between two years and the x values must be half-units (−1.5, −0.5, 0.5, 1.5) or rescaled.
What to do instead
Always recode: for odd n, x = …, −2, −1, 0, 1, 2, …; for even n, use x = −3, −1, 1, 3 (or half-units) so that Σx = 0 and the normal equations collapse to a = Σy/n and b = Σxy/Σx². State the origin and unit of x in the answer, and convert a forecast back to the original year label.
The candidate adds probabilities for 'at least one' events in binomial problems (P(X ≥ 1) = np) or uses the binomial formula for a Poisson situation where n is large and p is small without checking np.
The phrase 'at least one' invites the complement rule, but students reach for np because it appears in the mean; and the binomial–Poisson distinction is taught as 'Poisson for rare events' without the operational test n ≥ 50, p ≤ 0.05, np moderate.
What to do instead
For 'at least one', write P(X ≥ 1) = 1 − P(X = 0) = 1 − qⁿ first, and only then substitute. For distribution choice, record n, p and np at the top of the problem: if n is large, p small and np < 10, use Poisson with λ = np; otherwise binomial. Verify that probabilities sum to 1 over the support.
The candidate applies the normal distribution to a discrete binomial or Poisson variable without the continuity correction, so P(X ≥ 5) is computed as P(Z ≥ (5 − μ)/σ) instead of P(Z ≥ (4.5 − μ)/σ).
Normal approximation is taught with the standardisation formula z = (x − μ)/σ, and the ±0.5 adjustment is presented as a footnote; in a discrete-to-continuous transition the boundary shift is the whole point and is easily dropped.
What to do instead
Before standardising, draw a small number line marking the discrete values and shade the region, then move each boundary by 0.5 away from the shaded region. Write the corrected inequality explicitly (e.g. X ≥ 5 becomes X > 4.5). Check the approximation conditions np > 5 and nq > 5 before starting.