Kerala Plus Two Statistics (DHSE) common mistakes
Kerala Plus Two Statistics (DHSE) · Understand the traps and how to avoid them.
When computing the median of a grouped frequency distribution, the candidate writes N/2 as the target cumulative frequency but then substitutes the total frequency N (or N/2 again) into the interpolation formula instead of the cumulative frequency of the class preceding the median class, producing a median shifted by a full class width.
The formula is memorised as a string of symbols without a picture of the ogive. Because the median class is defined by the cumulative frequency just exceeding N/2, students conflate the three different frequency quantities (N, N/2, cf of preceding class) that appear in the same expression.
What to do instead
Before substituting, draw a three-column table: class, frequency f, cumulative frequency cf. Circle the first cf that is greater than or equal to N/2, mark its class as the median class, and label the cf of the row above it as C. Write the formula as Median = L + [(N/2 − C)/f] × h and substitute only after each symbol has a circled value in the table. Verify the answer lies inside the median class; if it does not, the substitution is wrong.
The candidate computes the standard deviation of a frequency distribution using the class midpoints as if they were raw observations, dividing by the number of classes or forgetting to weight by frequency, and reports a value far smaller than the actual spread.
The mechanics of the raw-data formula are over-learnt in earlier classes, and the frequency column is treated as decoration rather than as a weight. Students also forget that the mean in the deviation step must be the weighted mean, not the simple average of midpoints.
What to do instead
Always add a column f×x and compute the weighted mean first. Then build a column of deviations d = x − mean, a column f×d, and a column f×d². Compute variance as (Σfd²)/N and take the square root. As a sanity check, the standard deviation must lie between zero and roughly half the range of the data; if it does not, the weighting step has been skipped.
Given a bivariate frequency table or a pair of series, the candidate computes the correlation coefficient using the regression coefficients bxy and byx as r = bxy + byx or r = bxy − byx, or reports r from a regression slope without checking its sign against the scatter.
The identities r = ±√(bxy·byx) and the relation between the two slopes are compressed in revision notes, and the square root and sign rule get lost. Students also assume any two slopes can be multiplied, ignoring that both must be computed with the same designation of dependent variable.
What to do instead
Fix the convention first: write bxy as the slope of y on x and byx as the slope of x on y, and note that r has the sign of both slopes. Then use r = √(bxy × byx) only after confirming the product is non-negative and both slopes share a sign. Cross-check by computing r directly from the covariance formula, and confirm that |r| ≤ 1; if the computed value exceeds 1, the slopes have been interchanged.
When a regression line is requested, the candidate fits y on x when the question asks for the line used to predict x from y, or substitutes the means into the wrong normal equations, producing a line that passes through the point of averages but has the reciprocal slope.
The notation 'y on x' is read as 'y first' rather than as 'y is the dependent variable'. Because both lines pass through the mean point, a wrong choice still looks plausible on a sketch and the error survives checking.
What to do instead
Rewrite the question as a sentence: 'predict [which variable] from [which variable]'. The predicted variable is the dependent one. Write the normal equations for that choice explicitly (Σy = na + bΣx and Σxy = aΣx + bΣx² for y on x), solve for a and b, and confirm the fitted line passes through the mean point. State the dependent variable in the final answer line.
The candidate evaluates a conditional probability by multiplying the two events, writing P(A|B) = P(A)P(B) or P(A∩B) = P(A|B)P(B) with the conditional factor replaced by an unconditional one, and so reports an intersection probability as a conditional one.
The multiplication rule and the definition of conditional probability are learnt as separate formulas, and the conditioning event is not tracked once the symbols are on paper. Independence is also assumed implicitly whenever the word 'and' appears.
What to do instead
Start every such problem by writing the conditioning event with a vertical bar and reading it aloud as 'given that'. Use P(A|B) = P(A∩B)/P(B) as the master formula and rearrange only when needed. Never assume independence unless the question states it or the sampling is explicitly with replacement; if independence is used, write that assumption on the answer sheet.
In a Bayes' theorem problem, the candidate sets up the tree or table with the prior probabilities on the wrong branch, or omits one of the mutually exclusive and exhaustive causes, so the denominator sums to less than one and the posterior probabilities do not add to one.
The problem is read as a single conditional probability rather than as a reversal of conditioning. Students also forget that the denominator must be the total probability of the observed event across every cause.
What to do instead
Draw a two-stage tree: causes on the first branch with priors summing to 1, observed event on the second branch with conditional probabilities. Compute the denominator as the sum of the products along every path that ends in the observed event, and check that the posterior probabilities over all causes sum to 1. If they do not, a cause or a path has been dropped.
The candidate uses the binomial probability formula for a problem that is sampling without replacement from a small finite population, or applies the Poisson formula to counts with a fixed upper bound, without checking the underlying conditions.
Binomial and Poisson are the two formulas most heavily drilled, so pattern-matching on keywords like 'defective' or 'calls' replaces checking the assumptions of fixed number of trials, constant probability, and independent trials.
What to do instead
Before writing any distribution formula, list the four checks: fixed number of trials n, two outcomes per trial, constant probability p, independent trials. If sampling is without replacement from a finite lot, use the hypergeometric approach or state that the binomial is an approximation valid only when the lot is large relative to the sample. For Poisson, confirm the events occur in a fixed interval with a known mean rate and no upper bound on the count.
When computing the expectation and variance of a function of a random variable, the candidate writes E(X²) as [E(X)]² and then reports Var(X) = 0, or computes E(2X + 3) as 2E(X) without adding the constant.
The linearity of expectation is over-generalised to the square, and the constant term in a linear transformation is treated as a scale factor. The identity Var(X) = E(X²) − [E(X)]² is memorised but its two terms are not distinguished.
What to do instead
Build the distribution table with columns x, P(X = x), x·P(X = x), and x²·P(X = x). Compute E(X) from the third column and E(X²) from the fourth, then Var(X) = E(X²) − [E(X)]². For a linear transformation use E(aX + b) = aE(X) + b and Var(aX + b) = a²Var(X), writing the constant term explicitly each time.
In a hypothesis test, the candidate states the alternative hypothesis as the null, or compares the calculated test statistic with the critical value using the wrong tail, and so rejects the null hypothesis when the sample actually supports it.
The logic of proof by contradiction is unfamiliar, and the words 'null' and 'alternative' are treated as interchangeable labels. One-tailed and two-tailed critical values are also recalled as a single number.
What to do instead
Write both hypotheses as full sentences before any calculation: the null always contains equality and represents no effect; the alternative is what the investigator wants to establish. Decide the tail from the wording of the alternative (greater than, less than, or not equal to) and mark it on a sketch of the sampling distribution. Compare the calculated statistic with the critical value for that tail and state the decision in words as well as symbols.
The candidate treats the standard error of the mean as the population standard deviation, using σ in place of σ/√n when constructing a confidence interval or computing a test statistic, so the interval is too wide by a factor of √n.
The symbol σ is recognised but the sampling distribution of the mean is not visualised, and the square-root-of-n factor is dropped during hurried substitution. Small-sample and large-sample formulas are also mixed without checking n.
What to do instead
Write the sampling distribution explicitly before substituting: the sample mean has mean μ and standard deviation σ/√n. Compute the standard error as a separate labelled step and carry it into both the confidence interval and the test statistic. Check whether n is large enough for the normal approximation or whether the t distribution with n − 1 degrees of freedom is required, and state the choice.