Two Sections, One Shield: Which Group Is Really More Consistent?
At a Plus Two result-review meeting in a Kollam school, the statistics teacher places two summary sheets before the principal. Section A has 40 candidates whose mean score in Mathematics is 62 marks with a standard deviation of 8 marks. Section B has 50 candidates whose mean is 58 marks with a standard deviation of 6 marks. The principal wants one combined mean and one combined standard deviation for all 90 candidates, and she also wants to award a 'most consistent section' shield. A parent-teacher association member argues that giving 5 grace marks to every candidate of Section A will both raise its average and shrink its spread, and should therefore settle the shield question. The teacher suspects that the raw standard deviations cannot be compared directly because the two sections have different averages, and wants the argument set out with clear statistical reasoning rather than a single number.
1. Find the mean and standard deviation of the combined group of 90 candidates, showing the deviation method clearly.
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Combined mean
$$\bar{x}=\frac{n_A\bar{x}_A+n_B\bar{x}_B}{n_A+n_B}=\frac{40(62)+50(58)}{90}=\frac{2480+2900}{90}=\frac{5380}{90}=59.78$$
Combined variance - use the formula that allows for the shift of each group mean from the combined mean:
$$\sigma^2=\frac{n_A(\sigma_A^2+d_A^2)+n_B(\sigma_B^2+d_B^2)}{n_A+n_B},\quad d_A=\bar{x}_A-\bar{x},\ d_B=\bar{x}_B-\bar{x}$$
$$d_A=62-59.78=+2.22,\quad d_A^2=4.94$$ $$d_B=58-59.78=-1.78,\quad d_B^2=3.16$$
$$\sigma^2=\frac{40(64+4.94)+50(36+3.16)}{90}=\frac{40(68.94)+50(39.16)}{90}=\frac{2757.6+1958.0}{90}=52.40$$
$$\sigma=\sqrt{52.40}=7.24\text{ marks}$$
Check: the combined mean (59.78) must lie between 58 and 62, and the combined SD (7.24) between 6 and 8 - both hold, so the arithmetic is plausible. Note the combined SD is not the average of 8 and 6; ignoring the between-group deviations $d_A,d_B$ would give a wrong, smaller value.
2. Using an appropriate relative measure, decide which section deserves the consistency shield, and evaluate the PTA member's grace-mark proposal.
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Raw SDs can be compared only when the means are alike. Here they are not, so use the coefficient of variation:
$$CV_A=\frac{8}{62}\times100=12.90%,\qquad CV_B=\frac{6}{58}\times100=10.34%$$
$CV_B<CV_A$, so Section B is relatively more consistent and should get the shield. Its spread is smaller not merely in absolute marks but also as a proportion of its own mean.
Grace marks: adding 5 to every score is a change of origin ($y=x+5$). A constant shift changes the mean (to 67) but leaves every deviation unchanged, so the SD stays exactly 8 and the spread is untouched. The new $CV_A=\frac{8}{67}\times100=11.94%$, still above 10.34%. So the proposal raises the average but cannot make Section A more consistent; it is a change of location, not of dispersion.
The judgement to defend in the answer: consistency is about spread relative to level, and only a change of scale (not of origin) can alter a CV.
3. The school later converts Section A's marks to a 50-mark paper by multiplying every score by 0.5. State the new mean, SD and CV of Section A, and say whether the consistency ranking changes.
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Multiplying every score by 0.5 is a change of scale ($y=0.5x$):
- New mean $=0.5\times62=31$
- New SD $=|0.5|\times8=4$
- New $CV=\frac{4}{31}\times100=12.90%$ - unchanged
In general, for $y=ax+b$ with $a>0$: mean becomes $a\bar{x}+b$, SD becomes $a\sigma$, and the CV is invariant under any change of origin and scale (provided the mean stays positive).
So the ranking does not change: Section B is still relatively more consistent. The examinable point is that comparing means and SDs across papers of different maximum marks is meaningless without a scale-free measure - which is exactly why the CV exists.