NIMCET 2027 2014 Previous Year Questions (PYQs) – Page 8 of 12

Given: \(n=200\), reported mean \(=40\) ⇒ reported total $$S_{\text{reported}}=200\times 40=8000.$$

Two scores were misread: true \((43,35)\) but used as \((34,53)\).

Correct the total: $$S_{\text{correct}}=S_{\text{reported}}-\big(34+53\big)+\big(43+35\big) =8000-87+78=7991.$$

Corrected mean: $$\bar{x}_{\text{correct}}=\frac{S_{\text{correct}}}{n} =\frac{7991}{200}=39.955\;\approx\;39.96.$$

Answer: \(39.955\) (≈ \(39.96\)).

AP: \(a,\,a+d,\,a+2d,\,\dots,\,a+2nd\) has \(2n+1\) terms. The mean is the middle term \(a+nd\).

Mean deviation from the mean (MD): average of absolute deviations from \(a+nd\).

Symmetry gives pairs at distances \(kd\) for \(k=1,\dots,n\). Hence

Answer: \(\displaystyle \boxed{\frac{d\,n(n+1)}{2n+1}}\).