1. The middle term of $(1+x)^{2n}$ is $\binom{2n}{n}x^n$; factoring $(2n)!$ gives the stated form. (b) Median $\approx45.14$, MD about median $\approx14.83$, coefficient $\approx0.33$. A 5-mark either/or from the Algebra (Binomial Theorem) / Descriptive Statistics units of the Tamil Nadu HSC Class-11 Business Mathematics & Statistics syllabus. Both alternatives are solved. (a) Middle term of $(1+x)^{2n}$. Step 1 — Locate the middle term. $(1+x)^{2n}$ has $2n+1$ terms, so there is a single middle term, the $(n+1)$th: $T_{n+1}=\binom{2n}{n}x^n=\frac{(2n)!}{n!\,n!}\,x^n.$ Step 2 — Split $(2n)!$ into odd and even factors. $(2n)!=\big[1\cdot3\cdot5\cdots(2n-1)\big]\big[2\cdot4\cdot6\cdots(2n)\big].$ The even product $=2\cdot4\cdots2n=2^n(1\cdot2\cdots n)=2^n\,n!.$ Hence $(2n)!=\big[1\cdot3\cdot5\cdots(2n-1)\big]\,2^n\,n!.$ Step 3 — Substitute. $T_{n+1}=\frac{\big[1\cdot3\cdot5\cdots(2n-1)\big]2^n\,n!}{n!\,n!}\,x^n=\frac{1\cdot3\cdot5\cdots(2n-1)\,2^n x^n}{n!}.$ Hence proved.
2. Coefficient of mean deviation about the median. Step 1 — Cumulative frequencies ($N=150$).

Step 2 — Median. $\dfrac{N}{2}=75$ lies in the class $40\text{-}50$ (c.f. jumps $56\to93$). With $L=40,\ cf=56,\ f=37,\ h=10$:

$\text{Median}=L+\frac{\tfrac{N}{2}-cf}{f}\times h=40+\frac{75-56}{37}\times10=40+\frac{190}{37}=45.14.$

Step 3 — Mean deviation about the median. Using $|x-\text{Median}|$ with Median $=45.14$:

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