Statistics & Linear ProgrammingmediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Statistics & Linear Programming: Variable Takes Values Frequencies Respectively Mean Data Med (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
A variable XX takes values 0,0,2,6,12,20,…,n(n−1)0, 0, 2, 6, 12, 20, \ldots, n(n-1) with frequencies (n0),(n1),(n2),…,(nn)\binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n} respectively. If the mean of this data is 60, then its median is:
A5656correct
B4242
C7272
D9090
Solution
Step 1: Compute the mean using the identity r(r−1)(nr)=n(n−1)(n−2r−2)r(r-1)\dbinom{n}{r} = n(n-1)\dbinom{n-2}{r-2}
xˉ=∑r=0nr(r−1)(nr)2n=n(n−1)∑r=2n(n−2r−2)2n=n(n−1)⋅2n−22n=n(n−1)4\bar{x} = \frac{\displaystyle\sum_{r=0}^n r(r-1)\binom{n}{r}}{2^n} = \frac{n(n-1)\displaystyle\sum_{r=2}^n\binom{n-2}{r-2}}{2^n} = \frac{n(n-1)\cdot2^{n-2}}{2^n} = \frac{n(n-1)}{4}
 Step 2: Solve for nn
n(n−1)4=60  ⟹  n2−n−240=0  ⟹  (n−16)(n+15)=0  ⟹  n=16\frac{n(n-1)}{4} = 60 \implies n^2-n-240=0 \implies (n-16)(n+15)=0 \implies n=16
 Step 3: Locate the median Total frequency N=216=65536N = 2^{16} = 65536. The median is the mean of the (N/2)(N/2)th and (N/2+1)(N/2+1)th values, where N/2=32768N/2 = 32768. The cumulative frequency up to value r(r−1)r(r-1):
∑i=07(16i)=26333<32768<39203=∑i=08(16i)\sum_{i=0}^{7}\binom{16}{i} = 26333 < 32768 < 39203 = \sum_{i=0}^{8}\binom{16}{i}
 Both the 3276832768th and 3276932769th observations correspond to value r=8r=8, giving r(r−1)=56r(r-1) = 56. Step 4: Conclude
Median=56+562=56\text{Median} = \frac{56+56}{2} = 56
 Answer: (1)
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