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

X

takes values

0, 0, 2, 6, 12, 20, \ldots, n(n-1)

with frequencies

\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:

56

correct

42

72

90

Solution

Step 1: Compute the mean using the identity

r(r-1)\dbinom{n}{r} = n(n-1)\dbinom{n-2}{r-2}

\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

n

\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 = 2^{16} = 65536

. The median is the mean of the

(N/2)

th and

(N/2+1)

th values, where

N/2 = 32768

. The cumulative frequency up to value

r(r-1)

\sum_{i=0}^{7}\binom{16}{i} = 26333 < 32768 < 39203 = \sum_{i=0}^{8}\binom{16}{i}

Both the

32768

th and

32769

th observations correspond to value

r=8

, giving

r(r-1) = 56

. Step 4: Conclude

\text{Median} = \frac{56+56}{2} = 56

Answer: (1)

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