Question

A variable X takes values $0, 0, 2, 6, 12, 20, \dots, n(n-1)$ with frequencies $\binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \binom{n}{3}, \binom{n}{4}, \binom{n}{5}, \dots, \binom{n}{n}$ respectively. If the mean of this data is 60, then its median is :

• A. 56
• B. 42
• C. 72
• D. 90

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Mean of a frequency distribution is $\frac{\sum f_i x_i}{\sum f_i}$. Median is the value corresponding to the cumulative frequency $\frac{N}{2}$.
: Key Formula or Approach:
The $r$-th value is $x_r = r(r-1)$ for $r = 0, \dots, n$ with frequency $f_r = \binom{n}{r}$.
Using identity $r(r-1)\binom{n}{r} = n(n-1)\binom{n-2}{r-2}$.
Step 2: Detailed Explanation:
Sum of frequencies $\sum f_r = 2^n$.
$\sum f_r x_r = \sum_{r=2}^n \binom{n}{r} r(r-1) = n(n-1) \sum_{r=2}^n \binom{n-2}{r-2} = n(n-1) 2^{n-2}$.
Mean $= \frac{n(n-1) 2^{n-2}}{2^n} = \frac{n(n-1)}{4} = 60 \implies n(n-1) = 240 \implies n = 16$.
Total observations $= 2^{16}$.
Cumulative frequency reaches half at $r = 8$ (middle of binomial expansion).
Median value is $x_8 = 8(8-1) = 56$.
Step 3: Final Answer:
The median is 56.