Statistics
Mean of weighted data; binomial coefficients as frequencies
MMTS_Full_Test_11
Grade 12
Question:
Consider data on $X$ taking values $0, 2, 4, 8, \ldots, 2^n$ with frequencies ${}^nC_0, {}^nC_1, \ldots, {}^nC_n$ respectively. If the mean of this data is $\dfrac{728}{2^n}$, then $n$ is equal to
Step-by-Step Solution
Key Concept: Compute $\bar{x} = \dfrac{\sum f_i x_i}{\sum f_i} = \dfrac{\sum_{k=0}^n 2^k\,{}^nC_k}{2^n} = \dfrac{3^n-1}{2^n}$ using the binomial theorem $\sum 2^k{}^nC_k = 3^n$.
Mean $=\frac{3^n-1}{2^n}=\frac{728}{2^n} \Rightarrow 3^n=729 \Rightarrow n=6$.
Correct Answer: (D) 6