A variable X takes values 0, 0, 2, 6, 12, 20, | vedclass

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(A) The mean $\bar{X}$ is given by $\frac{\sum f_i x_i}{\sum f_i}$.Here, $\sum f_i = \sum_{k=0}^n {}^nC_k = 2^n$.The values are $x_k = k(k-1)$ for $k=

Vedclass · Accepted Answer

$56$

Vedclass · Answer

(A) The mean $\bar{X}$ is given by $\frac{\sum f_i x_i}{\sum f_i}$.Here, $\sum f_i = \sum_{k=0}^n {}^nC_k = 2^n$.The values are $x_k = k(k-1)$ for $k=0, 1, ..., n$.The sum $\sum_{k=0}^n k(k-1) {}^nC_k = n(n-1) 2^{n-2}$.Thus, $\bar{X} = \frac{n(n-1) 2^{n-2}}{2^n} = \frac{n(n-1)}{4} = 60$.$n^2 - n - 240 = 0 \implies (n-16)(n+15) = 0$. Since $n > 0$, $n = 16$.The total frequency is $2^{16} = 65536$. The median is the value at the $\frac{65536+1}{2} \approx 32768$th position.The distribution of frequencies follows the binomial expansion of $(1+1)^n$. The median of a binomial distribution $B(n, p)$ with $p=0.5$ is approximately $np = 16 	imes 0.5 = 8$. The value at $k=8$ is $8(8-1) = 56$.

$A$ variable $X$ takes values $0, 0, 2, 6, 12, 20, ..., n(n-1)$ with frequencies $^nC_0, ^nC_1, ^nC_2, ^nC_3, ^nC_4, ^nC_5, ..., ^nC_n$, respectively. If the mean of this data is $60$, then its median is:

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