If a variable takes the values 0, 1, 2,…,n with frequencies proporiotnal to binomial coefficient `n_(C_(0)), n_(C_(1)), n_(C_(2)),….n_(C_(n))`, then mean of distribution is

Options

A. `(n^(2) - 1)/(12)`
B. `(n)/(2)`
C. `(n)/(4)`
D. `(n)/(3)`

Answer

To find the mean of the distribution where a variable takes values \(0, 1, 2, \ldots, n\) with frequencies proportional to the binomial coefficients \(nC_0, nC_1, nC_2, \ldots, nC_n\), we can follow these steps:

### Step 1: Identify the values and frequencies
The values of the variable \(x_i\) are:
\[ x_i = 0, 1, 2, \ldots, n \]
The corresponding frequencies \(f_i\) are:
\[ f_i = nC_i \quad \text{for } i = 0, 1, 2, \ldots, n \]

### Step 2: Write the formula for the mean
The mean \( \bar{x} \) of a distribution is given by:
\[ \bar{x} = \frac{\sum_{i=0}^{n} x_i f_i}{\sum_{i=0}^{n} f_i} \]

### Step 3: Calculate the denominator
The denominator is the total frequency:
\[ \sum_{i=0}^{n} f_i = \sum_{i=0}^{n} nC_i \]
Using the binomial theorem, we know:
\[ \sum_{i=0}^{n} nC_i = 2^n \]

### Step 4: Calculate the numerator
The numerator is:
\[ \sum_{i=0}^{n} x_i f_i = \sum_{i=0}^{n} i \cdot nC_i \]
To compute this, we can use the property of derivatives of binomial expansion. The sum can be expressed as:
\[ \sum_{i=0}^{n} i \cdot nC_i = n \cdot \sum_{i=1}^{n} n-1C_{i-1} = n \cdot 2^{n-1} \]
This is derived from the fact that \(i \cdot nC_i = nC_{i-1} \cdot i\).

### Step 5: Substitute into the mean formula
Now we can substitute the values we calculated into the mean formula:
\[ \bar{x} = \frac{n \cdot 2^{n-1}}{2^n} \]

### Step 6: Simplify the expression
Simplifying this gives:
\[ \bar{x} = \frac{n}{2} \]

### Conclusion
Thus, the mean of the distribution is:
\[ \bar{x} = \frac{n}{2} \]