If a variable tikes values 0. 1. 2. 3...n with frequencies 1. C(n, 1) C(n, 2) C(n,3) respectively, then the arithmetic mean

Options

A. 2n
B. n+1
C. n
D. `(n)/(2)`

Answer

To find the arithmetic mean of a variable that takes values from 0 to n with corresponding frequencies given by \( C(n, 0), C(n, 1), C(n, 2), \ldots, C(n, n) \), we will follow these steps:

### Step 1: Identify the values and frequencies
The values of the variable \( x_i \) are:
- \( x_0 = 0 \)
- \( x_1 = 1 \)
- \( x_2 = 2 \)
- \( x_3 = 3 \)
- ...
- \( x_n = n \)

The corresponding frequencies \( f_i \) are:
- \( f_0 = C(n, 0) = 1 \)
- \( f_1 = C(n, 1) \)
- \( f_2 = C(n, 2) \)
- \( f_3 = C(n, 3) \)
- ...
- \( f_n = C(n, n) \)

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

### Step 3: Calculate the numerator \( \sum_{i=0}^{n} x_i f_i \)
We need to compute:
\[ \sum_{i=0}^{n} x_i f_i = 0 \cdot C(n, 0) + 1 \cdot C(n, 1) + 2 \cdot C(n, 2) + 3 \cdot C(n, 3) + \ldots + n \cdot C(n, n) \]
This can be expressed as:
\[ \sum_{i=1}^{n} i \cdot C(n, i) \]

### Step 4: Use the property of binomial coefficients
Using the identity:
\[ \sum_{i=0}^{n} i \cdot C(n, i) = n \cdot 2^{n-1} \]
we can simplify our sum:
\[ \sum_{i=1}^{n} i \cdot C(n, i) = n \cdot 2^{n-1} \]

### Step 5: Calculate the denominator \( \sum_{i=0}^{n} f_i \)
The denominator is the sum of all frequencies:
\[ \sum_{i=0}^{n} f_i = C(n, 0) + C(n, 1) + C(n, 2) + \ldots + C(n, n) = 2^n \]
This is derived from the binomial theorem.

### Step 6: Substitute into the mean formula
Now we can substitute the values we found into the mean formula:
\[ \bar{x} = \frac{n \cdot 2^{n-1}}{2^n} \]

### Step 7: Simplify the expression
Simplifying this gives:
\[ \bar{x} = \frac{n}{2} \]

### Final Answer
Thus, the arithmetic mean of the variable is:
\[ \bar{x} = \frac{n}{2} \]