To find the mean of a distribution where a variable takes the values 0, 1, 2, ..., n with frequencies proportional to the binomial coefficients \( \binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n} \), we can follow these steps:
### Step 1: Define the Mean
The mean \( \mu \) of a distribution can be calculated using the formula:
\[
\mu = \frac{\sum (x_i \cdot f_i)}{\sum f_i}
\]
where \( x_i \) are the values of the variable and \( f_i \) are the corresponding frequencies.
### Step 2: Identify the Values and Frequencies
In this case, the values \( x_i \) are \( 0, 1, 2, \ldots, n \) and the frequencies \( f_i \) are proportional to the binomial coefficients:
\[
f_0 = \binom{n}{0}, \quad f_1 = \binom{n}{1}, \quad f_2 = \binom{n}{2}, \ldots, \quad f_n = \binom{n}{n}
\]
### Step 3: Calculate the Total Frequency
The total frequency \( \sum f_i \) can be calculated as:
\[
\sum f_i = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \ldots + \binom{n}{n} = 2^n
\]
This is based on the binomial theorem, which states that the sum of the binomial coefficients for a given \( n \) is \( 2^n \).
### Step 4: Calculate the Weighted Sum
Now, we need to calculate the weighted sum \( \sum (x_i \cdot f_i) \):
\[
\sum (x_i \cdot f_i) = 0 \cdot \binom{n}{0} + 1 \cdot \binom{n}{1} + 2 \cdot \binom{n}{2} + \ldots + n \cdot \binom{n}{n}
\]
This can be simplified using the property of derivatives of the binomial expansion.
### Step 5: Use the Binomial Expansion
The binomial expansion states:
\[
(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k
\]
Differentiating both sides with respect to \( x \):
\[
n(1 + x)^{n-1} = \sum_{k=1}^{n} k \binom{n}{k} x^{k-1}
\]
Now, if we set \( x = 1 \):
\[
n(1 + 1)^{n-1} = \sum_{k=1}^{n} k \binom{n}{k}
\]
This simplifies to:
\[
n \cdot 2^{n-1} = \sum_{k=1}^{n} k \binom{n}{k}
\]
### Step 6: Substitute into the Mean Formula
Now substituting back into our mean formula:
\[
\mu = \frac{\sum (x_i \cdot f_i)}{\sum f_i} = \frac{n \cdot 2^{n-1}}{2^n} = \frac{n}{2}
\]
### Final Result
Thus, the mean of the distribution is:
\[
\mu = \frac{n}{2}
\]
