Calculating the Mean of the Given Distribution

The question provides a discrete distribution with values $x_i$ and corresponding frequencies $f_i$. We need to calculate the mean ($\bar{x}$) of this distribution.

Understanding the Distribution Data

The data given is:

• Values ($x_i$): $1, 2, 3, \dots, n$
• Frequencies ($f_i$): $1, 2^{-1}, 2^{-2}, \dots, 2^{-(n-1)}$

We can rewrite the frequencies as:

$ f_i = \frac{1}{2^{i-1}} $

Mean Calculation Formula

The mean ($\bar{x}$) of a discrete distribution is defined as the sum of the products of each value and its corresponding frequency:

$ \bar{x} = \sum_{i=1}^{n} x_i f_i $

Applying the Formula

Substitute the given $x_i$ and $f_i$ values into the formula:

$ \bar{x} = \sum_{i=1}^{n} i \cdot \frac{1}{2^{i-1}} $

Let's write out the terms of the sum:

$ \bar{x} = \left( 1 \cdot \frac{1}{2^{1-1}} \right) + \left( 2 \cdot \frac{1}{2^{2-1}} \right) + \left( 3 \cdot \frac{1}{2^{3-1}} \right) + \dots + \left( n \cdot \frac{1}{2^{n-1}} \right) $

$ \bar{x} = \left( 1 \cdot \frac{1}{2^0} \right) + \left( 2 \cdot \frac{1}{2^1} \right) + \left( 3 \cdot \frac{1}{2^2} \right) + \dots + \left( n \cdot \frac{1}{2^{n-1}} \right) $

$ \bar{x} = 1 + \frac{2}{2} + \frac{3}{4} + \dots + \frac{n}{2^{n-1}} $

Solving the Arithmetico-Geometric Series

The sum we need to calculate is an arithmetico-geometric series. Let $S_n$ represent this sum:

$ S_n = 1 + \frac{2}{2} + \frac{3}{4} + \dots + \frac{n}{2^{n-1}} $

Let $r = \frac{1}{2}$. Then the series can be written as:

$ S_n = \sum_{i=1}^{n} i \cdot r^{i-1} = 1 + 2r + 3r^2 + \dots + nr^{n-1} $

To find the sum $S_n$, we use a standard method. Multiply the entire series by $r$:

$ rS_n = r + 2r^2 + 3r^3 + \dots + (n-1)r^{n-1} + nr^n $

Now, subtract $rS_n$ from $S_n$:

$ S_n - rS_n = (1 + 2r + 3r^2 + \dots + nr^{n-1}) - (r + 2r^2 + \dots + nr^n) $

Group the terms:

$ S_n(1-r) = 1 + (2r-r) + (3r^2-2r^2) + \dots + (nr^{n-1} - (n-1)r^{n-1}) - nr^n $

$ S_n(1-r) = 1 + r + r^2 + \dots + r^{n-1} - nr^n $

The sum $1 + r + r^2 + \dots + r^{n-1}$ is a finite geometric series. Its sum is $\frac{1 - r^n}{1 - r}$.

$ S_n(1-r) = \frac{1 - r^n}{1 - r} - nr^n $

Now, substitute $r = \frac{1}{2}$. This means $1-r = \frac{1}{2}$:

$ S_n \left( \frac{1}{2} \right) = \frac{1 - \left(\frac{1}{2}\right)^n}{\frac{1}{2}} - n \left(\frac{1}{2}\right)^n $

Simplify the equation:

$ \frac{S_n}{2} = 2 \left( 1 - \frac{1}{2^n} \right) - \frac{n}{2^n} $

$ \frac{S_n}{2} = 2 - \frac{2}{2^n} - \frac{n}{2^n} $

$ \frac{S_n}{2} = 2 - \frac{n+2}{2^n} $

To find $S_n$, multiply both sides by 2:

$ S_n = 2 \left( 2 - \frac{n+2}{2^n} \right) $

$ S_n = 4 - \frac{2(n+2)}{2^n} $

Simplify the fraction:

$ S_n = 4 - \frac{n+2}{2^{n-1}} $

To express this as a single fraction with the denominator $2^{n-1}$:

$ S_n = \frac{4 \cdot 2^{n-1}}{2^{n-1}} - \frac{n+2}{2^{n-1}} $

$ S_n = \frac{2^2 \cdot 2^{n-1} - (n+2)}{2^{n-1}} $

$ S_n = \frac{2^{n+1} - n - 2}{2^{n-1}} $

Conclusion

The calculated mean ($\bar{x}$) of the distribution is:

$ \bar{x} = \frac{2^{n+1} - n - 2}{2^{n-1}} $