Calculating the Mean with Binomial Frequencies

The question asks for the mean of a set of numbers from 1 to 10, where each number $i$ is associated with a specific frequency. The frequencies are given by binomial coefficients ${9}C_{i-1}$.

The numbers ($x_i$) and their corresponding frequencies ($f_i$) are:

• Number 1 ($x_1=1$) has frequency ${9}C_{1-1} = {9}C_0$.
• Number 2 ($x_2=2$) has frequency ${9}C_{2-1} = ${9}C_1$.
• Number 3 ($x_3=3$) has frequency ${9}C_{3-1} = ${9}C_2$.
• ...
• Number 10 ($x_{10}=10$) has frequency ${9}C_{10-1} = ${9}C_9$.

In general, for a number $i$ (where $i$ goes from 1 to 10), its frequency is $f_i = {9}C_{i-1}$.

Formula for the Mean

The mean ($\bar{x}$) of a set of numbers with given frequencies is calculated using the formula:

$$ \bar{x} = \frac{\sum (x_i \times f_i)}{\sum f_i} $$

Here, $x_i$ represents the numbers (1, 2, ..., 10) and $f_i$ represents their frequencies (${9}C_0$, ${9}C_1$, ..., ${9}C_9$).

Calculating the Sum of Frequencies

The sum of the frequencies is the sum of binomial coefficients ${9}C_k$ for $k$ from 0 to 9:

$$ \sum_{i=1}^{10} f_i = \sum_{i=1}^{10} {9}C_{i-1} = {9}C_0 + {9}C_1 + {9}C_2 + \dots + {9}C_9 $$

This sum is equal to the sum of all terms in the binomial expansion of $(1+1)^9$, which is $2^9$.

$$ \sum f_i = 2^9 = 512 $$

Calculating the Sum of (Number $\times$ Frequency)

The sum of (number $\times$ frequency) is:

$$ \sum_{i=1}^{10} (x_i \times f_i) = \sum_{i=1}^{10} (i \times {9}C_{i-1}) $$

Let's write out the terms:

$$ (1 \times {9}C_0) + (2 \times {9}C_1) + (3 \times {9}C_2) + \dots + (10 \times {9}C_9) $$

We can use the property that $k \times {n}C_k = n \times {n-1}C_{k-1}$ for $k \ge 1$. The sum is $\sum_{i=1}^{10} i \times {9}C_{i-1}$. Let $k = i-1$. Then $i=k+1$, and the summation goes from $k=0$ to $k=9$. The sum becomes:

$$ \sum_{k=0}^{9} (k+1) \times {9}C_k = \sum_{k=0}^{9} (k \times {9}C_k) + \sum_{k=0}^{9} (1 \times {9}C_k) $$

The second part of the sum is $\sum_{k=0}^{9} {9}C_k = 2^9 = 512$, as calculated before.

For the first part, $\sum_{k=0}^{9} (k \times {9}C_k)$: The term for $k=0$ is $0 \times {9}C_0 = 0$. For $k \ge 1$, we use the property $k \times {9}C_k = 9 \times {8}C_{k-1}$. So, $\sum_{k=0}^{9} (k \times {9}C_k) = 0 + \sum_{k=1}^{9} (k \times {9}C_k) = \sum_{k=1}^{9} (9 \times {8}C_{k-1})$. Let $j=k-1$. When $k=1$, $j=0$. When $k=9$, $j=8$. The sum becomes $9 \sum_{j=0}^{8} {8}C_j$. The sum $\sum_{j=0}^{8} {8}C_j = {8}C_0 + {8}C_1 + \dots + {8}C_8 = 2^8$. So, $\sum_{k=0}^{9} (k \times {9}C_k) = 9 \times 2^8$.

Now, combine the two parts of the sum $\sum_{i=1}^{10} (i \times {9}C_{i-1})$:

$$ \sum (x_i \times f_i) = (9 \times 2^8) + 2^9 $$

Since $2^9 = 2 \times 2^8$, we have:

$$ \sum (x_i \times f_i) = 9 \times 2^8 + 2 \times 2^8 = (9+2) \times 2^8 = 11 \times 2^8 $$

Calculating the Mean

Now, substitute the sums into the mean formula:

$$ \bar{x} = \frac{\sum (x_i \times f_i)}{\sum f_i} = \frac{11 \times 2^8}{2^9} $$

$$ \bar{x} = \frac{11 \times 2^8}{2 \times 2^8} = \frac{11}{2} = 5.5 $$

The mean of the numbers with the given binomial frequencies is 5.5.

Comparing with Options

Let's compare the calculated mean with the given options:

The calculated mean, 5.5, matches Option 3.

Revision Table: Mean Calculation

Additional Information: Mean, Frequencies, and Binomials

The mean is a fundamental measure of central tendency. When dealing with data where values repeat (i.e., they have frequencies), the weighted mean formula is used. Each value is weighted by its frequency, meaning it contributes to the sum proportional to how many times it appears. The sum of the weighted values is then divided by the total sum of the frequencies (which is the total number of observations).

Binomial coefficients, ${n}C_k$, appear in probability, statistics, and algebra (binomial theorem). The sum of binomial coefficients for a fixed $n$ is always a power of 2, representing the total number of subsets of a set with $n$ elements.

The specific sum encountered, $\sum_{k=0}^{n} k \times {n}C_k$, is related to the expected value of a binomial distribution. The expected value of a binomial distribution $B(n, p)$ is $np$. The terms ${n}C_k$ are proportional to the probability mass function of a binomial distribution with $p=0.5$. If we consider a binomial distribution $B(n, 0.5)$, the probability of getting $k$ successes is ${n}C_k (0.5)^k (0.5)^{n-k} = {n}C_k (0.5)^n$. The expected value is $\sum_{k=0}^n k \times P(X=k) = \sum_{k=0}^n k \times {n}C_k (0.5)^n = (0.5)^n \sum_{k=0}^n k \times {n}C_k$. We know this expected value is $n \times 0.5$. So, $(0.5)^n \sum_{k=0}^n k \times {n}C_k = n \times 0.5$, which means $\sum_{k=0}^n k \times {n}C_k = n \times 0.5 / (0.5)^n = n \times 0.5^{1-n} = n \times 2^{n-1}$.

In our problem, the sum is $\sum_{i=1}^{10} i \times {9}C_{i-1}$. Let $j = i-1$, then $i=j+1$. The sum is $\sum_{j=0}^{9} (j+1) \times {9}C_j = \sum_{j=0}^{9} j \times {9}C_j + \sum_{j=0}^{9} {9}C_j$. We found $\sum_{j=0}^{9} {9}C_j = 2^9$. Using the property or the expected value result with $n=9$, $\sum_{j=0}^{9} j \times {9}C_j = 9 \times 2^{9-1} = 9 \times 2^8$. So the sum $\sum_{i=1}^{10} i \times {9}C_{i-1} = 9 \times 2^8 + 2^9 = 9 \times 2^8 + 2 \times 2^8 = 11 \times 2^8$. This confirms the detailed step-by-step calculation.