Answer :
To determine whether the sum of the differences [tex]\(\sum_{i=1}^n\left(x_i - \bar{x}\right)\)[/tex] is zero for any distribution consisting of [tex]\(n\)[/tex] observations, let's systematically break down the problem.
First, let's define the terms involved:
- [tex]\(x_i\)[/tex] represents the [tex]\(i\)[/tex]-th observation in the distribution.
- [tex]\(\bar{x}\)[/tex] represents the mean (average) of these observations.
The mean [tex]\(\bar{x}\)[/tex] is computed as:
[tex]\[ \bar{x} = \frac{1}{n} \sum_{i=1}^n x_i \][/tex]
Now, we are interested in the sum of the differences between each observation [tex]\(x_i\)[/tex] and the mean [tex]\(\bar{x}\)[/tex]:
[tex]\[ \sum_{i=1}^n (x_i - \bar{x}) \][/tex]
Let's expand this summation:
[tex]\[ \sum_{i=1}^n (x_i - \bar{x}) = (x_1 - \bar{x}) + (x_2 - \bar{x}) + \cdots + (x_n - \bar{x}) \][/tex]
This can be rewritten by factoring out the [tex]\(\bar{x}\)[/tex]:
[tex]\[ = \sum_{i=1}^n x_i - \sum_{i=1}^n \bar{x} \][/tex]
Since [tex]\(\bar{x}\)[/tex] is a constant (the mean of all [tex]\(x_i\)[/tex]), we can also express the second sum as follows:
[tex]\[ = \sum_{i=1}^n x_i - \sum_{i=1}^n \bar{x} = \sum_{i=1}^n x_i - n \bar{x} \][/tex]
We know from the definition of the mean [tex]\(\bar{x}\)[/tex] that:
[tex]\[ n \bar{x} = \sum_{i=1}^n x_i \][/tex]
Therefore,
[tex]\[ \sum_{i=1}^n (x_i - \bar{x}) = \sum_{i=1}^n x_i - n \bar{x} = \sum_{i=1}^n x_i - \sum_{i=1}^n x_i = 0 \][/tex]
This shows us that the sum of the differences [tex]\(\sum_{i=1}^n (x_i - \bar{x})\)[/tex] is always zero for any distribution consisting of [tex]\(n\)[/tex] observations. Thus, the correct answer is:
B. False
First, let's define the terms involved:
- [tex]\(x_i\)[/tex] represents the [tex]\(i\)[/tex]-th observation in the distribution.
- [tex]\(\bar{x}\)[/tex] represents the mean (average) of these observations.
The mean [tex]\(\bar{x}\)[/tex] is computed as:
[tex]\[ \bar{x} = \frac{1}{n} \sum_{i=1}^n x_i \][/tex]
Now, we are interested in the sum of the differences between each observation [tex]\(x_i\)[/tex] and the mean [tex]\(\bar{x}\)[/tex]:
[tex]\[ \sum_{i=1}^n (x_i - \bar{x}) \][/tex]
Let's expand this summation:
[tex]\[ \sum_{i=1}^n (x_i - \bar{x}) = (x_1 - \bar{x}) + (x_2 - \bar{x}) + \cdots + (x_n - \bar{x}) \][/tex]
This can be rewritten by factoring out the [tex]\(\bar{x}\)[/tex]:
[tex]\[ = \sum_{i=1}^n x_i - \sum_{i=1}^n \bar{x} \][/tex]
Since [tex]\(\bar{x}\)[/tex] is a constant (the mean of all [tex]\(x_i\)[/tex]), we can also express the second sum as follows:
[tex]\[ = \sum_{i=1}^n x_i - \sum_{i=1}^n \bar{x} = \sum_{i=1}^n x_i - n \bar{x} \][/tex]
We know from the definition of the mean [tex]\(\bar{x}\)[/tex] that:
[tex]\[ n \bar{x} = \sum_{i=1}^n x_i \][/tex]
Therefore,
[tex]\[ \sum_{i=1}^n (x_i - \bar{x}) = \sum_{i=1}^n x_i - n \bar{x} = \sum_{i=1}^n x_i - \sum_{i=1}^n x_i = 0 \][/tex]
This shows us that the sum of the differences [tex]\(\sum_{i=1}^n (x_i - \bar{x})\)[/tex] is always zero for any distribution consisting of [tex]\(n\)[/tex] observations. Thus, the correct answer is:
B. False
