Let's analyze the given options to determine which statement is true about the mean of a given dataset.
1. Option A: The sum of the deviations of individual observations from mean is non-zero.
- This statement is false. By definition, the sum of the deviations of individual observations from the mean is always zero. This is because the deviations (each observation minus the mean) balance out around the mean. Mathematically:
[tex]\[
\sum_{i=1}^{n} (x_i - \bar{x}) = 0
\][/tex]
where \(x_i\) are the individual observations and \(\bar{x}\) is the mean of these observations.
2. Option B: If a constant \(k\) is added to each data value, then the new mean is \(k\) times the old mean.
- This statement is false. When a constant \(k\) is added to each data value, the new mean is actually the old mean plus \(k\), not multiplied by \(k\). Mathematically, if \(\bar{x}\) is the original mean and we add a constant \(k\) to each data value (\(x_i\)), the new mean \(\bar{x'}\) will be:
[tex]\[
\bar{x'} = \bar{x} + k
\][/tex]
3. Option C: The mean of the sum of two population functions of equal number of observations is the sum of their means.
- This statement is true. If we have two populations \(X\) and \(Y\), each with \(n\) observations, and \(\bar{x}\) and \(\bar{y}\) are their respective means, then the mean of the combined population (which consists of adding the corresponding observations of \(X\) and \(Y\)) is \(\bar{x} + \bar{y}\). Mathematically:
[tex]\[
\overline{(X + Y)} = \bar{x} + \bar{y}
\][/tex]
4. Option D: The mean of constant times a population function is equal to the mean of the old population function.
- This statement is false. When a population function is multiplied by a constant \(c\), the new mean becomes \(c\) times the original mean, not equal to the original mean. Mathematically, if \(\bar{x}\) is the mean of the population \(X\), then the mean of the population \(cX\) is:
[tex]\[
\bar{cX} = c \cdot \bar{x}
\][/tex]
Given the analysis, Option C is the correct statement about the mean of a given data set.
