If the mean and the standard deviation of n observations x 1, x 2, _ _ _ _ _, x nbe x̅ and σ respectively, then the mean and standard deviation of -x 1, -x 2, -x 3, _ _ _ _ _, -x nare, respectively,:
-x̅, σ
Let the original set of \(n\) observations be \(x_1, x_2, \ldots, x_n\).
The mean of these observations is given by:
\begin{equation*} \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \end{equation*}
The standard deviation of these observations is given by:
\begin{equation*} \sigma = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2} \end{equation*}
We are asked to find the mean and standard deviation of a new set of observations: \(-x_1, -x_2, \ldots, -x_n\). Let's call this new set \(y_1, y_2, \ldots, y_n\), where \(y_i = -x_i\) for each \(i\).
The mean of the new observations, denoted by \(\bar{y}\), is given by:
\begin{equation*} \bar{y} = \frac{1}{n} \sum_{i=1}^{n} y_i \end{equation*}
Substitute \(y_i = -x_i\) into the formula:
\begin{equation*} \bar{y} = \frac{1}{n} \sum_{i=1}^{n} (-x_i) \end{equation*}
We can factor out the constant \(-1\) from the summation:
\begin{equation*} \bar{y} = - \frac{1}{n} \sum_{i=1}^{n} x_i \end{equation*}
We know that \(\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i\). So, substitute \(\bar{x}\) into the equation for \(\bar{y}\):
\begin{equation*} \bar{y} = -\bar{x} \end{equation*}
Thus, the mean of the new observations \(-x_1, -x_2, \ldots, -x_n\) is \(-\bar{x}\).
The standard deviation of the new observations, denoted by \(\sigma_y\), is given by:
\begin{equation*} \sigma_y = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (y_i - \bar{y})^2} \end{equation*}
Substitute \(y_i = -x_i\) and \(\bar{y} = -\bar{x}\) into the formula:
\begin{equation*} \sigma_y = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (-x_i - (-\bar{x}))^2} \end{equation*}
\begin{equation*} \sigma_y = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (-x_i + \bar{x})^2} \end{equation*}
The term \((-x_i + \bar{x})^2\) is the same as \((\bar{x} - x_i)^2\), which is also the same as \((x_i - \bar{x})^2\) because squaring makes the sign positive:
\begin{equation*} (-x_i + \bar{x})^2 = (\bar{x} - x_i)^2 = (-1(x_i - \bar{x}))^2 = (-1)^2 (x_i - \bar{x})^2 = 1 \cdot (x_i - \bar{x})^2 = (x_i - \bar{x})^2 \end{equation*}
So, substitute this back into the standard deviation formula:
\begin{equation*} \sigma_y = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2} \end{equation*}
This is the formula for the standard deviation of the original observations, \(\sigma\).
\begin{equation*} \sigma_y = \sigma \end{equation*}
Thus, the standard deviation of the new observations \(-x_1, -x_2, \ldots, -x_n\) is \(\sigma\). Standard deviation is always non-negative, so even though we are dealing with negative values, the spread around the mean remains the same as the original data.
For the new observations \(-x_1, -x_2, \ldots, -x_n\):
Comparing this with the given options, we find that the correct answer is \(-\bar{x}\) and \(\sigma\), respectively.
| Statistic | Original Observations (\(x_i\)) | New Observations (\(-x_i\)) |
|---|---|---|
| Mean | \(\bar{x}\) | \(-\bar{x}\) |
| Standard Deviation | \(\sigma\) | \(\sigma\) |
Let the original observations be \(x_1, x_2, \ldots, x_n\) with mean \(\bar{x}\) and standard deviation \(\sigma_x\). Consider a linear transformation \(y_i = a x_i + b\), where \(a\) and \(b\) are constants. The new observations are \(y_1, y_2, \ldots, y_n\).
| Statistic | Effect of Transformation \(y_i = a x_i + b\) | Formula |
|---|---|---|
| New Mean (\(\bar{y}\)) | Mean is affected by both multiplication (\(a\)) and addition (\(b\)). | \(\bar{y} = a \bar{x} + b\) |
| New Standard Deviation (\(\sigma_y\)) | Standard deviation is affected by multiplication (\(a\)) but not by addition (\(b\)). It is scaled by the absolute value of \(a\). | \(\sigma_y = |a| \sigma_x\) |
In this specific question, the transformation is \(y_i = -x_i\). This can be written in the form \(y_i = a x_i + b\) with \(a = -1\) and \(b = 0\).
This confirms our calculated results.
Measures of Central Tendency: These statistics describe the center of a dataset. The most common ones are:
Measures of Dispersion: These statistics describe the spread or variability of a dataset. The most common ones are:
Understanding how these measures change under data transformations is important in statistics.
Which one is parameter from population?
For a group of 5 male residents in a society, the mean and standard deviation of their ages are 63 years and 9 years, respectively. For a group of 4 female residents, these values are 54 years and 6 years, respectively. The variance of the combined group of male and female residents is:
Factory A and Factory B employ 476 and 524 employees. respectively. The average weekly salary of an employee in Factory A is $34.5 whereas for an employee in Factory B it is $28.5, The standard deviation in paying the individual salary has been recorded as $5 and $4.5 for Factory A and Factory B, respectively. Which factory has greater variability in paying individual salary?
Among the options for parameters, which option is correct for population?
The mean and standard deviation (s.d.) of runs scored by three batsmen A, B and C in 9 consecutive matches are given below,
| Batsman | Mean | s.d. |
| A | 40 | 4 |
| B | 64 | 8 |
| C | 72 | 12 |
Which of these statements on variation is INCORRECT?
In a class of 50 students, the marks obtained are
Marks | 15 | 30 | 37 | 40 | 45 | 48 |
No. of students | 2 | 8 | 14 | 12 | 10 | 4 |
Its median is
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Find the median if the given data set is:
3, 3, 7, 8, 12, 13, 16, 19
A machine produces 0, 1 or 2 defective pieces in a day with an associated probability of \(\frac{{1}}{{6}}\), \(\frac{{2}}{{3}}\) and \(\frac{{1}}{{6}}\) respectively. The mean value and the variance of the number of defective pieces produced by the machine in a day, respectively, are
The median of 7, 5, 8, x, 12, 17 is 10, then what is the value of x?