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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,:

This question was previously asked in
SSC CGL 2016 (Tier 1) Previous Year Question Paper (11-Sep-2016) (Shift 2)
The correct answer is

-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 \(i\).

Calculating the Mean of the New Observations

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 \(-\bar{x}\).

Calculating the Standard Deviation of the New Observations

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.

Summary of Results

For the new observations \(-x_1, -x_2, \ldots, -x_n\):

  • Mean: \(-\bar{x}\)
  • Standard Deviation: \(\sigma\)

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 \(\sigma\) \(\sigma\)

Revision Table: Effects of Linear Transformation on Mean and Standard Deviation

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 \(b = 0\).

  • New Mean: \(\bar{y} = a \bar{x} + b = (-1)\bar{x} + 0 = -\bar{x}\)
  • New Standard Deviation: \(\sigma_y = |a| \sigma_x = |-1| \sigma = 1 \cdot \sigma = \sigma\)

This confirms our calculated results.

Additional Information on Measures of Central Tendency and Dispersion

Measures of Central Tendency: These statistics describe the center of a dataset. The most common ones are:

  • Mean: The average of all observations. It is sensitive to extreme values.
  • Median: The middle value when the data is ordered. It is less sensitive to extreme values than the mean.
  • Mode: The value that appears most frequently in the dataset.

Measures of Dispersion: These statistics describe the spread or variability of a dataset. The most common ones are:

  • Range: The difference between the highest and lowest values. It is simple to calculate but only uses two values.
  • Variance: The average of the squared differences from the mean. It measures how far each number in the set is from the mean.
  • Standard Deviation: The square root of the variance. It is in the same units as the original data and is widely used because it relates to the shape of many distributions (like the normal distribution).

Understanding how these measures change under data transformations is important in statistics.

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