Shifting and Scaling Effects on Mean and Standard Deviation

In this post, we will explain the effects of shifting (addition or subtraction) and scaling (multiplication or division) of scores in the entire data set. The first part of this post gives you the fundamental ideas of what happens if a constant value is added, subtracted, multiplied or divided, and the second part explains the combined effects of these four operations to see the effects to the mean and the standard deviation.

Let’s take a look at the following scores.

\( \{1, 2, 3, 4, 5 \} \)

\( \text{Mean: } \displaystyle \mu = \frac{1+2+3+4+5}{5} = 3 \)

\( \text{Standard deviation: } \displaystyle \sigma = \sqrt{\frac{(1-3)^2 + (2-3)^2 + (3-3)^2 + (4-3)^2 + (5-3)^2}{5}} \approx 1.58 \)

Shifting data by addition

What happens to mean and standard deviation when we add a constant value to every score in the data set?

If we add \( 4 \) to each score, the new data set is \( \{ 5, 6, 7, 8, 9 \} \).

\( \begin{align} \displaystyle \text{Mean: } \frac{5+6+7+8+9}{5} &= 7 \\ &= 3 + 4 \\ &= \require{AMSsymbols} \color{green}{\mu + 4} \end{align} \)

\( \begin{align} \displaystyle \text{Standard deviation: } \sqrt{\frac{(5-7)^2 + (6-7)^2 + (7-7)^2 + (8-7)^2 + (9-7)^2}{5}} &\approx 1.58 \\ &= \color{green}{\sigma} \end{align} \)

What we notice is that adding \( a \) to the entire data set, the the new mean becomes \( \sigma + a \) and the standard division remains unchanged.

Shifting data by subtraction

If we subtract \( \color{green} {2} \) from each score, the new data set is \( \{ -1, 0, 1, 2, 3 \} \).

\( \begin{align} \displaystyle \text{Mean: } \frac{-1+0+1+2+3}{5} &= 1 \\ &= 3 – 2 \\ &= \color{green}{\mu – 2} \end{align} \)

\( \begin{align} \displaystyle \text{Standard deviation: } \sqrt{\frac{(-1-1)^2 + (0-1)^2 + (1-1)^2 + (2-1)^2 + (2-1)^2}{5}} &\approx 1.58 \\ &= \color{green}{\sigma} \end{align} \)

What we notice is that subtracting \( b \) to the entire data set, the the new mean becomes \( \mu – b \) and the standard division remains unchanged.

Scaling data by multiplication

If we multiply each score by \( \color{green}{10} \), the new data set is \( \{ 10, 20, 30, 40, 50 \} \).

\( \begin{align} \displaystyle \text{Mean: } \frac{10+20+30+40+50}{5} &= 30 \\ &= 10 \times 3 \\ &= \color{green}{10 \times \mu} \end{align} \)

\( \begin{align} \displaystyle \text{Standard deviation: } \sqrt{\frac{(10-30)^2 + (20-30)^2 + (30-30)^2 + (40-30)^2 + (50-30)^2}{5}} &\approx 15.8 \\ &= 10 \times 1.58 \\ &= \color{green}{10 \times \sigma} \end{align} \)

What we notice is that multiplying the entire data set by \( n \), the the new mean becomes \( \mu\times n \) and the new standard division is \( \sigma \times n \).

Scaling data by division

If we divide each score by \( \color{green}{10} \), the new data set is \( \{ 0.1, 0.2, 0.3, 0.4, 0.5 \} \).

\( \begin{align} \displaystyle \text{Mean: } \frac{0.1+0.2+0.3+0.4+0.5}{5} &= 0.3 \\ &= 3 \div 10 \\ &= \color{green}{\mu \div 10} \end{align} \)

\( \begin{align} \displaystyle \text{Standard deviation: } \sqrt{\frac{(0.1-0.3)^2 + (0.2-0.3)^2 + (0.3-0.3)^2 + (0.4-0.3)^2 + (0.5-0.3)^2}{5}} &\approx 0.158 \\ &= 1.58 \div 10 \\ &= \color{green}{\sigma \div 10} \end{align} \)

What we notice is that dividing the entire data set by \( n \), the the new mean becomes \( \mu\div n \) and the new standard division is \( \sigma \div n \).

Shifting and Scaling data

If we multiply by \( \color{green}{10} \) and add \( \color{green}{4} \) to each score, the new data set is \( \{ 14, 24, 35, 44, 54 \} \).

\( \begin{align} \displaystyle \text{Mean: } \frac{14+24+34+44+54}{5} &= 34 \\ &= 10 \times 3 + 4 \\ &= \color{green}{10 \times \mu + 4} \end{align} \)

\( \begin{align} \displaystyle \text{Standard deviation: } \sqrt{\frac{(14-34)^2 + (24-34)^2 + (34-34)^2 + (44-34)^2 + (54-34)^2}{5}} &\approx 15.8 \\ &= 10 \times 1.58 \\ &= \color{green}{10 \times \sigma} \end{align} \)

What we notice is that by multiplying the entire data set by \( n \) and adding \( a \), then the new mean becomes \( \mu \div n + a \), and the new standard division is \( \sigma \times n \).

Summary

Effects on the Mean

What happens to the mean if a constant is added to the entire data set?

Adding a constant, \( a \), to the entire data set results in adding the constant to the existing mean.
\( \mu_{\text{new}} = \mu + a \)

What happens to the mean if a constant is subtracted from the entire data set?

Subtracting a constant \( b \) from the entire data set results in subtracting the constant from the existing mean.
\( \mu_{\text{new}} = \mu – b \)

What happens to the mean if a constant is multiplied by the entire data set?

Multiplying a constant \( n \) to the entire data set results in multiplying the existing mean by the constant.
\( \mu_{\text{new}} = \mu \times n \)

What happens to the mean if a constant is divided into the entire data set?

Dividing the entire data set by a constant \( n \) results in dividing the existing mean by the constant.
\( \mu_{\text{new}} = \mu \div n \)

Effects on the Standard Deviation

What happens to the standard deviation if a constant is added to the entire data set?

We are adding a constant, \( a \), to the entire data set, resulting in the existing standard deviation being unchanged.
\( \sigma_{\text{new}} = \sigma \)

What happens to the standard deviation if a constant is subtracted from the entire data set?

Subtracting a constant \( b \) from the entire data set results in the existing standard deviation being unchanged.
\( \sigma_{\text{new}} = \sigma \)

What happens to the standard deviation if a constant is multiplied by the entire data set?

Multiplying a constant \( n \) by the entire data set results in multiplying the existing standard deviation by the constant.
\( \sigma_{\text{new}} = \sigma \times n \)

What happens to the standard deviation if a constant is divided into the entire data set?

Dividing the entire data set by a constant \( n \) results in dividing the existing standard deviation by the constant.
\( \sigma_{\text{new}} = \sigma \div n \)

Practice Questions

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