5.11: Effects of Linear Transformations
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Learning Objectives
- Compute the mean of a transformed variable
- Compute the variance of a transformed variable
This section covers the effects of linear transformations on measures of central tendency and variability. Let's start with an example we saw before in the section that defined linear transformation: temperatures of cities. Table \(\PageIndex{1}\) shows the temperatures of \(5\) cities.
| City | Degrees Fahrenheit | Degrees Centigrade |
|---|---|---|
|
Houston
Chicago Minneapolis Miami Phoenix |
54
37 31 78 70 |
12.22
2.78 -0.56 25.56 21.11 |
|
Mean
Median |
54.000
54.000 |
12.220
12.220 |
| Variance | 330.00 | 101.852 |
| SD | 18.166 | 10.092 |
Recall that to transform the degrees Fahrenheit to degrees Centigrade, we use the formula
\[C = 0.556F - 17.778\]
which means we multiply each temperature Fahrenheit by \(0.556\) and then subtract \(17.778\). As you might have expected, you multiply the mean temperature in Fahrenheit by \(0.556\) and then subtract \(17.778\) to get the mean in Centigrade. That is, \((0.556)(54) - 17.778 = 12.22\). The same is true for the median. Note that this relationship holds even if the mean and median are not identical as they are in Table \(\PageIndex{1}\).
The formula for the standard deviation is just as simple: the standard deviation in degrees Centigrade is equal to the standard deviation in degrees Fahrenheit times \(0.556\). Since the variance is the standard deviation squared, the variance in degrees Centigrade is equal to \(0.5562^2\) times the variance in degrees Fahrenheit.
To sum up, if a variable \(X\) has a mean of \(\mu\), a standard deviation of \(\sigma\), and a variance of \(\sigma ^2\), then a new variable \(Y\) created using the linear transformation
\[Y = bX + A\]
will have a mean of \(b\mu + A\), a standard deviation of \(b\sigma\), and a variance of \(b^2\sigma ^2\).
It should be noted that the term "linear transformation" is defined differently in the field of linear algebra. For details, follow this link .