Probability density function
The general results about the probability density function of \(Y = g(X)\) when \(X\) is a continuous random variable, will now be applied to the transformation \(Y = a + bX\).
Writing
\[ y = g(x) = a + bx \qquad x = h(y) = \frac{y-a}{b} \spaced{and} h'(y) = \frac 1 b \]the random variable \(Y\) has pdf
\[ f_Y(y) \;\;=\;\; f_X\left(h(y)\right) \times h'(y) \;\;=\;\; \frac 1 {\left|b\right|} f_X\left(\frac{y-a}{b}\right) \]Transforming a normal variable
The previous page gave the mean and variance of a linear function of \(X\), \(Y = (a + bX)\). If \(X\) has a normal distribution, we will now prove that \(Y\) also has a normal distribution.
Linear transformation of a normal variable
If \(a\) and \(b\) are constants and \(X \sim \NormalDistn(\mu, \sigma^2)\), the random variable \(Y = (a + bX)\) also has a normal distribution
\[ Y \;\;\sim\;\; \NormalDistn(a + b\mu,\; b^2 \sigma^2) \]The pdf of \(Y\) is
\[ \begin{align} f_Y(y) \;\;&=\;\; \frac 1 b f_X\left(\frac{y-a}{b}\right) \\ &=\;\; \frac 1 b \frac 1{\sqrt{2\pi}\;\sigma} e^{- \frac{\Large \left(\frac{y-a}b -\mu\right)^2}{\Large 2 \sigma^2}} \\ &=\;\; \frac 1{\sqrt{2\pi}\;b\sigma} e^{- \frac{\Large \left(y-a - b\mu\right)^2}{\Large 2 b^2\sigma^2}} \end{align} \]and this is the pdf of a \(\NormalDistn(a + b\mu,\; b^2 \sigma^2)\) distribution.
Standard normal distribution and z-scores
In particular, this result provides the distribution of a z-score.
Distribution of z-scores
If \(X \sim \NormalDistn(\mu, \sigma^2)\), the random variable \(Z = \dfrac {X-\mu} {\sigma} \) has a normal distribution with zero mean and standard deviation one,
\[ Z \sim \NormalDistn(0, 1) \]We can write
\[ Z = \left(-\frac {\mu} {\sigma} \right) + \left(\frac 1 {\sigma} \right) X \]The proof follows from applying the general result about the distribution of \(a + bX\) with
\[ a = -\frac {\mu} {\sigma} \spaced{and} b = \frac 1 {\sigma} \]Probabilities about z-scores can be found using computer software or tables.
Example
If the weight of a Fuji apple is
\[ X \;\; \sim \; \; \NormalDistn(\mu=180, \sigma=10) \]then
\[ P(X \lt 162) = P(Z \lt \frac {162 - 180} {10}) = P(Z \lt -1.8) \]This can be evaluated in Excel by typing
=NORM.S.DIST(-1.8, TRUE)
in a spreadsheet cell to get the probability 0.0359.