<p>If the normal distribution has&nbsp;\(\mu=0\) and \(\sigma^2=1\), it is called a <strong>standard</strong> normal distribution and is denoted as&nbsp;\(Z \sim N(0,1)\). The probability density function (pdf) for the standard normal distribution is given by</p>
<p>\[\mathrm{f}(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\]</p>
<p>In general, any normal distribution \(X \sim N(\mu,\sigma^2)\) can be transformed into the standard normal \(Z \sim N(0,1)\) by using the transformation</p>
<p>\[Z = \frac{X-\mu}{\sigma}\]</p>
<p>For example, if&nbsp;\(X \sim N(\mu, \sigma^2)\) and we want to find \(P(x_1 \leq X \leq x_2)\), then&nbsp;\(X\) must first be transformed to \(Z \sim N(0, 1)\). Then&nbsp;\(P(x_1 \leq X \leq x_2) = P(z_1 \leq Z \leq z_2)\) where&nbsp;\(z_1 = \frac{x_1 - \mu}{\sigma}\) and \(z_2 = \frac{x_2 - \mu}{\sigma}\).</p>
<p>Why do we need to standardize?  The purpose for standardization is two-fold:</p>
<ol class="num-list"><li><div class="list-item-paragraph">Standardizing allows us to have a more meaningful understanding of the data.</div></li><li><div class="list-item-paragraph">When the mean, \(\mu\), or/and variance, \(\sigma^2\) is unknown, standardizing allows us to use the inverse standard normal function in GDC to calculate the unknown normal distribution parameters.</div></li></ol>
<p><strong>Note:</strong> Some advantages of using \(Z \sim N(0,1)\):</p>
<ol class="i-list"><li><div class="list-item-paragraph">Symmetry about \(0\).</div></li><li><div class="list-item-paragraph">We can express \(Z\) in terms of \(\mu\) and \(\sigma\)</div></li><li><div class="list-item-paragraph">If&nbsp;\(P(Z &gt; z_1) = P(Z &lt; z_2)\), we know that&nbsp;\(z_1 = -z_2\). (try sketching the curve to see!)</div></li><li><div class="list-item-paragraph">If&nbsp;\(P(Z &gt; z_1) &gt; 0.5\) we can conclude that&nbsp;\(z_1 &lt; 0\).</div></li></ol>

If the normal distribution has \(\mu=0\) and \(\sigma^2=1\), it is called a **standard** normal distribution and is denoted as \(Z \sim N(0,1)\). The probability density function (pdf) for the standard normal distribution is given by

\[\mathrm{f}(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\]

In general, any normal distribution \(X \sim N(\mu,\sigma^2)\) can be transformed into the standard normal \(Z \sim N(0,1)\) by using the transformation

For example, if \(X \sim N(\mu, \sigma^2)\) and we want to find \(P(x_1 \leq X \leq x_2)\), then \(X\) must first be transformed to \(Z \sim N(0, 1)\). Then \(P(x_1 \leq X \leq x_2) = P(z_1 \leq Z \leq z_2)\) where \(z_1 = \frac{x_1 - \mu}{\sigma}\) and \(z_2 = \frac{x_2 - \mu}{\sigma}\).

Why do we need to standardize?  The purpose for standardization is two-fold:

Standardizing allows us to have a more meaningful understanding of the data.

When the mean, \(\mu\), or/and variance, \(\sigma^2\) is unknown, standardizing allows us to use the inverse standard normal function in GDC to calculate the unknown normal distribution parameters.

**Note:** Some advantages of using \(Z \sim N(0,1)\):

We can express \(Z\) in terms of \(\mu\) and \(\sigma\)

If \(P(Z > z_1) = P(Z < z_2)\), we know that \(z_1 = -z_2\). (try sketching the curve to see!)

The standard normal distribution

The standard normal distribution