The Standard Normal Distribution (HSC SSCE Mathematics Advanced): Revision Notes

The Standard Normal Distribution

Introduction to the standard normal distribution

The standard normal distribution is a special type of continuous probability distribution that plays a central role in statistics. It belongs to a family of distributions called Gaussian or normal distributions, which are characterised by their distinctive bell-shaped curves.

The random variable for the standard normal distribution is denoted by $Z$ (rather than $X$), and its values are written as $z$ (rather than $x$).

The probability density function

The probability density function (PDF) of the standard normal distribution is denoted by $\phi(z)$, where $\phi$ is the Greek letter phi (corresponding to Latin f).

The formula for the PDF is:

$\phi(z) = \frac{e^{-\frac{1}{2}z^2}}{\sqrt{2\pi}}$

Properties of the bell curve

The graph of $y = \phi(z)$ has several important characteristics:

Shape and symmetry:

• The y-intercept is $\phi(0) = \frac{1}{\sqrt{2\pi}} \approx 0.4$ (because $e^0 = 1$)
• The maximum value occurs at $z = 0$, making this the mode
• The function is even, meaning it has line symmetry about the y-axis
• Replacing $z$ with $-z$ leaves the equation unchanged: $\phi(-z) = \phi(z)$

Domain and range:

• The function is defined for all real values of $z$
• The function is always positive: $\phi(z) > 0$ for all $z$

Asymptotic behaviour:

• As $z \to \infty$ or $z \to -\infty$, the function approaches zero
• The z-axis acts as a horizontal asymptote in both directions

Inflection points:

• The curve has points of inflection at $z = -1$ and $z = 1$
• At these points, the y-coordinate is $\frac{e^{-\frac{1}{2}}}{\sqrt{2\pi}} \approx 0.24$

Mean and variance

The standard normal distribution has two defining parameters:

• Mean: $\mu = 0$
• Variance: $\sigma^2 = 1$
• Standard deviation: $\sigma = 1$

These values are reflected in the graph:

• The mean $\mu = 0$ corresponds to the centre of symmetry and the maximum turning point
• The inflection points at $z = -1$ and $z = 1$ are each one standard deviation from the mean

The cumulative distribution function

While the PDF $\phi(z)$ tells us the probability density at each point, to calculate actual probabilities we need the cumulative distribution function (CDF), denoted by $\Phi(z)$ (uppercase phi).

The CDF is defined as:

$\Phi(z) = \int_{-\infty}^{z} \phi(t) \, dt$

This represents the probability that $Z \leq z$, which equals the area under the PDF curve from $-\infty$ to $z$.

Important properties of the CDF:

• The graph of $y = \Phi(z)$ is S-shaped (sigmoid)
• It has horizontal asymptotes: $y = 0$ on the left and $y = 1$ on the right
• It has point symmetry about the point $(0, 0.5)$
• At $z = 0$, exactly half the area lies to the left: $\Phi(0) = 0.5$

Using z-tables to find probabilities

Reading the z-table

Here is a standard z-table showing values of $\Phi(z)$ for positive z-values:

The table only shows positive z-values because we can use symmetry to find negative values.

Calculating probabilities

To find probabilities for the standard normal distribution, we combine the CDF values with two key principles:

Basic probability calculations

Right-tail probability: $P(Z \geq a) = 1 - \Phi(a)$

This works because the total area is 1, so the area to the right equals 1 minus the area to the left.

Left-tail probability: $P(Z \leq -a) = P(Z \geq a) = 1 - \Phi(a)$

This uses the symmetry property: the area to the left of $-a$ equals the area to the right of $a$.

Central probability from zero: $P(0 \leq Z \leq a) = \Phi(a) - \Phi(0) = \Phi(a) - 0.5$

This subtracts the left-half area from the total area up to $a$.

Central probability symmetric about zero: $P(-a \leq Z \leq a) = \Phi(a) - \Phi(-a)$

Alternatively, using symmetry: $P(-a \leq Z \leq a) = 2 \times P(0 \leq Z \leq a)$

Worked example: Finding probabilities

Working with negative z-values

When dealing with negative z-values, always use the symmetry of the normal curve.

The empirical rule (68-95-99.7 rule)

A particularly useful approximation for normal distributions is the empirical rule, which tells us what proportion of data falls within 1, 2, or 3 standard deviations of the mean.

For the standard normal distribution, where $\sigma = 1$:

• Approximately 68% of values fall within $[-1, 1]$: $P(-1 \leq Z \leq 1) \approx 0.6827 \approx 68\%$
• Approximately 95% of values fall within $[-2, 2]$: $P(-2 \leq Z \leq 2) \approx 0.9545 \approx 95\%$
• Approximately 99.7% of values fall within $[-3, 3]$: $P(-3 \leq Z \leq 3) \approx 0.9973 \approx 99.7\%$

Using the empirical rule in practice

Quartiles and percentiles

Quartiles and percentiles can be found from the graph of the CDF or by using interpolation on z-tables.

Key values for the standard normal distribution

Finding quartiles and percentiles

The IQR criterion for outliers

Using the IQR criterion, an outlier is any value outside:

$[Q_1 - 1.5 \times \text{IQR}, Q_3 + 1.5 \times \text{IQR}]$

For the standard normal distribution:

$Q_3 + 1.5 \times \text{IQR} \approx 0.67 + 1.5(1.35) \approx 2.70$

So an outlier lies outside $[-2.70, 2.70]$

The probability of being an outlier:

$P(Z < -2.70 \text{ or } Z > 2.70) = 2 \times P(Z > 2.70) \approx 0.007$

This means approximately 7 in 1000 scores (or just under 1%) would be classified as outliers.

Summary