The normal distribution is synonymous with a bell-shaped curve. It is a symmetric, unimodal; since it is a probability distribution, the area beneath the curve is 1. The standard normal distribution with mean 0 and standard deviation 1, denoted \\(N(\mu=0,\sigma=1)\\), is depicted below.

Figure 1

The standard normal distribution is the prototypical bell-shaped curve.

The standard normal distribution is the prototypical bell-shaped curve.

Many phenomena are normally distributed, with a typical example being the height of human beings. Below, the heights of adult US males are depicted. Notice that the mean is no longer 0, but 70.9 in. The major tick-marks along the horizontal axis are separated by 2.75 in. The height is said to be normally distributed with (population) mean 70.9 and standard deviation 2.75.

Figure 2

The height of US adult males is normally distributed \(N(\mu=70.9, \sigma=2.75)\).

The height of US adult males is normally distributed \(N(\mu=70.9, \sigma=2.75)\).

When writing \(N(\mu, \sigma)\), the greek letters \(\mu, \sigma\) are called distribution parameters. These parameters completely determine the location of the peak and the spread of the curve.

Z-Scores

We will use a calculator (or table) to compute probabilities associated with the normal distribution. To utilize these tools, we will first find the Z-score, which computes the number of standard deviations an observation falls above or below the mean:

\begin{align} Z=\frac{x-\mu}{\sigma} \label{Z-score}\tag{1} \end{align}

Here lowercase \(x\) represents an observation and \(\mu, \sigma\) are given parameters.

Example 1

Let \(X\) represent a random variable that is normally distributed with mean 10 and standard deviation 5:

  • (a) Find the Z-score of an observation \(x_1=17.3\);
  • (b) Interpret your answer from part (a).

(a) We compute the Z-score by using formula \( (\ref{Z-score}) \):

\begin{align} Z = \frac{17.3 - 10}{5} = \frac{7.3}{5}= 1.46 \end{align}

(b) The observation \(x_1\) lies 1.46 standard deviations to the right of the population mean.

Calculate 1

For the data above,

  • (a) find the Z-score for an observation \(x_2=2\);
  • (b) Interpret your answer.

Reveal Answer

(a) We compute the Z-score by using formula \( (\ref{Z-score}) \):

\begin{align} Z = \frac{2 - 10}{5} = \frac{-8}{5}= -1.6 \end{align}

(b) The observation \(x_1\) lies 1.6 standard deviations below (or to the left) of the population mean.

Question 1

Which observation \(x_1\) or \(x_2\) is a more unusual occurrence?

Reveal Answer

Observation \(x_2\) is a more unusual occurrence as it lies further from the mean.

Example 2

Find the probability \(P(Z < 1.73)\). Using the calculator below, we find \(P(Z < 1.73) = .958\)

Normal Probabilities
1. Press 2nd VARS to enter DISTR
2. Select 2:normalcdf(
3. Type lower bound
4. Type comma
5. Type upper bound
6. Enter closing )
7. Press ENTER

Remark 1

There is a small amount of area in the tails of the distribution. We may use -4 and 4 instead of \(-\infty\) and \(\infty\) in the calculator.

Figure 3

The probability \(P(Z<1.73)\) is shaded.

The probability \(P(Z<1.73)\) is shaded.

Example 3

Find the probability \(P(Z > .53)\). Using the calculator, we find \(P(Z > .53) = .298\).

Example 4

Find the probability \(P(|Z| < 1)\). We may rewrite this probability as \(P(-1 < Z < 1) = .682\).

Remark 2

68% of all observations fall within 1 standard deviation from the mean.

Figure 5

The probability \( P(|Z| < 1) \) is shaded from Example 4.

The probability \( P(|Z| < 1) \) is shaded from Example 4.

Example 5

Find the probability \(P(|Z| > 1)\).

Key: We wish to find the area of the unshaded region in Figure 5. We may rewrite this probability as

\[P( |Z| > 1) = 1 - P(|Z| < 1 )= 1 - .682 = .318\]

Exercise 1

Compute the following probabilities:

(a) \(P(|Z|<2)\)

(b) \(P(|Z|<3)\)

Reveal Answer

\begin{align} P(|Z|<2) &= .954 \\
P(|Z|<3) &= .997 \end{align}

68-95-99.7 Rule

68-95-99.7 Rule: 68% of all observations fall within 1 standard deviation, 95% of all observations fall within 2 standard deviations, and 99.7% of all observations fall within 3 standard deviations.

Example 6

(a) Find the height of an individual in the \(75^{th}\) percentile.

(b) Find the height of a student whose height is 16% from the tallest.

Inverse Normal
1. Press 2nd VARS to enter DISTR
2. Select 3:invNorm(
3. Enter percentile as decimal
4. Enter closing )
5. Press ENTER
**Remark**: Enter percentile as a decimal in **Area**, and select **Paste** and **ENTER**.

Example 6 (a) Solution

We find that the \(75^{th}\) percentile corresponds to the Z-score \(Z=.674\). To find the height, we work backwards using \( (\ref{Z-score}) \):

\begin{align} .674 &= \frac{x - \mu}{\sigma}= \frac{x - 70.9}{2.75} \end{align}

We solve for \(x\):

\begin{align} (2.75)\times.674 &= x - 70.9 \\
1.854 &= x - 70.9 \\
x &= 70.9 + 1.854 = 72.754 \approx 6 \; {\tt ft} \; 1 \; {\tt in} \end{align}

Exercise 6b

Calculate the height of a student whose height is 16% from the tallest.

  • Alternately, we may say that the student is in the \(1-.16 = .84 = 84^{th}\) percentile.

Reveal Answer

We find \(Z = .994\). We solve for \(x\):

\begin{align} (2.75)\times .994 &= x - 70.9 \\
2.734 &= x - 70.9 \\
x &= 70.9 + 2.734 = 73.634 \approx 6 \; {\tt ft} \; 2 \; {\tt in} \end{align}

Z-score of \( z_\alpha \)

Figure 6

 \(  z_{\alpha} \) is the \(Z\)-score such that the area to the right is \( \alpha \), i.e. \( P(Z>z_{\alpha})=\alpha\).

\( z_{\alpha} \) is the \(Z\)-score such that the area to the right is \( \alpha \), i.e. \( P(Z>z_{\alpha})=\alpha\).

Figure 7

\( z_{\alpha/2} \) is the \(Z\)-score such that the area to the right is \( \alpha/2 \). In particular, \( P(|Z|>z_{\alpha/2})=\alpha \).

\( z_{\alpha/2} \) is the \(Z\)-score such that the area to the right is \( \alpha/2 \). In particular, \( P(|Z|>z_{\alpha/2})=\alpha \).