Study Notes: The Normal Probability Distribution and Standardization

The Normal Probability Distribution

Introduction to the Normal Distribution

The normal distribution is a fundamental probability distribution in statistics, characterized by its symmetric, bell-shaped curve. It is widely used to model real-world phenomena such as heights, test scores, and measurement errors.

• Probability Density Function (PDF): Describes the likelihood of a random variable taking on a particular value.

• Notation: $X \sim N(\mu, \sigma^2)$, where $\mu$ is the mean and $\sigma^2$ is the variance.

• Properties: Symmetric about the mean, total area under the curve equals 1.

Standard Normal Distribution

The standard normal distribution is a special case of the normal distribution with mean $\mu = 0$ and variance $\sigma^2 = 1$. The random variable is denoted by $Z$.

• Definition: $Z \sim N(0, 1)$

• Probability Density Function:

$f(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2},\quad -\infty < z < \infty$

• Standard Normal Curve: The graph of $f(z)$.

Standardizing a Normal Distribution

Any normal random variable $X$ can be transformed into a standard normal variable $Z$ using the following formula:

$Z = \frac{X - \mu}{\sigma}$

• This process is called standardization.

• After standardization, $Z$ has mean 0 and standard deviation 1.

Empirical Rule (68-95-99.7 Rule)

The empirical rule provides a quick estimate of the spread of data in a normal distribution:

• Approximately 68% of data falls within 1 standard deviation of the mean.

• Approximately 95% falls within 2 standard deviations.

• Approximately 99.7% falls within 3 standard deviations.

For $X \sim N(\mu, \sigma^2)$:

• $P(\mu - \sigma < X < \mu + \sigma) \approx 0.68$

• $P(\mu - 2\sigma < X < \mu + 2\sigma) \approx 0.95$

• $P(\mu - 3\sigma < X < \mu + 3\sigma) \approx 0.997$

Properties of the Normal Distribution

General vs. Standard Normal

Property	General Normal ($X \sim N(\mu, \sigma^2)$)	Standard Normal ($Z \sim N(0, 1)$)
Total Area	1	1
Symmetry	About $\mu$	About $0$
Mean = Median = Mode	$\mu$	0

Empirical Rule Table

Interval	General Normal	Standard Normal
1 SD	$P(\mu - \sigma < X < \mu + \sigma) = 0.68$	$P(-1 < Z < 1) = 0.68$
2 SD	$P(\mu - 2\sigma < X < \mu + 2\sigma) = 0.95$	$P(-2 < Z < 2) = 0.95$
3 SD	$P(\mu - 3\sigma < X < \mu + 3\sigma) = 0.997$	$P(-3 < Z < 3) = 0.997$

Calculating Z-scores

Definition and Interpretation

A z-score measures how many standard deviations a data value $x$ is from the mean $\mu$:

$z = \frac{x - \mu}{\sigma}$

• If $x = \mu + \sigma$, then $z = 1$

• If $x = \mu + 2\sigma$, then $z = 2$

• If $x < \mu$, then $z$ is negative

Example: IQ scores with $\mu = 100$, $\sigma = 15$, $x = 120$:

$z = \frac{120 - 100}{15} = 1.33$

Interpretation: The score is 1.33 standard deviations above the mean.

Using the Standard Normal Distribution

Transforming Between X and Z

Given $X \sim N(\mu, \sigma^2)$ and $Z \sim N(0, 1)$:

• $Z = \frac{X - \mu}{\sigma}$

• $X = \sigma Z + \mu$

Probabilities for intervals can be computed by converting $X$ values to $Z$-scores.

Computing Areas Using the Z-table

The Z-table provides the area under the standard normal curve to the left of a specified $z$-score.

• To find $P(Z < z)$, look up $z$ in the table.

• To find $P(Z > z)$, use $1 - P(Z < z)$.

• To find $P(z_1 < Z < z_2)$, compute $P(Z < z_2) - P(Z < z_1)$.

Example: Area to the left of $z = 1.33$ is $0.9082$.

Examples of Area Calculations

• Area to the left: $z = -0.38$, area = $0.3520$

• Area to the right: $z = 1.25$, area = $1 - 0.8944 = 0.1056$

• Area between: $z_1 = -1.02$, $z_2 = 2.94$, area = $0.9984 - 0.1539 = 0.8445$

Generalization for Any Normal Variable

• Convert $x$ to $z$-score: $z = \frac{x - \mu}{\sigma}$

• Use Z-table to find area to the left or right.

• For intervals, subtract areas: $P(z_1 < Z < z_2) = P(Z < z_2) - P(Z < z_1)$

Computing Normal Probabilities

Step-by-Step Procedure

1. Convert $X$ values to $Z$-scores.

2. Use the Z-table to find the corresponding probabilities.

Remarks:

• Z-table covers $-3.40 \leq z \leq 3.40$

• Table entries are for $P(Z < z)$

• Rows: ones and tenths; Columns: hundredths

Types of Probability Calculations

• Left-tail: $P(Z < z)$

• Right-tail: $P(Z > z) = 1 - P(Z < z)$

• Interval: $P(z_1 < Z < z_2) = P(Z < z_2) - P(Z < z_1)$

Example Table: Z-table (Excerpt)

z	0.00	0.01	0.02	0.03	0.04	0.05
1.3	0.9032	0.9049	0.9066	0.9082	0.9099	0.9115
1.6	0.9452	0.9463	0.9474	0.9484	0.9495	0.9505

Additional info: Table values are inferred from the slides and standard Z-tables.

Percentiles of the Normal Distribution

Definition and Calculation

The p-th percentile $x_p$ of a normal random variable $X$ is the value such that:

$P(X \leq x_p) = p$

• To find $x_p$, first find the corresponding $z_p$ from the Z-table such that $P(Z \leq z_p) = p$.

• Then, convert back to $X$ using $x_p = \mu + z_p \sigma$.

Example: For $p = 0.95$, $z_p = 1.645$ (average of 1.64 and 1.65 from the table).

Special Cases:

• Median ($p = 0.5$): $z = 0$

• First quartile ($p = 0.25$): $z \approx -0.674$

• Third quartile ($p = 0.75$): $z \approx 0.674$

Procedure for Finding Cutoff Points

1. Draw the normal curve and shade the area for the given percentile.

2. Use the Z-table to find the $z$-score corresponding to the shaded area.

3. Calculate the cutoff value: $x = \mu + z \sigma$

Example: For body temperature, mean $98.2^\circ$F, $\sigma = 0.73^\circ$F, lowest 3% cutoff:

$z = -1.88$ (from Z-table for 0.03 area to the left)

$x = 98.2 + (-1.88 \times 0.73) = 96.8^\circ$F$

Continuous Random Variables: Probability Interpretation

For any continuous random variable, the probability of observing a specific value is zero:

$P(a < X < b) = P(a \leq X < b) = P(a < X \leq b) = P(a \leq X \leq b)$

This is because the area under a single point is zero; only intervals have nonzero probability.

Summary Table: Types of Normal Probability Calculations

Type	Formula	Example
Left-tail	$P(Z < z)$	$P(Z < 1.33) = 0.9082$
Right-tail	$P(Z > z) = 1 - P(Z < z)$	$P(Z > 1.25) = 1 - 0.8944 = 0.1056$
Interval	$P(z_1 < Z < z_2) = P(Z < z_2) - P(Z < z_1)$	$P(-1.02 < Z < 2.94) = 0.9984 - 0.1539 = 0.8445$

Practice Problems

• Given: Heights in a sixth-grade class are normally distributed with $\mu = 55$ inches, $\sigma = 2.5$ inches.

• Billy is 50 inches tall. Find: His z-score and the percent of students taller than Billy.

• Catherine is 56.08 inches tall. Find: The proportion of students shorter than her.

• Diana is taller than 80% of the class. Find: Her height.

Additional info: These exercises reinforce the application of z-scores and percentiles in normal distributions.