Continuous Random Variables and the Normal Distribution

Continuous Random Variables

A continuous random variable can assume any value within one or more intervals.
Examples:
- Height of a person.
- Time taken to complete an exam.
- Price of a house.
Continuous random variables can assume infinite, uncountable values over an interval.

Continuous Probability Distribution

Deals with frequency and relative frequency distributions.
Example: Height of female students.
Table showing height intervals, frequencies, and relative frequencies.

Properties of Continuous Probability Distribution

The probability that X assumes a value in any interval lies in the range 0 to 1.
The total probability of all (mutually exclusive) intervals within which X can assume a value is 1.
Illustrative probability distribution curve.

Probability Density Function

Denoted as f(x), it has the following properties:
1. $f(x) > 0$ for all values of X.
2. The total area under the probability density function f(x) over all values of X within its range is equal to 1.
3. The probability that X lies between two values is the area under the density function graph between the two values.
4. The probability that a continuous random variable X assumes a single value is always zero:
  $P(a) = 0$ and $P(b) = 0 => P(a ≤ X ≤ b) = P(a < X < b)$
  $\int_{a}^{b} f(x)dx = P(a < X < b)$

Probability as Area

The shaded area under the curve between points a and b represents the probability that $X$ lies between a and b.
$P(a ≤ X ≤ b)$ : Shaded area under the curve represents the probability that X is between a and b
$P(a < X < b)$ : Probability of X lying between a and b.
Note that the probability of any single value is zero.

Cumulative Density Function

The cumulative density (distribution) function, $F(x)$ , expresses the probability that $X$ does not exceed the value of $x$ .
- $F(x) = P(X ≤ x)$
If $a$ and $b$ are two possible values of $X$ , with $a < b$ , the probability that $X$ lies between $a$ and $b$ is:
- $P(a < X < b) = F(b) - F(a)$

Normal Probability Distribution

A bell-shaped curve with the following characteristics:
1. The total area under the curve is 1.
2. The curve is symmetric about the mean, where mean = median = mode.
3. Asymptotic, meaning the tails get closer and closer to the horizontal axis but never touch it.
- $P(X < m) = 0.5$
- $P(X > m) = 0.5$

The Normal Distribution

Approximates probability distributions of a wide range of random variables.
Distributions of sample means approach a normal distribution given a “large” sample size.
Leads to good business decisions for a number of applications.

Many Normal Distributions

Different normal distributions are obtained by varying the parameters $\mu$ and $\sigma$ .

Parameters of the Normal Probability Distribution

The mean ( $\mu$ ) and the standard deviation ( $\sigma$ ) are the parameters of the normal distribution.
Each different set of values of $\mu$ and $\sigma$ gives a different normal distribution.
The value of $\mu$ determines the center of a normal distribution curve on a horizontal axis.
The value of $\sigma$ gives the spread of a normal distribution curve.
Changing $\mu$ shifts the distribution left or right; changing $\sigma$ increases or decreases the spread.
Notation: Given the mean $\mu$ and variance $\sigma^2$ , the normal distribution is defined as:
- $X \sim N(\mu, \sigma^2)$

The Normal Distribution Shape - Example

Example: Ages of employees in Industries A, B, and C are normally distributed.

Normal Probability Density Function

The formula for the normal probability density function $f(x)$ is:
- $f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2\sigma^2}}$
- Where:
  - $e$ = the mathematical constant approximated by 2.71828
  - $\pi$ = the mathematical constant approximated by 3.14159
  - $\mu$ = the population mean
  - $\sigma^2$ = the population variance
  - $x$ = any value of the continuous variable, $- \infty < x < \infty$

Finding Normal Probabilities

The probability for a range of values is measured by the area under the curve.
- $P(a < X < b) = F(b) - F(a)$
Illustrative examples showing the area under the curve between points a and b.

The Standard Normal Distribution

Any normal distribution can be transformed into the standardized normal distribution (Z), with a mean of 0 and standard deviation/variance of 1.
Transformation formula:
- $Z = \frac{X - \mu}{\sigma}$
- $Z \sim N(0,1)$
Need to transform X units into Z units by subtracting the mean of X and dividing by its standard deviation

The Standard Normal Distribution

The units on the horizontal axis of the standard normal curve are denoted by Z and called Z-values or Z-scores.
A specific value of Z gives the distance between the mean and the point represented by Z in terms of the standard deviation.

Revisiting the Empirical Rule

Presentation of probabilities within standard deviations of the mean
- $P(-1 ≤ Z ≤ 1)$
- $P(-2 ≤ Z ≤ 2)$
- $P(-3 ≤ Z ≤ 3)$

Examples

Example 1: Find the area under the standard normal curve between Z = 0 and Z = 1.95.
- $=.9744 - .5$
Example 2: Find the area under the standard normal curve between Z = -1.95 and Z = 0.
- $=.5 - .0256$
Example 3: Find the area under the standard normal curve between Z = -1.95 and Z = 1.95.
- $=.9744 - .5=.5 - .0256$
- $\implies (.9744 -.5)+ (.5 -.0256) = .9744 -.0256$

Examples

Example 4: Find the following probabilities for the standard normal curve:
- a) $P (Z < 1 ) = .8413$
- b) $P (1 < Z < 1.42) = P(Z <1.4) - P(Z <1) = .9222 -.8413$
- c) $P (Z > 1.42) = 1- P(Z <1.42) = 1-.9222$

Finding Normal Probabilities

$P(a < X < b)$ : Transforming X values to Z values
- $P(\frac{a - \mu}{\sigma} < Z < \frac{b - \mu}{\sigma}) = F(\frac{b - \mu}{\sigma}) - F(\frac{a - \mu}{\sigma})$
- $Z = \frac{x - m}{\sigma}$

The Standard Normal Table

Example: Let X be a normal random variable with a mean equal to 40 and a standard deviation to 10. Find probabilities.
- a) $P (X < 60 ) = P(Z < 2) = .9772$ where $Z = \frac{60 - 40}{10} = 2$
- b) $P (X < 20) = P(Z < -2) = .0228$ where $Z = \frac{20 - 40}{10} = -2$
- c) $P (20 < X < 60 ) = P(-2 < Z < 2) = P(Z < 2) – P(Z < -2) = .9772 - .0228$
Illustrative example: Z values in standard normal table.

Applications of the Normal Distribution

Example: The lifespan of a calculator follows a normal distribution with a mean of 54 months and a standard deviation of 8 months. Calculators malfunctioning within 36 months are replaced. What percentage of calculators are expected to be replaced?
- $P(X < 36 ) = P(Z < -2.25) = 1.22\%$
- Where $Z = \frac{36 - 54}{8} = -2.25$
Note that the distribution is the same, only the scale has changed. We can express the problem in original units (X) or in standardized units (Z).

Determining X values when an area under the normal distribution curve is known

Example: The lifespan of a calculator has a normal distribution with a mean of 54 months and a standard deviation of 8 months. What should the warranty period be if the company does not want to replace more than 1% of all calculators sold?
- $Z = \frac{X - \mu}{\sigma} \implies Z = -2.33$
- $X = \mu + Z*\sigma$
- $X = 54 + (-2.33) * 8 = 35$ weeks

Determining the X values when an area under the normal distribution curve is known

Suppose the examination scores are normally distributed with a mean of 76 and a standard deviation of 15. The top 15% of the students receive A’s and the bottom 10% receive F’s. Find the minimum score needed to receive an A and the minimum score needed to pass (not to receive an F).
- $Z = \frac{X - \mu}{\sigma}$

The Normal Approximation of the Binomial Distribution

The binomial probability distribution is applied to discrete random variables:
1. There are n identical trials.
2. Each trial has only two possible outcomes.
3. The probability of two outcomes remains constant.
4. The trials are independent.
The Normal distribution is used as an approximation to the binomial distribution when $np ≥ 5$ and $nq ≥ 5$ .

The Normal Approximation of the Binomial Distribution

The normal distribution applies to a continuous random variable.
The binomial distribution applies to a discrete random variable.
The second step in applying the normal approximation to the binomial distribution is to convert the discrete random variable to a continuous random variable by making the correction of continuity.
Continuity correction factor: The addition of 0.5 and/or subtraction of 0.5 from the value(s) of X when a normal distribution is used as an approximation to a binomial distribution, where X is the number of successes in n trial is called the continuity correction factor.

Use the normal approximation to the binomial distribution

Example: An International Revenue Oversight Board survey found that 80% of taxpayers said that it was very important for the Internal Revenue Service to ensure that high-income tax payers do not cheat on their tax returns.
- a) For the sample of 100 taxpayers, what is the probability that 90 taxpayers say that it is very important to ensure that high-income tax payers do not cheat on their tax returns?
- b) For the sample of 100 taxpayers, what is the probability that at least 90 taxpayers say that it is very important to ensure that high-income tax payers do not cheat on their tax returns?

Solution

a) Binomial Distribution:
Normal Distribution:
- $P(X = 90)$
- $P(89.5 ≤ X ≤ 90.5 ) = P(\frac{89.5 - 80}{4} ≤ Z ≤ \frac{90.5 - 80}{4}) = P(2.375 ≤ Z ≤ 2.625)$
- $\mu = (100)(0.80) = 80$
- $\sigma = \sqrt{(100)(0.80)(0.20)} = 4$
b) Binomial Distribution:
Normal Distribution:
- $P(X ≥ 90)$
- $P(X ≥ 89.5 ) = P(Z ≥ \frac{89.5 - 80}{4}) = P(Z ≥ 2.375) = 1 - P(Z < 2.375)$
- $\mu = (100)(0.80) = 80$
- $\sigma = \sqrt{(100)(0.80)(0.20)} = 4$

How to Do It in Excel?

Open blank worksheet.
Select Formulas
Click on fx (function wizard).
Select Statistical category.
Select the NORM.DIST function.
Fill in the requested information in the template.
True indicates cumulative probabilities.
Click OK.
- Example Result: $\mu= 22.14$ and $\sigma = 6.09$ $\implies P(X ≤ 30) = .902$