MTH-361A | Spring 2025 | University of Portland
March 10, 2025
The normal r.v.
A normal r.v. is a type of continuous r.v. whose probability distribution follows the normal distribution, also known as the Gaussian distribution. The normal distribution is characterized by two parameters, \(\mu\) as the mean and \(\sigma^2\) as the variance: \[X \sim \text{N}(\mu,\sigma^2)\]
Sample Space:
Parameters
The Normal PDF
The normal r.v. \(X \sim \text{N}(\mu,\sigma^2)\) has infinite possible outcomes (or infinite sized sample space) where \(\mu\) is the mean and \(\sigma^2\) is the variance with PDF given as \[f(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}, \ -\infty < x < \infty\]
Normal Expected Value
Normal Distribution with Expected Value
Normal R.V.
Let \(\mu=10\) and \(s=2.24\) be the mean and standard deviation respectively.
\[ \begin{aligned} \text{R.V. } & \longrightarrow X \sim \text{N}\left(10,2.24^2\right) \\ \text{PDF } & \longrightarrow f(x) = \frac{1}{\sqrt{2 \pi (2.24)^2}} e^{-\frac{(x-10)^2}{2(2.24)^2}} \\ \text{for } & x \in (-\infty,\infty) \\ \text{expected value} & \longrightarrow \text{E}(X) = 10 \end{aligned} \]
In general, the expected value of the normal r.v. is given by \[\text{E}(X) = \mu,\] which is the center of the normal distribution.
Key idea the Central Limit Theorem (CLT). Image source: Medium–AI/Data Science Digest
Watch the 1st 20 minutes of this YouTube video: https://www.youtube.com/watch?v=zeJD6dqJ5lo&t=564s
Normal Distribution
Example:
What is \(P(7 \le X \le 14)\) for \(X \sim \text{N}(10,2.24)\)? \[ \begin{aligned} P(7 \le X \le 13) & = \int_7^{14} f(x) \ dx \\ & = \int_7^{14} \frac{1}{\sqrt{2 \pi (2.24)^2}} e^{-\frac{(x-10)^2}{2(2.24)^2}} \ dx \\ & = P(X \le 13) - P(X \le 7) \\ P(7 \le X \le 13) & \approx 0.8727 \end{aligned} \]
Using R:
## [1] 0.8726884
\(\star\) Note that the Normal PDF is symmetrical but the interval probability we just computed was not.
Normal Distribution
Example:
What is \(P(7.7639 \le X \le 12.2361)\) for \(X \sim \text{N}(10,2.24)\)? \[ \begin{aligned} P(7.7639 \le X \le 12.2361) & = P(X \le 12.2361) - P(X \le 7.7639) \\ P(7.7639 \le X \le 12.2361) & \approx 0.6818 \end{aligned} \]
Using R:
## [1] 0.6818462
\(\star\) Note that \(P(7.7639 \le X \le 12.2361) = P(10-2.24 \le X \le 10+2.24) \approx 0.6818\).
Normal Distribution
Example:
What is \(P(141.3397 \le X \le 158.6603)\) for \(X \sim \text{N}(150,8.66)\)? \[ \begin{aligned} P(141.3397 \le X \le 158.6603) & = P(X \le 158.6603) - P(X \le 141.3397) \\ P(141.3397 \le X \le 158.6603) & \approx 0.6827 \end{aligned} \]
Using R:
## [1] 0.6827063
\(\star\) Note that \(P(141.3397 \le X \le 158.6603) = P(150-8.66 \le X \le 150+8.66) \approx 0.6827\).
Normal Distribution
Example:
What is \(P(5.5279 \le X \le 14.4721)\) for \(X \sim \text{N}(10,2.24)\)? \[ \begin{aligned} P(5.5279 \le X \le 14.4721) & = P(X \le 14.4721) - P(X \le 5.5279) \\ P(5.5279 \le X \le 14.4721) & \approx 0.9541 \end{aligned} \]
Using R:
## [1] 0.9541176
\(\star\) Note that \(P(5.5279 \le X \le 14.4721) = P(10-2 \times 2.24 \le X \le 10+2 \times 2.24) \approx 0.9541\).
\(\dagger\) Is \(P(141.3397 \le X \le 158.6603) = P(150-2 \times 8.66 \le X \le 150+2 \times 8.66) \approx 0.95\)?
1st standard deviation from the mean
\[P(\mu - \sigma \le X \le \mu + \sigma) \approx 0.68\]
2nd standard deviation from the mean
\[P(\mu - 2\sigma \le X \le \mu + 2\sigma) \approx 0.95\]
3rd standard deviation from the mean
\[P(\mu - 3\sigma \le X \le \mu + 3\sigma) \approx 0.997\]
The Normal PDF satisfies the probability axioms
\[P(\mu - \infty \le X \le \mu + \infty) \approx 1\]
\(\star\) Key Idea: Because of the axiom that the sum of the probabilities for all outcomes in the sample space is equal to 1, the total area under the Normal PDF is always 1.
The standard normal distribution is when \(\mu=0\) and \(s=1\) or \(Z \sim \text{N}(0,1)\).
The transformation formula (the z-score)
Standardized scores that measure how many standard deviations a value is from the mean. \[Z = \frac{X - \mu}{\sigma}\]
The Standard Normal PDF
Using the z-score transformation, the normal PDF reduces to \[f(z) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{z^2}{2}}, \ -\infty < z < \infty\]
The standard normal distribution, \(Z \sim \text{N}(0,1)\).
\(\star\) Key Idea: The standard normal distribution is that it is a normal distribution with a mean of 0 and a standard deviation of 1. It serves as a reference distribution, allowing any normally distributed variable to be standardized.
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