91Ó°ÊÓ

In statistics, when we talk about independent variables, we're referring to variables that do not influence each other. They operate separately and changes in one variable do not affect the other.
In the context of our bivariate normal distribution problem, we have two variables: \(X\) and \(Y\). These variables have a correlation coefficient \(\rho = 0\), which indicates that they are independent.

• This means that knowing the value of one variable gives no information about the other variable.
• Each variable follows its individual normal distribution pattern without any interaction.

Understanding this concept is crucial because it simplifies calculations. For instance, the probability of a joint event (where both \(X\) and \(Y\) are within certain ranges) can simply be the product of the individual probabilities, if the variables are independent.

Standardization is a method used to transform a variable's distribution to the standard normal distribution, which has a mean of zero and a standard deviation of one. This transformation allows us to use the standard normal distribution tables, making probability calculations more accessible.
The formula for standardization is: \[Z = \frac{X - \mu}{\sigma}\]where

• \(\mu\) is the mean of the distribution,
• \(\sigma\) is the standard deviation,
• and \(Z\) is the standardized value.

In our exercise, both \(X\) and \(Y\) were standardized. For example, to find the probability \(P(2.95 < X < 3.05)\), we converted the endpoints of \(X\)'s range to \(Z\)-scores using the standardization formula. This process allowed us to refer to a standard normal distribution table and determine the probability of \(Z\) falling between those scores.

Joint probability refers to the probability of two events happening at the same time. In statistical terms, it's the likelihood of both \(X\) and \(Y\) falling within specified ranges simultaneously.
In the case of independent variables, calculating joint probability becomes straightforward. You simply multiply the individual probabilities of each event. For our problem, given that \(X\) and \(Y\) are independent with \(\rho = 0\), we could find:

• The probability \(P(2.95 < X < 3.05, 7.60 < Y < 7.80)\) by multiplying \(P(2.95 < X < 3.05)\) with \(P(7.60 < Y < 7.80)\).

When variables are not independent, the joint probability requires more complex calculations that take into account the correlation between variables.

Normal distribution tables, also known as Z-tables, are tools that help us find the probabilities associated with standardized normal variables. The tables provide the probability that a standard normal random variable \(Z\) is less than or equal to a given value.
These tables are especially useful after standardizing a normal variable. Once we have the \(Z\)-scores, we can use the table to find probabilities easily.
Some points to consider when using normal distribution tables:

• It often provides cumulative probabilities up to certain \(Z\)-scores.
• To find the probability between two \(Z\)-scores, find the cumulative probability for each and subtract the smaller from the larger.

In our bivariate distribution exercise, after standardizing the \(X\) and \(Y\) values, these tables were used to find the corresponding probabilities, greatly simplifying our task of finding the likelihood of \(X\) and \(Y\) falling within specific ranges.