For a Calculus I class, the final exam score Y and the...

For a Calculus I class, the final exam score $Y$ and the average $X$ of the four earlier tests have a bivariate normal distribution with mean $\mu_{1}=73,$ standard deviation $\sigma_{1}=12,$ mean $\mu_{2}=70$ , standard deviation $\sigma_{2}=15 .$ The correlation is $\rho=.71 .$ Determine $\begin{array}{ll}{\text { (a) }} & {\mu_{Y|X=x}} \\ {\text { (b) }} & {\sigma_{Y|X=x}^{2}} \\ {\text { (c) }} & {\sigma_{Y|X=x}}\end{array}$ (d) $P(Y>90 | X=80),$ i.e., the probability that the final exam score exceeds 90 given that the average of the four earlier tests is 80

For a Calculus I class, the final exam score $Y$ and the average $X$ of the four earlier tests have a
bivariate normal distribution with mean $\mu_{1}=73,$ standard deviation $\sigma_{1}=12,$ mean $\mu_{2}=70$ , standard deviation $\sigma_{2}=15 .$ The correlation is $\rho=.71 .$ Determine

$\begin{array}{ll}{\text {
(a) }} & {\mu_{Y|X=x}} \\ {\text {
(b) }} & {\sigma_{Y|X=x}^{2}} \\ {\text {
(c) }} & {\sigma_{Y|X=x}}\end{array}$
(d) $P(Y>90 | X=80),$ i.e., the probability that the final exam score exceeds 90 given that the average of the four earlier tests is 80

Key Concepts

Correlation Coefficient

The correlation coefficient quantifies the degree of linear relationship between two variables. In the context of bivariate normal distributions, it not only indicates the strength and direction of the association but also affects the formulas for the conditional mean and variance. This measure is crucial in understanding how variability in one variable is related to variability in the other, and it directly influences the predictive power of one variable over the other.

Standardization of Normal Variables

Standardization is the process of converting a normal random variable into a standard normal variable by subtracting its mean and dividing by its standard deviation. This transformation is fundamental for calculating probabilities, as it allows one to use standard normal distribution tables or software to find the likelihood of events. It also aids in comparing scores from different normal distributions on a common scale.

Bivariate Normal Distribution

The bivariate normal distribution describes the joint behavior of two normally distributed variables with a specified relationship. In this distribution, each variable is normally distributed by itself, and every linear combination of the two variables is also normally distributed. This framework allows for a structured way to analyze how two related variables interact and depend on one another.

Conditional Distribution in Bivariate Normal

Within the bivariate normal framework, the conditional distribution of one variable given the other is itself normally distributed. Its mean is computed via a linear regression relationship that involves the means of the two variables, the correlation between them, and their standard deviations, while its variance is reduced from the marginal variance by a factor that depends on the square of the correlation. This concept is key in determining expected values and variability when one variable is fixed.

Transcript

00:01 Okay, so for this example, we're looking at a biovariant distribution, a biobariant normal distribution.

00:09 And what we're going to do is denote that our y is going to be for the calculus one class.

00:16 And the x is denoted for the other calculus class.

00:20 And we're given specific set of data, we're given the average, which is 73.

00:26 The average for the second is 70 we're given the standard deviation sigma of 1 is 12 and the standard deviation sigma of 2 is 15 and we're also given the correlation coefficient is 0 .71 so this is going to be our row value given as our row value which is the correlation so what do we do the first thing for part a we are asked to find specifically, let's write that down, the mean, y.

01:02 So the y is denoted for the calculus one class.

01:06 So we're going to find the mean for the calculus one class, given that we have the x for the other classes, so uppercase x, and at a specific value of lowercase x.

01:22 So we're going to find the mean for the calculation.

01:25 1 class, given the other calculus classes, add a specific value x for what are we looking at specifically? we're looking at final exam scores, at a specific value of a final exam score for the other calculus classes.

01:46 So let's find that and the equation we'll use for that is simply going to be the following.

01:52 This would equal to our mu -sub -2.

01:56 And the two is going to be for the second, so it's going to be for other calculus classes, plus our row, which is our correlation coefficient, we take our sigma sub 2, we take our x value minus mu sub 1, and this is divided by our sigma 1.

02:20 So now we just plug in what we know, so our mu sub 2 is 70, this is giving.

02:25 Us 70 plus our row value is going to be 0 .71 then we take that multiply by our sigma sub 2 15 and close this out then in parentheses we have x so keep this x we're not given a value for this lowercase x that's what we're trying to essentially arrive at a function in terms of x at the end of it all and our mu sub 1 is going to be 73 and our sigma sub 1 is 12.

03:00 So now we can reduce this further.

03:02 You leave the 70.

03:04 If you do the math for this portion here, it's going to be plus 10 .65.

03:11 And now we get x minus 73 divided by 12.

03:17 So this is going to be our mean.

03:21 So our mean for the calculus 1 class.

03:26 With respect to the other classes at a specific value x for the final exam score for the other classes.

03:38 So this is the final equation that we will use.

03:44 So this is for part a...