Question

Let \(X\) and \(Y\) have a bivariate normal distribution with respective parameters \(\mu_{x}=2.8, \mu_{y}=110, \sigma_{x}^{2}=0.16, \sigma_{y}^{2}=100\), and \(\rho=0.6 .\) Using \(R\), compute: (a) \(P(106

Short answer

The probabilities \(P(106<Y<124)\) and \(P(106<Y<124 | X=3.2)\) can be calculated using the parameters of the bivariate normal distribution and the standard normal distribution table or functions provided in R. The calculated probabilities should be in the range of 0 to 1.

Step-by-step solution

Understand bivariate normal distribution

A bivariate normal distribution involves two variables X and Y. It is defined by five parameters: the means \(\mu_{x}, \mu_{y}\) of X and Y, the variances \(\sigma_{x}^{2}, \(\sigma_{y}^{2}\) of X and Y, and the correlation coefficient \(\rho\) between X and Y. With these parameters, the Hi and Lo (lower limit and upper limit) of the normal probability distribution of Y can be calculated.

Calculate the probabilities P(106

To calculate \(P(106<Y<124)\), we need to standardize the limits of the range to conform with the standard normal distribution. The standard value \(z_{\text{hi}}=(\text{upper limit} - \mu_y) / \sigma_y \) and \(z_{\text{lo}}=(\text{lower limit} - \mu_y) / \sigma_y\). Using the given parameters, calculate \(z_{\text{hi}}=(124 - 110) / 10 = 1.4\) and \(z_{\text{lo}}=(106 - 110) / 10 = -0.4\). Look up these z-values in the z-table (or use R's pnorm function) to find the probability \(P(106<Y<124) = P(-0.4<Z<1.4)\)

Calculate the conditional probability P(106

For a conditional probability where X has a specific value x, the mean and variance of Y change. The new mean is \(E[Y|X=x] = \mu_y + \rho * \(\frac{\sigma_y}{\sigma_x}\) * (X - \mu_x) \) and the new variance is \(Var[Y|X=x] = \sigma_y^{2} * (1-\rho^{2})\(. Using the given parameters, calculate the new \(\mu_{y|x}\) and \(\sigma_{y|x}^{2}\). Then proceed to calculate the z-values and lookup the probability as in step 2.