Problem 32 - Applied Statistics and Probability for Engineers

Question

The systolic and diastolic blood pressure values (mm Hg) are the pressures when the heart muscle contracts and relaxes (denoted as \(Y\) and \(X,\) respectively). Over a collection of individuals, the distribution of diastolic pressure is normal with mean 73 and standard deviation \(8 .\) The systolic pressure is conditionally normally distributed with mean \(1.6 x\) when \(X=x\) and standard deviation of \(10 .\) Determine the following: (a) Conditional probability density function \(f_{Y \mid 73}(y)\) of \(Y\) given \(X=73\) (b) \(P(Y<115 \mid X=73)\) (c) \(E(Y \mid X=73)\) (d) Recognize the distribution \(f_{X Y}(x, y)\) and identify the mean and variance of \(Y\) and the correlation between \(X\) and \(Y\)

Step-by-Step Solution

Answer

(a) \(N(116.8, 100)\), (b) \(P \approx 0.4286\), (c) \(E(Y | X=73) = 116.8\), (d) Bivariate normal distribution with variance 263.84.

Step 1: Conditional PDF

Given that the systolic pressure is conditionally normally distributed with mean \(1.6x\) for \(X=x\) and a standard deviation of \(10\), the conditional probability density function \(f_{Y|X}(y)\) when \(X = 73\) is a normal distribution \(N(1.6 \times 73, 10^2)\). Thus, \(f_{Y|73}(y)\) is a normal distribution with mean \(1.6 \times 73 = 116.8\) and variance \(10^2 = 100\).

Step 2: Calculate the Probability

To find \(P(Y < 115 \mid X=73)\), we use the standard normal distribution. Calculate the Z-score with the formula \(z = \frac{115 - 116.8}{10}\), which simplifies to \(z = \frac{-1.8}{10} = -0.18\). Using standard normal distribution tables, find \(P(Z < -0.18)\). This corresponds to about 0.4286.

Step 3: Expected Value Estimate

The expected value \(E(Y \mid X=73)\) of a normal distribution is simply its mean value. Since we have a conditional mean of \(1.6x\) with \(x=73\), we have \(E(Y \mid X=73) = 116.8\).

Step 4: Joint Distribution Recognition

Recognizing the joint distribution, \(f_{XY}(x, y)\) implies a bivariate normal distribution because both \(X\) and the conditional distribution of \(Y\) are normally distributed. \(X\) is \(N(73, 8^2)\) and the conditional \(Y|X=x\) is \(N(1.6x, 10^2)\). The mean of \(Y\) will be \(E(1.6X) = 1.6 \times 73 = 116.8\). Variance of \(Y\) involves both \(Var(1.6X) + Var(Y|X) = (1.6 \times 8)^2 + 10^2 = 163.84 + 100 = 263.84\). The correlation \(\rho_{XY}\) can be determined from the linear relationship; here, provided \(Cov(X,Y) = 1.6 \times Var(X) = 1.6 \times 8^2 = 102.4\), thus \(\rho_{XY} = \frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}} = \frac{102.4}{\sqrt{8^2 \times 263.84}}\).

Key Concepts

Bivariate Normal Distribution

In statistics, a bivariate normal distribution is a probability distribution that describes the joint behavior of two continuous random variables. This distribution is an extension of the univariate normal distribution to two dimensions. When we say that two variables, like systolic and diastolic blood pressure, follow a bivariate normal distribution, we imply that:

• Each variable is normally distributed on its own.
• The relationship between the two variables is linear.
• The joint probability distribution can be described by parameters such as the mean, variance, and covariance.

In the context of this exercise, the diastolic pressure, denoted by \(X\), follows a normal distribution with a mean of 73 and a standard deviation of 8. The systolic pressure, denoted by \(Y\), is conditionally normally distributed given \(X\). This means that for any fixed \(X=x\), \(Y\) follows a normal distribution with its parameters determined by \(x\). Understanding the joint distribution as bivariate helps us analyze interrelated measurements like blood pressure effectively.

Expected Value

The expected value of a random variable is a fundamental concept that signifies the average or mean value that the variable takes on over a large number of samples. In other words, it is the center of the distribution. For a normal distribution, like the ones we are examining here, the expected value is the mean of the distribution.

• For the systolic pressure given diastolic pressure \(X=73\), the expected value \(E(Y | X=73)\) is simply the mean of the conditional distribution of \(Y\).
• In our case, this is calculated as \(1.6 \times 73 = 116.8\).

This expected value provides us with the average systolic pressure one might anticipate when the diastolic pressure is known to be 73. It serves as a key summary measure for the conditional distribution, simplifying data interpretation.

Z-score Calculation

Z-scores are a way of standardizing values within a normal distribution to make them easier to interpret. A Z-score indicates how many standard deviations an element is from the mean. Calculating a Z-score transforms a normal distribution into the standard normal distribution, which has a mean of 0 and a standard deviation of 1.

• To find the Z-score in this problem, we use the formula \(z = \frac{(Y - \mu)}{\sigma}\), where \(Y\) is the score you're comparing, \(\mu\) is the distribution mean, and \(\sigma\) is the standard deviation.
• For \(Y < 115\) given \(X = 73\), substitute \(Y = 115\), \(\mu = 116.8\), and \(\sigma = 10\) for the calculation: \(z = \frac{115 - 116.8}{10} = -0.18\).

This Z-score allows us to use standard normal distribution tables (or software) to find the probability of a random variable being less than a certain value, which is helpful in assessing probabilities when distributions are known.

Variance and Covariance Analysis

Understanding variance and covariance is crucial when working with joint or conditional distributions. Variance measures the spread of a random variable around its mean, whereas covariance assesses the linear relationship between two variables.

• The variance of a normal distribution quantifies how much the values deviate from the mean on average. For our case with \(Y\), we have \(Var(Y) = Var(1.6X) + Var(Y|X)\).
• Given the data, this results in \( (1.6 \times 8)^2 + 10^2 = 263.84\).
• Covariance \(Cov(X,Y)\) provides insight into how \(X\) and \(Y\) move together. In our scenario, covariance is determined by \(1.6 \times Var(X) = 102.4\).
• The correlation coefficient \(\rho_{XY}\) is derived from covariance and shows the strength and direction of the linear relationship between two variables: \(\rho_{XY} = \frac{Cov(X,Y)}{\sqrt{Var(X)Var(Y)}}\).

These concepts allow us to understand and quantify the interdependencies in bivariate distributions, enabling us to make predictions about one variable based on the other.