Problem 54 - Applied Statistics and Probability for Engineers

Question

In an acid-base titration, a base or acid is gradually added to the other until they have completely neutralized each other. Let \(X\) and \(Y\) denote the milliliters of acid and base needed for equivalence, respectively. Assume that \(X\) and \(Y\) have a bivariate normal distribution with \(\sigma_{X}=5 \mathrm{~mL}, \sigma_{Y}=2 \mathrm{~mL}\), \(\mu_{X}=120 \mathrm{~mL}, \mu_{Y}=100 \mathrm{~mL},\) and \(\rho=0.6\) Determine the following: (a) Covariance between \(X\) and \(Y\) (b) Marginal probability distribution of \(X\) (c) \(P(X<116)\) (d) Conditional probability distribution of \(X\) given that \(Y=102\) (e) \(P(X<116 \mid Y=102)\)

Step-by-Step Solution

Answer

(a) 6 mL²; (b) X ~ N(120, 25); (c) 0.2119; (d) X|Y=102 ~ N(123, 16); (e) 0.0401.

Step 1: Covariance Calculation

To calculate the covariance between two variables in a bivariate normal distribution, we use the formula: \[ \text{Cov}(X,Y) = \rho \sigma_X \sigma_Y \] Given \( \rho = 0.6 \), \( \sigma_X = 5 \) and \( \sigma_Y = 2 \), the covariance is: \[ \text{Cov}(X,Y) = 0.6 \times 5 \times 2 = 6 \]

Step 2: Finding the Marginal Distribution of X

The marginal distribution of a variable in a bivariate normal distribution is itself a normal distribution. Since we are given that \(X\) is normal, the marginal distribution of \(X\) is:\[ X \sim N(\mu_X, \sigma_X^2) \]Given \(\mu_X = 120\) and \(\sigma_X = 5\), the distribution is: \[ X \sim N(120, 25) \]

Step 3: Calculate Probability P(X < 116)

Given that \(X\sim N(120, 25)\), we need to find \(P(X < 116)\). First, convert 116 to a standard normal variable using:\[ Z = \frac{X - \mu_X}{\sigma_X} = \frac{116 - 120}{5} = -0.8 \]Now, find the probability corresponding to \(Z = -0.8\) using standard normal distribution tables or a calculator. The probability \(P(Z < -0.8)\) is approximately 0.2119.

Step 4: Conditional Distribution of X Given Y = 102

The conditional distribution of \(X\) given \(Y = y\) in a bivariate normal distribution is also normal with mean and variance given by:\[ \mu_{X|Y} = \mu_X + \rho \frac{\sigma_X}{\sigma_Y}(y - \mu_Y) \]\[ \sigma_{X|Y}^2 = \sigma_X^2(1-\rho^2) \]Substitute \(y = 102\), \(\mu_X = 120\), \(\mu_Y = 100\), \(\rho = 0.6\), \(\sigma_X = 5\), and \(\sigma_Y = 2\):Mean:\[ \mu_{X|Y} = 120 + 0.6 \frac{5}{2} (102 - 100) = 120 + 3 = 123 \]Variance:\[ \sigma_{X|Y}^2 = 25(1-0.36) = 25 \times 0.64 = 16 \]So, \(X|Y=102 \sim N(123, 16)\).

Step 5: Calculate Conditional Probability P(X < 116 | Y = 102)

With \(X|Y=102 \sim N(123, 16)\), we need to find \(P(X < 116 | Y = 102)\). First, convert 116 to a standard normal variable using:\[ Z = \frac{116 - 123}{\sqrt{16}} = \frac{-7}{4} = -1.75 \]Look up \(P(Z < -1.75)\) using standard normal distribution tables or a calculator. The probability is approximately 0.0401.

Key Concepts

Covariance Calculation

Covariance is a measure used to express how two random variables change together. When dealing with bivariate normal distributions, it can indicate the degree of linear relationship between the variables. The formula for calculating the covariance between two such variables, \( X \) and \( Y \), is determined by the correlation coefficient \( \rho \) and the standard deviations of \( X \) and \( Y \). The formula is:

• \[ \text{Cov}(X,Y) = \rho \sigma_X \sigma_Y \]

From the given problem, it is known that \( \rho = 0.6 \), \( \sigma_X = 5 \), and \( \sigma_Y = 2 \). After substituting these values, the covariance is calculated as follows:

• \[ \text{Cov}(X,Y) = 0.6 \times 5 \times 2 = 6 \]

This result means that there is a moderate positive association between the variables, suggesting as one increases, the other tends to increase as well.

Marginal Distribution

In probability theory, the marginal distribution is the probability distribution of a subset of a collection of random variables. In a bivariate normal distribution, the marginal distribution of a single variable, like \( X \), is itself a normal distribution. If you have a set of joint random variables, marginal distributions are those distributions obtained by simply removing the other variables from consideration. For our bivariate case here, the marginal distribution of \( X \) is as follows:

• \[ X \sim N(\mu_X, \sigma_X^2) \]

Given the problem parameters \( \mu_X = 120 \) and \( \sigma_X = 5 \), the marginal distribution for \( X \) is concluded to be:

• \[ X \sim N(120, 25) \]

This indicates that \( X \) is distributed normally with a mean of 120 and a variance of 25.

Conditional Probability Distribution

Conditional probability distribution in a bivariate normal distribution considers one variable while we know some value of the other. Specifically, when you want to understand the distribution of \( X \) given a certain value of \( Y \), it is still a normal distribution but modified by the information about \( Y \). The mean and variance of this conditional distribution take into account the relationship between the variables, calculated with the formulas:

• \[ \mu_{X|Y} = \mu_X + \rho \frac{\sigma_X}{\sigma_Y}(y - \mu_Y) \]
• \[ \sigma_{X|Y}^2 = \sigma_X^2(1-\rho^2) \]

Inserting the values from the problem with \( Y = 102 \), the mean is found as:

• \[ \mu_{X|Y} = 120 + 0.6 \frac{5}{2} (102 - 100) = 123 \]

And the variance becomes:

• \[ \sigma_{X|Y}^2 = 25(1-0.36) = 16 \]

Hence, the distribution \( X|Y=102 \sim N(123, 16) \) suggests that, knowing \( Y \), \( X \) behaves differently than its marginal distribution.

Standard Normal Distribution

A standard normal distribution is a normal distribution that has been standardized, meaning it has a mean of 0 and a variance of 1. Standardizing a normal distribution involves transforming the original variable \( X \) into a new variable \( Z \) using the formula:

• \[ Z = \frac{X - \mu}{\sigma} \]

This transformation is crucial for computing probabilities using standard normal distribution tables. For instance, in the problem, you converted \( X < 116 \) to a standard normal variable using:

• \[ Z = \frac{116 - 120}{5} = -0.8 \]

It allows using the tables or a calculator to find \( P(Z < -0.8) \), providing the probability of \( X \) being less than a certain value. Likewise, for the conditional probability, \( X < 116 | Y = 102 \) was transformed:

• \[ Z = \frac{116 - 123}{4} = -1.75 \]

Standard normal distribution helps simplify the calculation by focusing purely on relative distances and probabilities.