Problem 49 - Applied Statistics and Probability for Engineers

Question

Based on the number of voids, a ferrite slab is classified as either high, medium, or low. Historically, \(5 \%\) of the slabs are classified as high, \(85 \%\) as medium, and \(10 \%\) as low. A sample of 20 slabs is selected for testing. Let \(X, Y,\) and \(Z\) denote the number of slabs that are independently classified as high, medium, and low, respectively. (a) What are the name and the values of the parameters of the joint probability distribution of \(X, Y,\) and \(Z ?\) (b) What is the range of the joint probability distribution of \(X\), \(Y,\) and \(Z ?\) (c) What are the name and the values of the parameters of the marginal probability distribution of \(X ?\) (d) Determine \(E(X)\) and \(V(X)\). Determine the following: (e) \(P(X=1, Y=17, Z=3)\) (f) \(P(X \leq 1, Y=17, Z=3)\) (g) \(P(X \leq 1)\) (h) \(E(Y)\) (i) \(P(X=2, Z=3 \mid Y=17)\) (j) \(P(X=2 \mid Y=17)\) (k) \(E(X \mid Y=17)\)

Step-by-Step Solution

Answer

(a) Multinomial with parameters \( n=20, p_1=0.05, p_2=0.85, p_3=0.10 \). (b) \( X+Y+Z=20 \) with \( X, Y, Z \geq 0 \). (c) Binomial \( n=20, p=0.05 \). (d) \( E(X)=1, V(X)=0.95 \). (e) Use multinomial PMF. (f) Sum of joint probabilities. (g) Use Binomial CDF. (h) 17. (i) Ratio of joint probabilities. (j) Ratio of probabilities for \( X \) and \( Y \). (k) Based on conditional configuration.

Step 1 - Determine Distribution Name and Parameters

Since the classification probabilities are given and a fixed number of slabs (20) is selected, the counts \( X \), \( Y \), and \( Z \) follow a Multinomial distribution. The parameters for this distribution are \( n=20 \) (number of trials) and \( p_1=0.05 \), \( p_2=0.85 \), \( p_3=0.10 \), which are the probabilities of a slab being high, medium, and low, respectively.

Step 2 - Determine the Range of Distribution

The distribution needs to satisfy \( X+Y+Z=20 \) with \( X, Y, Z \geq 0 \). Thus, \( X \) can range from 0 to 20, \( Y \) from 0 to 20, and \( Z \) from 0 to 20 as long as these sums up to 20.

Step 3 - Determine Marginal Distribution of X

The marginal distribution of \( X \) is a Binomial distribution because it counts the number of successes (high classification) out of 20 trials. Hence, it is \( X \sim \text{Binomial}(n=20, p=0.05) \).

Step 4 - Calculate Expectation and Variance of X

The expected value and variance for a Binomial distribution \( X \sim \text{Binomial}(n,p) \) are \( E(X) = np \) and \( V(X) = np(1-p) \). For \( X \), \( E(X) = 20 \times 0.05 = 1 \) and \( V(X) = 20 \times 0.05 \times 0.95 = 0.95 \).

Step 5 - Calculate Joint Probability P(X=1, Y=17, Z=3)

The joint probability mass function for multinomial distribution is \( P(X=x, Y=y, Z=z) = \frac{20!}{x!y!z!} p_1^x p_2^y p_3^z \). For \( X=1, Y=17, Z=2 \), \( \frac{20!}{1!17!2!} \times 0.05^1 \times 0.85^{17} \times 0.10^2 \). Calculate this value to find the probability.

Step 6 - Cumulative Probability P(X ≤ 1, Y=17, Z=3)

This is the sum of probabilities for two scenarios: \( P(X = 0, Y = 17, Z = 3) \) and \( P(X = 1, Y = 17, Z = 3) \). Compute separately and sum them together.

Step 7 - Cumulative Probability P(X ≤ 1)

\( P(X \leq 1) \) refers to the probability of either 0 or 1 was classified as high. Use the binomial distribution and sum \( P(X=0) + P(X=1) \) where each term is obtained from \( P(X=x) = \binom{n}{x} p^x (1-p)^{n-x} \) with parameters \( n=20, p=0.05 \).

Step 8 - Calculate Expectation E(Y)

Since \( Y \) is marginally Binomial with parameters \( n=20, p=0.85 \), its expectation is calculated as \( E(Y) = 20 \times 0.85 = 17 \).

Step 9 - Conditional Probability P(X=2, Z=3 | Y=17)

Given \( Y=17 \), find \( P(X=2, Z=3 | Y=17) = P(X=2, Z=3, Y=17)/P(Y=17) \). Compute the numerator using the multinomial PMF and \( P(Y=17) \) is found by considering all \( X+Z=3 \).

Step 10 - Conditional Probability P(X=2 | Y=17)

Use the relation \( P(X=2 | Y=17) = \frac{P(X=2, Y=17)}{P(Y=17)} \). Obtain each term similar to benchmarks set in Step 9.

Step 11 - Conditional Expectation E(X | Y = 17)

Given \( Y = 17 \), find the expected value \( E(X | Y = 17) \) based on configurations \( Z \) can take (1 to 3) using conditional probabilities of \( X \) given specific \( Z \) counts.

Key Concepts

Joint Probability Distribution

In probability and statistics, a joint probability distribution provides us with the probability of different outcomes for multiple random variables occurring at the same time. When dealing with multiple discrete random variables like the number of slabs classified as high, medium, or low, as in our exercise, the multinomial distribution is particularly valuable. This is because it deals with scenarios where there are more than two categories. The exercise uses a multinomial distribution to describe how a sample of 20 slabs is classified. Here's why:

• You have a fixed number of trials, which in this case refers to the 20 slabs.
• Each slab can be classified into one of three categories - high, medium, or low.
• The events are independent, meaning the classification of one slab does not affect the others.

Thus, the joint probability of exactly X high, Y medium, and Z low classifications is calculated using the multinomial formula. This joint probability considers all possible outcomes that can result in the exact count of slabs in each category, summing probabilities where the sum of the probabilities equals one. With the multinomial distribution, calculating joint probabilities involves factorials and the category probabilities raised to the power of their counts. Understanding the joint probability distribution helps us know not just individual event probabilities, but also how they can occur together in a multi-variable scenario.

Binomial Distribution

The binomial distribution is a special case of the multinomial distribution where there are only two categories: success or failure. In our context, when we focus on one classification, like the number of high slabs out of 20, it fits the binomial model. A binomial distribution is characterized by:

• A fixed number of trials (in the exercise, it's 20 slabs).
• Each trial has two possible outcomes, which in a broader context are generally termed as "success" or "failure".
• The probability of "success" is constant for each trial.

For instance, the number of slabs classified as high is a binomial distribution described as Binomial(n=20, p=0.05). Here, "high" is considered the success term, making its binomial parameters easy to compute: - **Expectation**: Given by E(X) = np, representing the average number of high slabs expected. - **Variance**: V(X) = np(1-p), showing how much the actual count may vary around the expectation. With these parameters, students can ascertain the likelihood of different outcomes. Understanding the binomial distribution offers a fundamental insight into probability scenarios with dichotomous outcomes, making it a foundational concept in probability theory.

Conditional Probability

Conditional probability refers to the probability of an event occurring given that another event has already occurred. In situations where outcomes are not independent, like in our slab classification problem, conditional probabilities are instrumental in updating our beliefs about one event based on what we know about another.To explore this, the exercise employs conditional probability to answer questions like: "What is the probability of having 2 high and 3 low slabs given we already know there are 17 medium slabs?"This is denoted as \( P(X=2, Z=3 \mid Y=17) \) and calculated using:\[P(A \mid B) = \frac{P(A \cap B)}{P(B)}\]Where:- \( P(X=2, Z=3, Y=17) \) is the joint probability of 2 high, 3 low, and 17 medium slabs.- \( P(Y=17) \) is the probability of having 17 medium slabs, found by summing over all combinations where X+Z sums to 3.Conditional probability allows you to compute likelihoods in dependent scenarios, enabling more accurate predictions and enhancing understanding of interconnected probabilistic events.

Expectation and Variance

Expectation and variance are key concepts in probability and statistics that give us insights into the behavior of random variables:- **Expectation (or Expected Value)** is the average value expected from a random variable. It is equivalent to the weighted average of all possible outcomes. In our slab classification scenario, the expectation can be perceived as the average number of slabs we expect to classify as high, medium, or low. When dealing with binomial or multinomial random variables like these, expectation is calculated using: - For a binomial distribution: \( E(X) = np \) - For a multinomial outcome, expectation for each category aligns similarly.- **Variance** measures the spread of the random variable outcomes, quantifying the deviation from the expected value. A low variance indicates that the results are close to the expected value, while high variance shows more spread. - For the binomial distribution, variance is \( V(X) = np(1-p) \). - In the multinomial situation, each category's variance can be individually computed similarly.Understanding expectation and variance helps predict how outcomes will behave on average and whether they will fluctuate significantly. This can influence decisions like sample size and classification thresholds, proving crucial for statistical inference and real-world applications.