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A box contains four slips of paper marked \(1,2,3,\) and 4. Two slips are selected without replacement. List the possible values for each of the following random variables: a. \(x=\) sum of the two numbers b. \(y=\) difference between the first and second numbers (first - second) c. \(z=\) number of slips selected that show an even number d. \(w=\) number of slips selected that show a 4

Short Answer

Expert verified
So, the possible values for random variables \(x, y, z\), and \(w\) are: \(x\) = {3, 4, 5, 6, 7}, \(y\) = {-1, -2, -3, 1, 2, 3}, \(z\) = {0, 1, 2}, and \(w\) = {0, 1} respectively. This is also considering the values in their absolute values ignoring signs where necessary (like \(y\)).

Step by step solution

01

Preparing all possible 2 slip combinations

In this scenario, 2 slips can be chosen from the set {1, 2, 3, and 4} without replacement. The possible combinations for 2 slips are: (1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), and (4, 3).
02

Determining values for random variable \(x\)

Given that \(x\) is the sum of the two numbers on the selected slips, we calculate the sum for each possible combination. Thus, the possible values for \(x\) are: 3, 4, 5, 3, 5, 6, 4, 5, 7, 5, 6, and 7.
03

Determining values for random variable \(y\)

Given that \(y\) is the difference between the first and the second numbers (first second), we calculate the difference for each possible pair. So, the possible values for \(y\) are: -1, -2, -3, 1, -1, -2, 2, 1, -1, 3, 2, and 1.
04

Determining values for random variable \(z\)

Given that \(z\) is the number of slips selected that show an even number, we count the number of even numbers in each pair. Thus, the possible values for \(z\) are: 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 2, and 2.
05

Determining values for random variable \(w\)

Given that \(w\) is the number of slips selected that show a 4, we count the number of 4's in each pair. So, the possible values for \(w\) are: 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, and 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Variables in Probability
When learning about probability, it's essential to understand what random variables are. A random variable is a numerical description of the outcome of a statistical experiment. In our exercise, each slip of paper chosen from the box represents a possible outcome and defining variables like x, y, z, and w to describe certain aspects of these outcomes (like their sum or their difference) is a way of assigning numerical values to these random and uncertain events.

In probability, random variables can be discrete or continuous. When dealing with discrete random variables, like in our exercise, we're looking at a countable number of distinct outcomes. The possible values for the random variables are determined by analyzing all possible outcomes - such as the sum of two numbers, the difference between the numbers, the count of even numbers, or the presence of a specific number, like 4.
Sum of Two Numbers in Probability
The sum of two numbers is a common random variable in probability problems involving more than one stage or selection. In our example, we're dealing with the sum of two numbers on slips of paper selected without replacement. To find all possible sums (x), we simply add the numbers on each pair of slips.

This action is crucial since it influences the probability of different sums occurring. For instance, some sums will appear more frequently depending on the number of combinations that result in that particular sum, thereby affecting their probabilities. Understanding these possible sums is not only important for direct questions but also for interpreting other statistical measures, like the expected value of the sum.
Difference Between Numbers in Probability
To explore the concept of the difference between numbers in probability, like the random variable y in our problem, it's important to analyze all possible pairs' outcomes, keeping the order in mind. It's because the difference is not commutative - the order of subtraction matters.

Calculating the difference by subtracting the second number from the first in each pair gives us positive, negative, or zero values. This variable relates to the situations where the order of selection changes the outcome, such as when tracking the changes in scores between two events or measuring the lead in a race. The different outcomes again affect the overall probability of these differences occurring.
Counting Even Numbers in Probability
Counting particular types of numbers, such as the even numbers in a set, is another interesting aspect of probability. When asked about the random variable z, which represents the number of slips showing an even number, we simply identify each pair where an even number appears.

Identifying such patterns in outcomes can help in predicting probabilities in more complex scenarios and is a common method used in statistical problem-solving. Even numbers have special properties that are often utilized in probability games and experiments, such as having certain symmetrical relationships with odd numbers.
Probability of Choosing a Specific Number
Lastly, understanding the probability of choosing a specific number is crucial for many probability puzzles and real-life forecasting problems. In our context, the random variable w signifies the occurrence of the number 4. To solve this, we note down the frequency of the number 4 appearing in all the possible pairs.

This approach can be extended to any specific number or event in various situations – from drawing a particular card from a deck to selecting a winning lottery number. Knowing how to calculate these probabilities forms the basis for studying more complex probability distributions and is fundamental for anyone delving into the field of statistics.

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Most popular questions from this chapter

Consider the following 10 observations on the lifetime (in hours) for a certain type of power supply: $$ \begin{array}{lllll} 152.7 & 172.0 & 172.5 & 173.3 & 193.0 \\ 204.7 & 216.5 & 234.9 & 262.6 & 422.6 \end{array} $$ Construct a normal probability plot, and comment on whether it is reasonable to think that the distribution of power supply lifetime is approximately normal. (The normal scores for a sample of size 10 are -1.539,-1.001,-0.656 , \(-0.376,-0.123,0.123,0.376,0.656,1.001,\) and \(1.539 .)\)

Example 6.14 gave the probability distributions shown below for \(x=\) number of flaws in a randomly selected glass panel from Supplier 1 \(y=\) number of flaws in a randomly selected glass panel from Supplier 2 for two suppliers of glass used in the manufacture of flat screen TVs. If the manufacturer wanted to select a single supplier for glass panels, which of these two suppliers would you recommend? Justify your choice based on consideration of both center and variability. $$ \begin{array}{lcccclcccc} \boldsymbol{x} & 0 & 1 & 2 & 3 & \boldsymbol{y} & 0 & 1 & 2 & 3 \\ \boldsymbol{p}(\boldsymbol{x}) & 0.4 & 0.3 & 0.2 & 0.1 & \boldsymbol{p}(\boldsymbol{y}) & 0.2 & 0.6 & 0.2 & 0 \end{array} $$

Let \(x\) denote the IQ of an individual selected at random from a certain population. The value of \(x\) must be a whole number. Suppose that the distribution of \(x\) can be approximated by a normal distribution with mean value 100 and standard deviation 15. Approximate the following probabilities. (Hint: See Example 6.32 ) a. \(P(x=100)\) b. \(P(x \leq 110)\) c. \(P(x<110)\) (Hint: \(x<110\) is the same as \(x \leq 109\).) d. \(P(75 \leq x \leq 125)\)

Suppose \(x=\) the number of courses a randomly selected student at a certain university is taking. The probability distribution of \(x\) appears in the following table: $$ \begin{array}{lccccccc} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ p(x) & 0.02 & 0.03 & 0.09 & 0.25 & 0.40 & 0.16 & 0.05 \end{array} $$ a. What is \(P(x=4)\) ? b. What is \(P(x \leq 4)\) ? c. What is the probability that the selected student is taking at most five courses? d. What is the probability that the selected student is taking at least five courses? More than five courses? e. Calculate \(P(3 \leq x \leq 6)\) and \(P(3 < x < 6)\). Explain in words why these two probabilities are different.

An appliance dealer sells three different models of freezers having \(13.5,15.9,\) and 19.1 cubic feet of storage space. Let \(x=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(x\) has the following probability distribution: $$ \begin{array}{lrrr} x & 13.5 & 15.9 & 19.1 \\ p(x) & 0.2 & 0.5 & 0.3 \end{array} $$ a. Calculate the mean and standard deviation of \(x\). (Hint: See Example 6.15\()\) b. Give an interpretation of the mean and standard deviation of \(x\) in the context of observing the outcomes of many purchases.

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