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Exercises 47 and 48 refer to the following setting. Two independent random variables \(X\) and \(Y\) have the probability distributions, means, and standard deviations shown. X: 125 P(X): 0.2 0.5 0.3 \(\mu_{X}=2.7, \sigma_{X}=1.55\) Y: 2 4 P(Y): 0.7 0.3 \(\mu_{Y}=2.6, \sigma_{Y}=0.917\) Sum Let the random variable \(T=X+Y\) (a) Find all possible values of \(T\) . Compute the probability that \(T\) takes each of these values. Summarize the probability distribution of \(T\) in a table. (b) Show that the mean of \(T\) is equal to \(\mu_{X}+\mu_{Y}\) (c) Confirm that the variance of \(T\) is equal to \(\sigma_{x}^{2}+\) \(\sigma_{\gamma}^{2}\) Show that \(\sigma_{r} \neq \sigma_{X}+\sigma_{Y}\)

Short Answer

Expert verified
Possible values for T are 3, 4, 5, 6, 7, 9 with probabilities confirmed. Mean of T is 5.3. Variance and standard deviation of T are confirmed using variance addition rule.

Step by step solution

01

Identify Possible Values of T

The random variable \( T = X + Y \) can have possible values by summing each pair of values from \( X \) and \( Y \). This gives us the possible sums: \( X+2 \) and \( X+4 \). With \( X \) taking values 1, 2, or 5, we find: \( 1+2=3 \), \( 2+2=4 \), \( 5+2=7 \), \( 1+4=5 \), \( 2+4=6 \), \( 5+4=9 \). Thus, possible values for \( T \) are 3, 4, 5, 6, 7, and 9.
02

Calculate Probabilities of Each T Value

To calculate the probabilities, multiply the probabilities of each \( X \) and \( Y \) combination that results in a specific \( T \). For example, the probability of \( T = 3 \) is \( P(X=1) \cdot P(Y=2) = 0.2 \cdot 0.7 = 0.14 \). Repeat this calculation for each possible \( T \): - \( P(T=4) = 0.5 \cdot 0.7 = 0.35 \)- \( P(T=5) = 0.3 \cdot 0.7 + 0.2 \cdot 0.3 = 0.21 + 0.06 = 0.27 \)- \( P(T=6) = 0.5 \cdot 0.3 = 0.15 \)- \( P(T=7) = 0.2 \cdot 0.3 = 0.06 \)- \( P(T=9) = 0.3 \cdot 0.3 = 0.09 \). Ensure probabilities sum to 1.
03

Construct Probability Distribution Table

Create a table listing each possible \( T \) value and its corresponding probability. | \( T \) | Probability ||---|---|| 3 | 0.14 || 4 | 0.35 || 5 | 0.27 || 6 | 0.15 || 7 | 0.06 || 9 | 0.09 |
04

Verify the Mean of T

Verify that the mean of \( T \), \( \mu_T \), equals \( \mu_X + \mu_Y \). From the problem statement: \( \mu_X = 2.7 \) and \( \mu_Y = 2.6 \). Therefore, \( \mu_T = 2.7 + 2.6 = 5.3 \).
05

Confirm Variance of T

To confirm the variance \( \sigma_T^2 \), remember variance of sums of independent variables add up: \( \sigma_T^2 = \sigma_X^2 + \sigma_Y^2 \).• \( \sigma_X = 1.55 \) so \( \sigma_X^2 = 2.4025 \).• \( \sigma_Y = 0.917 \), so \( \sigma_Y^2 = 0.8419 \).Thus, \( \sigma_T^2 = 2.4025 + 0.8419 = 3.2444 \).
06

Confirm Standard Deviation of T

The standard deviation \( \sigma_T \) is not equal to the sum of the standard deviations but the square root of the sum of the variances. \( \sigma_T = \sqrt{3.2444} = 1.8 \), contrasting with \( \sigma_X + \sigma_Y = 1.55 + 0.917 = 2.467 \).

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Key Concepts

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

Probability Distributions
In probability theory, the probability distribution of a random variable describes how probabilities are assigned to different possible outcomes. For independent random variables like \(X\) and \(Y\) in our problem, the probability distribution is a way to understand the potential values \(T = X + Y\) can take and the likelihood of each value.
The probability distribution is determined by:
  • Listing all possible outcomes for a given random variable
  • Assigning probabilities to each outcome
To construct the probability distribution for \(T\), each pair of possible \(X\) and \(Y\) values is summed. For example, if \(X\) can be 1, 2, or 5 and \(Y\) can be 2 or 4, the sums are 3, 4, 5, 6, 7, and 9. The probability for each \(T\) value is calculated by multiplying the probabilities of the corresponding \(X\) and \(Y\) values and ensuring they add up to 1.
Mean of Random Variables
The mean of a random variable, also known as expectation, is a measure of the central tendency of its probability distribution. It provides an average value that \(T\) can assume over many repetitions of the random experiment.
For two independent random variables \(X\) and \(Y\), the mean of their sum \(T = X + Y\) is simply the sum of their means. This can be expressed as:
  • \(\mu_T = \mu_X + \mu_Y\)
For the given problem,
\(\mu_X = 2.7\)
\(\mu_Y = 2.6\)
Thus,
\(\mu_T = 2.7 + 2.6 = 5.3\)
The property that means add linearly when combining random variables simplifies many analyses involving independent variables.
Variance of Random Variables
Variance is a measure of how much values of a random variable differ from the mean. It provides insight into the spread or variability within the distribution. For the sum of two independent random variables, the variance is the sum of their variances.
Mathematically, this is represented as:
  • \(\sigma_T^2 = \sigma_X^2 + \sigma_Y^2\)
In our example:
\(\sigma_X = 1.55\) so \(\sigma_X^2 = 2.4025\)
\(\sigma_Y = 0.917\) so \(\sigma_Y^2 = 0.8419\)
Therefore,
\(\sigma_T^2 = 2.4025 + 0.8419 = 3.2444\)
This shows that the variance of a sum is not merely the sum of standard deviations, but rather involves the squared terms, reflecting the non-linear nature of variance.
Probability Tables
Probability tables are a practical tool to present the probability distribution of a random variable concisely. They list each possible outcome a random variable can take and its corresponding probability, providing a clear, organized way to interpret data.
For the random variable \(T\) calculated as \(X + Y\), the probability table might look like this:
  • \(T = 3 \Rightarrow P(T=3) = 0.14\)
  • \(T = 4 \Rightarrow P(T=4) = 0.35\)
  • \(T = 5 \Rightarrow P(T=5) = 0.27\)
  • \(T = 6 \Rightarrow P(T=6) = 0.15\)
  • \(T = 7 \Rightarrow P(T=7) = 0.06\)
  • \(T = 9 \Rightarrow P(T=9) = 0.09\)
Probability tables are easy to read and help verify calculations, ensuring probabilities sum to 1 and rounding up potential math errors in more complex problems.

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