Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
Certainly, let's solve the problem step-by-step:
### Given Problem
Let [tex]\( X_1 \)[/tex] and [tex]\( X_2 \)[/tex] be a random sample of size 2 from a distribution with the probability mass function:
[tex]\[ f(x) = \begin{cases} \frac{1}{4}, & \text{for } x = 1, 2, 3, 4 \\ 0, & \text{elsewhere} \end{cases} \][/tex]
We need to find:
a) The probability distribution of [tex]\( T = X_1 + X_2 \)[/tex].
b) The mean and variance of [tex]\( T = X_1 + X_2 \)[/tex].
c) The probability that [tex]\( T > 3 \)[/tex].
### a) The Probability Distribution of [tex]\( T = X_1 + X_2 \)[/tex]
First, we list all possible values of [tex]\( X_1 \)[/tex] and [tex]\( X_2 \)[/tex]:
[tex]\[ X_1, X_2 \in \{1, 2, 3, 4\} \][/tex]
By summing [tex]\( X_1 \)[/tex] and [tex]\( X_2 \)[/tex], we can find the possible values of [tex]\( T = X_1 + X_2 \)[/tex] and their respective probabilities. The possible values of [tex]\( T \)[/tex] range from [tex]\( 2 \)[/tex] (i.e., [tex]\( 1+1 \)[/tex]) to [tex]\( 8 \)[/tex] (i.e., [tex]\( 4+4 \)[/tex]).
All combinations and their counts:
[tex]\[ \begin{aligned} &E(2): 1+1 &\Rightarrow \text{Probability: } \frac{1}{4} \cdot \frac{1}{4} = 0.0625 \\ &E(3): 1+2, 2+1 &\Rightarrow \text{Probability: } \frac{1}{4} \cdot \frac{1}{4} \times 2 = 0.125 \\ &E(4): 1+3, 2+2, 3+1 &\Rightarrow \text{Probability: } \frac{1}{4} \cdot \frac{1}{4} \times 3 = 0.1875 \\ &E(5): 1+4, 2+3, 3+2, 4+1 &\Rightarrow \text{Probability: } \frac{1}{4} \cdot \frac{1}{4} \times 4 = 0.25 \\ &E(6): 2+4, 3+3, 4+2 &\Rightarrow \text{Probability: } \frac{1}{4} \cdot \frac{1}{4} \times 3 = 0.1875 \\ &E(7): 3+4, 4+3 &\Rightarrow \text{Probability: } \frac{1}{4} \cdot \frac{1}{4} \times 2 = 0.125 \\ &E(8): 4+4 &\Rightarrow \text{Probability: } \frac{1}{4} \cdot \frac{1}{4} = 0.0625 \\ \end{aligned} \][/tex]
So the probability distribution of [tex]\( T \)[/tex] is:
[tex]\[ \{2: 0.0625, 3: 0.125, 4: 0.1875, 5: 0.25, 6: 0.1875, 7: 0.125, 8: 0.0625\} \][/tex]
### b) The Mean and Variance of [tex]\( T = X_1 + X_2 \)[/tex]
#### Mean of [tex]\( T \)[/tex]:
The mean (expected value) [tex]\( E(T) \)[/tex] can be calculated as:
[tex]\[ E(T) = \sum_{t=2}^{8} t \cdot P(T=t) \][/tex]
Substituting the values:
[tex]\[ E(T) = 2 \cdot 0.0625 + 3 \cdot 0.125 + 4 \cdot 0.1875 + 5 \cdot 0.25 + 6 \cdot 0.1875 + 7 \cdot 0.125 + 8 \cdot 0.0625 = 5.0 \][/tex]
#### Variance of [tex]\( T \)[/tex]:
First, compute [tex]\( E(T^2) \)[/tex]:
[tex]\[ E(T^2) = \sum_{t=2}^{8} t^2 \cdot P(T=t) \][/tex]
Substituting the values:
[tex]\[ E(T^2) = 2^2 \cdot 0.0625 + 3^2 \cdot 0.125 + 4^2 \cdot 0.1875 + 5^2 \cdot 0.25 + 6^2 \cdot 0.1875 + 7^2 \cdot 0.125 + 8^2 \cdot 0.0625 = 27.5 \][/tex]
Now, use the formula for variance:
[tex]\[ \text{Var}(T) = E(T^2) - \left(E(T)\right)^2 = 27.5 - 5^2 = 27.5 - 25 = 2.5 \][/tex]
So, the mean and variance of [tex]\( T \)[/tex] are:
[tex]\[ \text{Mean} = 5.0, \quad \text{Variance} = 2.5 \][/tex]
### c) The Probability that [tex]\( T > 3 \)[/tex]
To find [tex]\( P(T > 3) \)[/tex]:
[tex]\[ P(T > 3) = P(T=4) + P(T=5) + P(T=6) + P(T=7) + P(T=8) \][/tex]
Substituting the values:
[tex]\[ P(T > 3) = 0.1875 + 0.25 + 0.1875 + 0.125 + 0.0625 = 0.8125 \][/tex]
So, the probability that [tex]\( T > 3 \)[/tex] is:
[tex]\[ P(T > 3) = 0.8125 \][/tex]
### Summary
a) The probability distribution of [tex]\( T \)[/tex] is:
[tex]\[ \{2: 0.0625, 3: 0.125, 4: 0.1875, 5: 0.25, 6: 0.1875, 7: 0.125, 8: 0.0625\} \][/tex]
b) The mean and variance of [tex]\( T \)[/tex] are:
[tex]\[ \text{Mean} = 5.0, \quad \text{Variance} = 2.5 \][/tex]
c) The probability that [tex]\( T > 3 \)[/tex] is:
[tex]\[ P(T > 3) = 0.8125 \][/tex]
### Given Problem
Let [tex]\( X_1 \)[/tex] and [tex]\( X_2 \)[/tex] be a random sample of size 2 from a distribution with the probability mass function:
[tex]\[ f(x) = \begin{cases} \frac{1}{4}, & \text{for } x = 1, 2, 3, 4 \\ 0, & \text{elsewhere} \end{cases} \][/tex]
We need to find:
a) The probability distribution of [tex]\( T = X_1 + X_2 \)[/tex].
b) The mean and variance of [tex]\( T = X_1 + X_2 \)[/tex].
c) The probability that [tex]\( T > 3 \)[/tex].
### a) The Probability Distribution of [tex]\( T = X_1 + X_2 \)[/tex]
First, we list all possible values of [tex]\( X_1 \)[/tex] and [tex]\( X_2 \)[/tex]:
[tex]\[ X_1, X_2 \in \{1, 2, 3, 4\} \][/tex]
By summing [tex]\( X_1 \)[/tex] and [tex]\( X_2 \)[/tex], we can find the possible values of [tex]\( T = X_1 + X_2 \)[/tex] and their respective probabilities. The possible values of [tex]\( T \)[/tex] range from [tex]\( 2 \)[/tex] (i.e., [tex]\( 1+1 \)[/tex]) to [tex]\( 8 \)[/tex] (i.e., [tex]\( 4+4 \)[/tex]).
All combinations and their counts:
[tex]\[ \begin{aligned} &E(2): 1+1 &\Rightarrow \text{Probability: } \frac{1}{4} \cdot \frac{1}{4} = 0.0625 \\ &E(3): 1+2, 2+1 &\Rightarrow \text{Probability: } \frac{1}{4} \cdot \frac{1}{4} \times 2 = 0.125 \\ &E(4): 1+3, 2+2, 3+1 &\Rightarrow \text{Probability: } \frac{1}{4} \cdot \frac{1}{4} \times 3 = 0.1875 \\ &E(5): 1+4, 2+3, 3+2, 4+1 &\Rightarrow \text{Probability: } \frac{1}{4} \cdot \frac{1}{4} \times 4 = 0.25 \\ &E(6): 2+4, 3+3, 4+2 &\Rightarrow \text{Probability: } \frac{1}{4} \cdot \frac{1}{4} \times 3 = 0.1875 \\ &E(7): 3+4, 4+3 &\Rightarrow \text{Probability: } \frac{1}{4} \cdot \frac{1}{4} \times 2 = 0.125 \\ &E(8): 4+4 &\Rightarrow \text{Probability: } \frac{1}{4} \cdot \frac{1}{4} = 0.0625 \\ \end{aligned} \][/tex]
So the probability distribution of [tex]\( T \)[/tex] is:
[tex]\[ \{2: 0.0625, 3: 0.125, 4: 0.1875, 5: 0.25, 6: 0.1875, 7: 0.125, 8: 0.0625\} \][/tex]
### b) The Mean and Variance of [tex]\( T = X_1 + X_2 \)[/tex]
#### Mean of [tex]\( T \)[/tex]:
The mean (expected value) [tex]\( E(T) \)[/tex] can be calculated as:
[tex]\[ E(T) = \sum_{t=2}^{8} t \cdot P(T=t) \][/tex]
Substituting the values:
[tex]\[ E(T) = 2 \cdot 0.0625 + 3 \cdot 0.125 + 4 \cdot 0.1875 + 5 \cdot 0.25 + 6 \cdot 0.1875 + 7 \cdot 0.125 + 8 \cdot 0.0625 = 5.0 \][/tex]
#### Variance of [tex]\( T \)[/tex]:
First, compute [tex]\( E(T^2) \)[/tex]:
[tex]\[ E(T^2) = \sum_{t=2}^{8} t^2 \cdot P(T=t) \][/tex]
Substituting the values:
[tex]\[ E(T^2) = 2^2 \cdot 0.0625 + 3^2 \cdot 0.125 + 4^2 \cdot 0.1875 + 5^2 \cdot 0.25 + 6^2 \cdot 0.1875 + 7^2 \cdot 0.125 + 8^2 \cdot 0.0625 = 27.5 \][/tex]
Now, use the formula for variance:
[tex]\[ \text{Var}(T) = E(T^2) - \left(E(T)\right)^2 = 27.5 - 5^2 = 27.5 - 25 = 2.5 \][/tex]
So, the mean and variance of [tex]\( T \)[/tex] are:
[tex]\[ \text{Mean} = 5.0, \quad \text{Variance} = 2.5 \][/tex]
### c) The Probability that [tex]\( T > 3 \)[/tex]
To find [tex]\( P(T > 3) \)[/tex]:
[tex]\[ P(T > 3) = P(T=4) + P(T=5) + P(T=6) + P(T=7) + P(T=8) \][/tex]
Substituting the values:
[tex]\[ P(T > 3) = 0.1875 + 0.25 + 0.1875 + 0.125 + 0.0625 = 0.8125 \][/tex]
So, the probability that [tex]\( T > 3 \)[/tex] is:
[tex]\[ P(T > 3) = 0.8125 \][/tex]
### Summary
a) The probability distribution of [tex]\( T \)[/tex] is:
[tex]\[ \{2: 0.0625, 3: 0.125, 4: 0.1875, 5: 0.25, 6: 0.1875, 7: 0.125, 8: 0.0625\} \][/tex]
b) The mean and variance of [tex]\( T \)[/tex] are:
[tex]\[ \text{Mean} = 5.0, \quad \text{Variance} = 2.5 \][/tex]
c) The probability that [tex]\( T > 3 \)[/tex] is:
[tex]\[ P(T > 3) = 0.8125 \][/tex]
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
