Answer :
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]
