To solve for the values related to the exponential random variable \( X \) with parameter \( \lambda = 2 \), we can use the properties of the exponential distribution.
a) \( E[(3X + 1)^2] \)
To find \( E[(3X + 1)^2] \), we first expand the expression:
\[ (3X + 1)^2 = 9X^2 + 6X + 1 \]
Now, we can use the linearity of expectation:
\[ E[(3X + 1)^2] = E[9X^2] + E[6X] + E[1] \]
Since expectations can be factored out:
\[ E[(3X + 1)^2] = 9E[X^2] + 6E[X] + 1 \]
Next, we calculate \( E[X] \) and \( E[X^2] \):
- The mean of an exponential random variable \( E[X] \) is given by \( \frac{1}{\lambda} \):
\[ E[X] = \frac{1}{2} = 0.5 \]
- The second moment \( E[X^2] \) can be calculated using the formula \( E[X^2] = \frac{2}{\lambda^2} \):
\[ E[X^2] = \frac{2}{2^2} = \frac{2}{4} = 0.5 \]
Now substituting these back into the equation for \( E[(3X + 1)^2] \):
\[ E[(3X + 1)^2] = 9(0.5) + 6(0.5) + 1 = 4.5 + 3 + 1 = 8.5 \]
Thus, the answer for part a) is:
\[ \boxed{8.5} \]
b) \( P(1 \leq X \leq 2) \)
To find the probability \( P(1 \leq X \leq 2) \) for an exponential random variable, we can use the cumulative distribution function (CDF):
The CDF for an exponential random variable \( X \) is given by:
\[ F(x) = 1 - e^{-\lambda x} \]
For our case, \( \lambda = 2 \):
- First, calculate \( P(X \leq 2) \):
\[ P(X \leq 2) = F(2) = 1 - e^{-2 \times 2} = 1 - e^{-4} \]
- Next, calculate \( P(X \leq 1) \):
\[ P(X \leq 1) = F(1) = 1 - e^{-2 \times 1} = 1 - e^{-2} \]
- Therefore, the probability \( P(1 \leq X \leq 2) \) is:
\[ P(1 \leq X \leq 2) = P(X \leq 2) - P(X \leq 1) = (1 - e^{-4}) - (1 - e^{-2}) = e^{-2} - e^{-4} \]
Thus, the answer for part b) is:
\[ \boxed{e^{-2} - e^{-4}} \]