Let \( X \) be a uniform random variable on the interval \([1, 3]\).
(a) P(a <= X <= b)
The probability that \( X \) is between \( a \) and \( b \) for a uniform distribution can be calculated using the formula for the probability of the event occurring within the bounds of the distribution:
\[ P(a \leq X \leq b) = \frac{\text{Length of the interval } [a, b]}{\text{Length of the interval } [1, 3]} \]
The length of the interval \([1, 3]\) is:
\[ 3 - 1 = 2 \]
The length of the interval \([a, b]\) is:
\[ b - a \]
Thus, the probability is:
\[ P(a \leq X \leq b) = \frac{b - a}{2} \]
(b) E[X]
The expected value \( E[X] \) for a uniform distribution on the interval \([1, 3]\) is calculated as the midpoint of the interval:
\[ E[X] = \frac{1 + 3}{2} = 2 \]
(c) E[X³]
To find \( E[X^3] \) for a uniform distribution on the interval \([1, 3]\), we use the formula:
\[ E[X^n] = \frac{1}{b-a} \int_a^b x^n , dx \]
Here, \( n = 3 \), \( a = 1 \), and \( b = 3 \). So, we calculate:
\[ E[X^3] = \frac{1}{3-1} \int_1^3 x^3 , dx = \frac{1}{2} \int_1^3 x^3 , dx \]
First, we evaluate the integral:
\[ \int x^3 , dx = \frac{x^4}{4} \]
Now, we compute the definite integral from 1 to 3:
\[ \int_1^3 x^3 , dx = \left[ \frac{x^4}{4} \right]_1^3 = \frac{3^4}{4} - \frac{1^4}{4} = \frac{81}{4} - \frac{1}{4} = \frac{80}{4} = 20 \]
Then substituting back into the expression for \( E[X^3] \):
\[ E[X^3] = \frac{1}{2} \cdot 20 = 10 \]
Summary of Answers:
(a) \( P(a \leq X \leq b) = \frac{b - a}{2} \)
(b) \( E[X] = 2 \)
(c) \( E[X^3] = 10 \)