Question

Let X be uniform on the interval [1, 3] Suppose that 1 < a < b < 3 Then,

(a) P(a <= X <= b) =

(Your answer to part (a) should be an algebraic expression involving a and b)

(b) E[X] =

(c) EX³] =

Answer 1

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) \( E[X^3] = 10 \)