Questions: Question If X ~ U(4.5,18.5) is a continuous uniform random variable, what is P(X<11) ? Select the correct answer below: 1/28 3/28 13/28 9/14 19/28

Solution Steps

The mean $E(X)$ of a uniform distribution $U(a, b)$ is given by the formula:

$$E(X) = \frac{a + b}{2}$$

Substituting the values $a = 4.5$ and $b = 18.5$:

$$E(X) = \frac{4.5 + 18.5}{2} = \frac{23}{2} = 11.5$$

The variance $\text{Var}(X)$ of a uniform distribution is calculated using the formula:

$$\text{Var}(X) = \frac{(b - a)^2}{12}$$

Substituting the values:

$$\text{Var}(X) = \frac{(18.5 - 4.5)^2}{12} = \frac{(14)^2}{12} = \frac{196}{12} = 16.3333$$

The standard deviation $\sigma(X)$ is the square root of the variance:

$$\sigma(X) = \sqrt{\text{Var}(X)} = \sqrt{16.3333} \approx 4.0415$$

The cumulative distribution function $F(x; a, b)$ for a uniform distribution is given by:

$$F(x; a, b) = \frac{x - a}{b - a}, \quad a \leq x \leq b$$

To find $P(X < 11)$, we calculate:

$$P(4.5 \leq X \leq 11) = F(11) - F(4.5)$$

Calculating $F(11)$:

$$F(11) = \frac{11 - 4.5}{18.5 - 4.5} = \frac{6.5}{14} \approx 0.4643$$

And since $F(4.5) = 0$:

$$P(4.5 \leq X \leq 11) = 0.4643 - 0 = 0.4643$$

Final Answer