Question

Let \( X \) be a random variable with distribution function \( F \), such that \[ \lim_{h \to 0^-} F(3 + h) = \frac{1}{4} \quad {and} \quad F(3) = \frac{3}{4}. \] {Then } \( 16 \, {Pr}(X = 3) \) equals _________ (answer in integer).

Hint:

The probability that a random variable takes a specific value can be computed as the difference between the value of the cumulative distribution function at that point and the limit from the left.

Answer 1

We know that the probability of \( X \) taking a specific value is given by: \[ {Pr}(X = 3) = F(3) - \lim_{h \to 0^-} F(3 + h). \] Substitute the given values: \[ {Pr}(X = 3) = \frac{3}{4} - \frac{1}{4} = \frac{1}{2}. \] Now, we compute \( 16 \times {Pr}(X = 3) \): \[ 16 \times {Pr}(X = 3) = 16 \times \frac{1}{2} = 8. \] Thus, the value of \( 16 \, {Pr}(X = 3) \) is \( \boxed{8} \).