Using cumulative distribution function to find probabilities

Question

An investment firm offers its customers municipal bonds that mature after varying numbers of years. Given that the cumulative distribution of T, the number of years to maturity for a randomly selected bond is:

$$F(t) = \begin{cases} 0, & t<1 \\ \frac 14, & 1\le t<3 \\ \frac 12, &3\le t<5 \\ \frac34, & 5\le t<7 \\ 1, &t\ge 7\end{cases}$$

Find $P(T=5)$, $P(T>3)$, and $P(1.4 < T <6)$.

I am unsure how to use the cdf for $P(t=5)$ because it is right between the given ranges. Is the answer $0+1/4+1/2 = 3/4$?

Answer 1

By defintion, the cdf $F$ of a random variable $X$ satisfies $F(t) = P(X\leq t)$. (Based on your post, it sounds like you might have $F$ and $P$ reversed in that equation.) Thus $$P(T = 5) = P(T \leq 5) - P(T < 5) = \frac{3}{4}- \frac{1}{2} = \frac{1}{4},$$ since $P(T \leq 5 - \epsilon) = F(5 - \epsilon) = \frac{1}{2}$ for small $\epsilon > 0$. For the latter two statements, use the fact that $P(X > t) = 1 - P(X\leq t)$ and $P(s < X < t) = P(X < t) - P(X \leq s)$.

Answer 2

(a) $P(T = 5) = F(5) − F(4) = \frac34 − \frac12 = \frac14$

(b) $P(T > 3) = 1 − F(3) = 1 − \frac12 = \frac12$

(c) $P(1.4 < T < 6) = F(6) − F(1.4) = \frac34 − \frac14 = \frac12$