Unusual Maximum function Came across this in my notes and not sure how to interpret it; Any help would be greatly appreciated. 
X       Probability
-7          0.04
5.5         0.96

t = max{x: P(X<x) <= 0.05}

The notes state the following: 
P(X<-7)=0 and P(X<5.5)=0.04. Therefore t=5.5

Can anybody help me with how this was evaluated?
 A: $X$ only take two values with positive probabilities, which are $-7$ and $5.5$.
For any values that are less than $-7$, the probability of $X$ taking such a value is $0$ as both values that take positive probability is bigger than that. 
For values between $-7$ (inclusive) and $5.5$ exclusive, $P(X<x)=P(X=-7).$
$$P(X < x) = \begin{cases} 0 & , x\le  -7 \\
 0.04 &, -7 < x \le 5.5 \\
1 &, x > 5.5\end{cases}$$
$t$ is the biggest value of $x$ such that $P(X<x) \le 0.5$, from the computation above, such biggest value satisfies $P(X<t)=0.04$, and $t$ can be chosen to be $5.5$.
A: You can write down the cdf:
$$F_X(x)=\begin{cases}0, \ x<-7 \\ 0.04, -7\leq x <5.5 \\ 1, \ x\geq 5.5 \end{cases}$$
Now you are looking for the maximum value of $x$ where $F_x(x)\leq 0.05$. This is at $P(X<5.5)=0.04$
A: You can only get $X=-7$ with a probability of $0.04$, or $X=5.5$ with a probability of $0.96$. So what is the probability of getting a number less then $-7$? It is zero. What's the probability of getting a number strictly between $-7$ and $5.5$? Once again, it is zero. What's the probability of getting a number less then $5.5$? It is exactly $0.04$, since you need to add $P(X<-7)=0$, $P(-7<X<5.5)=0$ and $P(X=-7)=0.04$. So $P(X<5.5)=0.04$. How about if we go from "less" to "less or equal" in the last expression? We need to add $P(X=5.5)=0.96$ so $P(X\le 5.5)=1$.
So now we can find an $x$ such as if $X<x$ we have $P(X)<0.05$. So if $x$ is say $4$, then $X<4$, therefore $P(X)=0.04<0.05$. The largest value for $x$ is $5.5$. If $X<x=5.5$ then $P(X)=0.04<0.05$. Let's make $x$ just slightly greater than $5.5$. Then $P(X<x)=1$.
