How to calculate the value of random variable in uniform distribution if given the probability?

Question

Jack always arrives 5 minutes before meetings, Peter always arrives from 0 to 10 minutes late into meetings. Jack is tired waiting for Peter and he decided that next time he's also going to be late. How much time does Jack have to be late for a meeting such that the probability of Peter being in the meeting is 0.85.

I think that the amount of time Peter is late for a meeting ($X$) obeys uniform distribution, then:

$$P(X=k) = \frac{1}{b-a+1}$$

where $a=1$, $b=10$.

We could say that $b-a=m$, $\quad m$ being the time Peter is late for a meeting. Then could:

$$m=\frac{1}{0.85}-1 = 0.176$$ ?

Even if this is correct how can this be, isn't uniform distribution always the same ($\frac{1}{10}$ in our case)?

Answer 1

Let $X$ be Peter's delay. Then $X$ is uniformly distributed on $[0,10]$, with cumulative distribution function $$P(X \le x) = F(x) = \frac{x}{10}$$ for $0 \le x \le 10$.

Jack's problem is to pick a time $x$ such that the probability that Peter has arrived is $0.85$; that is, he seeks $x$ such that $$P(X \le x) = \frac{x}{10} = 0.85$$ Solving for $x$, you find that Jack should arrive 8.5 minutes late at the meeting.

Answer 2

I understand from the question that the amount of time Peter is late is r.v. with $\mathcal{U}[0,10]$ distribution. This has pdf $f(x)=\frac{1}{10}$ and cdf $F(x)=\frac{x}{10}$ for $x\in[0,10]$ (and zero for $x<0$ and unity for $x>10$).

With this, your question is asking for $x$ such that $F(x)=0.85$.