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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)?

2 Answers 2

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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.

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    Thanks, but I don't understand why you have $x$ in the nominator. why is your formula different from the one I mentioned in my post? – 2017-02-02
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    Perhaps there is a confusion between the uniform distribution over the continuous interval $[a,b]$ (mine) and the uniform distribution over the discrete set of integers between $a$ and $b$ (yours). Time is usually modelled as continuous (i.e., Peter may arrive any time), so I am working with the first case, where the density function is $f(x) = 1/10$. – 2017-02-02
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    Indeed I didn't realize it's a continuous variable. In the class we were given this formula in order to calculate the $cdf$ of uniform distribution $F(x) = \frac{x-a}{b-a} \quad \Rightarrow x = 8.65$ ($a=1$ and $b=10$). Can you please confirm your calculation is correct? – 2017-02-02
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    Got it! in our case $a=0$ not $a=1$! – 2017-02-02
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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 $F(x)=0.85$.