Simon goes to his University class every weekday by using a train that leaves at 8:00 a.m. We assume that the duration of the journey, in minutes, is a random variable following the uniform distribution in the interval [58, 63]. Further, suppose that from the train platform he needs a 15-minute

Question

Simon goes to his University class every weekday by using a train that leaves at 8:00 a.m. We assume that the duration of the journey, in minutes, is a random variable following the uniform distribution in the interval [58, 63]. Further, suppose that from the train platform he needs a 15-minute walk to enter the classroom and that his class starts precisely at 9:15a.m.

(i) What is the probability that Simon arrives in time for his class?

(ii) What is the probability that he arrives at the class at least two minutes after it has begun?

(iii) Find Simon’s expected arrival time in the class.

Accepted Answer

Since the only source of uncertainty is the duration of the train journey, we define a random variable X, which represents the number of minutes that this journey lasts. Then X ∼ [latex]\mathcal{V}[/latex][58, 63], so that X has density function

[latex]f(x)= \begin{cases} \frac{1}{63 − 58}, \quad 58 \leq x \leq 63,\ 0, \qquad \quad ext{elsewhere}, \end{cases}= \begin{cases} \frac{1}{5}, \quad 58 \leq x \leq 63,\ 0, \quad ~ ext{elsewhere}, \end{cases} [/latex]

while the corresponding distribution function of X is given by

[latex]F(t)= \begin{cases} 0, \qquad ~~~~ t < 58,\ \frac{t-58}{63-58}, \quad 58 ≤ x ≤ 63,\ 1, \qquad ~~~~ t > 63. \end{cases} [/latex]

(i) For Simon to be in time (i.e. before 9:15 a.m.) for his class, he must arrive at the platform by 9:00 a.m., which means that the train journey lasts for 60 minutes at most. The probability for this is

[latex]P(X\leq60)={\frac{60-58}{63-58}}={\frac{2}{5}}.[/latex]

(ii) Now, we seek the probability that Simon arrives at the classroom after 9:17 a.m., which in turn implies that he arrives at the platform after 9:02 a.m. This happens if the train journey takes 62 minutes or more, with associated probability

[latex]P(X\gt 62)=1-F(62)=1-{\frac{62-58}{63-58}}={\frac{1}{5}},[/latex]

i.e. there is a 20% chance that he arrives at least two minutes late for the class.

(iii) The expected value of X, the journey time, by Proposition 7.1 is

[latex]E(X)={\frac{58+63}{2}}=60.5~\mathrm{minutes}.[/latex]

This implies that Simon is expected to arrive at the train platform of the University station half a minute after 9:00 a.m. and, as a result, his expected arrival time at the classroom is half a minute after the beginning of the class (9:15 a.m.).

Question 7.1: Simon goes to his University class every weekday by using a ......

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