TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Bura)/Übungen 2020W/HW04.2

Train

Suppose a train is delayed by approximately 60 minutes. Assuming that a uniform distributionon (0, 60) was used to model the delayed time (in minutes), compute the probability that train will reach its final destination in the interval between 57 and 60 minutes. Calculate the mean and standard deviation of the delayed time.

Lösungsvorschlag von Friday

--Friday Sa 30 Jan 2021 16:46:53 CET

We can compute the probability that the train will reach the destiantion between 57 and 60 minutes with a single line of R:

> 1 - punif(57, min=0, max=60)
    [1] 0.05

Next, we can calculate the expectation by just inserting the variables into the formula:

$${\begin{aligned}a&=0\\b&=60\\E(X)&={\frac {a+b}{2}}={\frac {60}{2}}=30\end{aligned}}$$

To calculate the standard deviation ${\textstyle \sigma }$ we first need to know that ${\textstyle \sigma ^{2}=Var(X)}$.

$${\begin{aligned}\sigma ^{2}&=Var(X)={\frac {(b-a)^{2}}{12}}={\frac {60^{2}}{12}}=300\\\sigma &={\sqrt {300}}=10*{\sqrt {3}}\end{aligned}}$$