Question

The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 00 and 77 minutes. find the probability that a randomly selected passenger has a waiting time greater than greater than 3.253.25 minutes.

Answer 1

Uniform distribution between 0 and 7 minutes.
Need to know probability of waiting greater than 3.25 minutes.

P(T>3.25) = (7-3.25)/(7-0) = 0.536

Answer: the probability that waiting time exceeds 3.25 minutes is 0.536

Answer 2

Answer: 0.5357

Step-by-step explanation:

If a continuous random variable x is distributed uniformly in interval [a,b] , then the probability density function is given by :-

$f(x)=\dfrac{1}{b-a}$

Given : The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 7 minutes.

Probability density function=$f(x)=\dfrac{1}{7-0}=\dfrac{1}{7}$

Now, the probability that a randomly selected passenger has a waiting time greater than greater than 3.253.25 minutes :-

$\int^{7}_{3.25}\ f(x)\ dx\\\\= \dfrac{1}{7}\int^{7}_{3.25}\ dx\\\\ \dfrac{1}{7}[x]^{7}_{3.25}\\\\=\dfrac{1}{7}[7-3.25]=\dfrac{3.75}{7}=0.535714285\approx0.5357$

Hence, the required probability = 0.5357