Question

Suppose that your bus arrives at your bus stop uniformly between 9:05am and 9:10am. LetX= timeyou wait for the bus. Thus we have thatX∼Unif(5,10).(a) Give the PDF and CDF forX.(b) What is the expected time that the bus will arrive?(c) Suppose you slept in a little and can make it to the bus stop at 9:07. What is the probability thatyou will have missed the bus?

Answer 1

Answer:

(a) Shown below

(b) The expected time it takes for the bus to arrive is 7.5 minutes.

(c) The probability that the bus was missed if I arrived at the bus stop at 9:07 AM is 0.40.

Step-by-step explanation:

Let X = waiting time for the bus.

It is provided that the bus arrives at the bus stop uniformly between 9:05 AM and 9:10 AM.

The random variable X is uniformly distributed with parameters a = 5 and b = 10.

The probability density function of a Uniform distribution is:

$f_{X}(x)=\frac{1}{b-a};\ a<X<b,\ a<b$

(a)

The probability density function of the random variable X is:

$f_{X}(x)=\left \{ {{\frac{1}{10-5};\ a<X<b,\ a<b \atop {0;\ otherwise}} \right.$

The cumulative distribution function of the random variable X is:

          $=0;\ for\ x <a$

$F_{X}(x)=\frac{x-a}{b-a}=\frac{x-5}{10-5};\ x\epsilon[a, b]$

          $=1;\ for\ x>b$

(b)

The expected value of Uniform distribution is:

$E(X)=\frac{1}{2}(a+b)$

Compute the expected value of X as follows:

$E(X)=\frac{1}{2}(a+b)$

         $=\frac{1}{2}\times (5+10)$

         $=\frac{1}{2}\times 15$

         $=7.5$

Thus, the expected time it takes for the bus to arrive is 7.5 minutes.

(c)

If I reach the bus stop at 9:07 AM and I have missed the bus, then this implies that the bus arrived between 9:05 AM and 9:07 AM.

Compute the probability that the bus arrived between 9:05 AM and 9:07 AM as follows:

$P(5<X<7)=\int\limits^{7}_{5}{\frac{1}{10-5}}\, dx$

                      $=\frac{1}{5}\times \int\limits^{7}_{5}{1}\, dx$

                      $=\frac{1}{5}\times |x|^{7}_{5}$

                      $=\frac{1}{5}\times (7-5)$

                      $=\frac{1}{5}\times 2$

                      $=0.40$

Thus, the probability that the bus was missed if I arrived at the bus stop at 9:07 AM is 0.40.