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

(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.

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)=(1)/(b-a);\ a<X<b,\ a<b`

(a)

The probability density function of the random variable X is:

`f_(X)(x)=\left \{ {{(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)=(x-a)/(b-a)=(x-5)/(10-5);\ x\epsilon[a, b]`

`=1;\ for\ x>b`

(b)

The expected value of Uniform distribution is:

`E(X)=(1)/(2)(a+b)`

Compute the expected value of X as follows:

`E(X)=(1)/(2)(a+b)`

`=(1)/(2)* (5+10)`

`=(1)/(2)* 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){(1)/(10-5)}\, dx\\`

`=(1)/(5)* \int\limits^(7)_(5){1}\, dx`

`=(1)/(5)* |x|^(7)_(5)`

`=(1)/(5)* (7-5)`

`=(1)/(5)* 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.