Question

The shuttle bus from your parking lot and your office building operates on a 15 minute schedule. You arrive at the parking lot at a random time during the bus’s cycle, that is, the time you have to wait for the bus is uniformly distributed over the interval from 0 to 15. What is the standard deviation of your waiting time? What is the probability that you will have to wait more than 2 standard deviations?

Answer 1

Answer:

(a) The standard deviation of your waiting time is 4.33 minutes.

(b) The probability that you will have to wait more than 2 standard deviations is 0.4227.

Step-by-step explanation:

Let X = the waiting time for the bus at the parking lot.

The random variable X is uniformly distributed with parameters a = 0 to b = 15.

The probability density function of X is given as follows:

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

(a)

The standard deviation of a Uniformly distributed random variable is given by:

$SD=\sqrt{\frac{(b-a)^{2}}{12}}$

Compute the standard deviation of the random variable X as follows:

$SD=\sqrt{\frac{(b-a)^{2}}{12}}$

      $=\sqrt{\frac{(15-0)^{2}}{12}}$

      $=\sqrt{\frac{225}{12}}$

      $=\sqrt{18.75}\\=4.33$

Thus, the standard deviation of your waiting time is 4.33 minutes.

(b)

The value representing 2 standard deviations is:

$X=2\times SD=2\times4.33=8.66$

Compute the value of P (X > 8.66) as follows:

$P(X>8.66)=\int\limits^{10}_{8.66} {\frac{1}{15-0}}\, dx$

                    $=\frac{1}{15}\times \int\limits^{10}_{8.66} {1}\, dx$

                    $=\frac{1}{15}\times |x|^{15}_{8.66}$

                    $=\frac{15-8.66}{15}$

                    $=0.4227$

Thus, the probability that you will have to wait more than 2 standard deviations is 0.4227.