# Solved-The waiting time at a bus stop is uniformly distributed between 1 and 12 minute

#### ByDr. Raju Chaudhari

Oct 4, 2020

The waiting time at a bus stop is uniformly distributed between 1 and 12 minute.

(a) What is the probability density function?
(b) What is the probability that the rider waits 8 minutes or less?
(c) What is the expected wait time?
(d) What is variance?

## Solution

If $X\sim U(a,b)$ then the probability density function of $X$ is $f(x) =\dfrac{1}{b-a}, a\leq x \leq b$.

The mean of $X$ is $E(X) =\dfrac{(a+b)}{2}$ and variance of $X$ is $V(X) =\dfrac{(b-a)^2}{12}$.

The waiting time at a bus stop is uniformly distributed between 1 and 12 minute. That is $X\sim U(1,12)$.

(a) The probability density function of $X$ is

 \begin{aligned} f(x) & = \frac{1}{12-1}, 1\leq x \leq 12\\ & = \frac{1}{11}, 1\leq x \leq 12\\ &= 0.0909, 1\leq x \leq 12. \end{aligned}

(b) The probability that the rider waits 8 minutes or less

 \begin{aligned} P(X\leq 8) & = \int_1^8 f(x) \; dx\\ & = \frac{1}{11}\int_1^8 \; dx\\ & = \frac{1}{11} \big[x \big]_1^8\\ &= \frac{1}{11}\big[ 8-1\big]\\ &= \frac{7}{11}\\ &= 0.6364. \end{aligned}

(c) The expected wait time is $E(X) =\dfrac{a+b}{2} =\dfrac{1+12}{2} =6.5$

(d) The variance of $X$ is $V(X) =\dfrac{(b-a)^2}{12} =\dfrac{(12-1)^2}{12} =10.08$