Question

a bus comes by every 12 minutes. the times from when a person arives at the busstop until the bus arrives follows a uniform distribution from 0 to 12 minutes. a person arrives at the bus stop at a randomly selected time. round to 4 decimal places where possible.
a. the mean of this distribution is
b. the standard deviation is
c. the probability that the person will wait more than 8 minutes is
d. suppose that the person has already been waiting for 2.1 minutes. find the probability that the persons total waiting time will be between 4.7 and 5.4 minutes
e. 11% of all customers wait at least how long for the train?
hint:
written hint
helpful videos: probability +, conditional probability + conditional probability + percentiles +

Explanation:

Step1: Recall mean formula for uniform distribution

For a uniform distribution $U(a,b)$, the mean $\mu=\frac{a + b}{2}$. Here $a = 0$ and $b=12$, so $\mu=\frac{0 + 12}{2}=6$.

Step2: Recall standard - deviation formula for uniform distribution

The standard deviation $\sigma=\sqrt{\frac{(b - a)^2}{12}}$. Substituting $a = 0$ and $b = 12$, we get $\sigma=\sqrt{\frac{(12-0)^2}{12}}=\sqrt{12}\approx3.4641$.

Step3: Calculate probability of waiting more than 8 minutes

The probability density function of $U(0,12)$ is $f(x)=\frac{1}{b - a}=\frac{1}{12}$ for $x\in[0,12]$. The probability $P(X>8)=\int_{8}^{12}\frac{1}{12}dx=\frac{12 - 8}{12}=\frac{4}{12}\approx0.3333$.

Step4: Calculate conditional probability

Since the uniform distribution has the memory - less property, the remaining waiting time is still uniformly distributed on $[0,12 - 2.1]=[0,9.9]$. The probability density function of the remaining waiting time is $f(x)=\frac{1}{9.9}$ for $x\in[0,9.9]$. Then $P(4.7-2.1

Step5: Find the 11th percentile

Let $x$ be the waiting time. We want to find $x$ such that $P(X\geq x)=0.11$, or $P(X

Answer:

a. 6.0000
b. 3.4641
c. 0.3333
d. 0.0707
e. 10.6800