Question: A bus comes by every 15 minutes. The times from when a person arrives at the busstop until the bus arrives follows a Uniform distribution from 0 to 15 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 5 minutes is
d. Suppose that the person has already been waiting for 1 minutes. Find the probability that the Person's total waiting time will be between 3.2 and 4.3 minutes
e. 57% of all customers wait at least how long for the train?
1. **Problem Statement:**
A bus arrives every 15 minutes, and the waiting time $X$ for a person arriving randomly at the bus stop follows a Uniform distribution on the interval $[0,15]$ minutes.
2. **Uniform Distribution Properties:**
For a Uniform distribution $X \sim U(a,b)$:
- Mean: $$\mu = \frac{a+b}{2}$$
- Variance: $$\sigma^2 = \frac{(b-a)^2}{12}$$
- Standard deviation: $$\sigma = \sqrt{\sigma^2}$$
- Probability for interval $[x_1,x_2]$: $$P(x_1 \leq X \leq x_2) = \frac{x_2 - x_1}{b - a}$$
3. **Given:**
$$a=0, \quad b=15$$
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### a. Mean of the distribution
$$\mu = \frac{0 + 15}{2} = \frac{15}{2} = 7.5$$
### b. Standard deviation
$$\sigma^2 = \frac{(15 - 0)^2}{12} = \frac{225}{12} = 18.75$$
$$\sigma = \sqrt{18.75} \approx 4.3301$$
### c. Probability that the person waits more than 5 minutes
$$P(X > 5) = 1 - P(X \leq 5) = 1 - \frac{5 - 0}{15 - 0} = 1 - \frac{5}{15} = 1 - \frac{1}{3} = \frac{2}{3} \approx 0.6667$$
### d. Conditional probability: Given the person has already waited 1 minute, find $P(3.2 \leq X \leq 4.3)$ total waiting time
Since the person has waited 1 minute, the remaining waiting time is $X - 1$ given $X > 1$.
We want:
$$P(3.2 \leq X \leq 4.3 \mid X > 1) = \frac{P(3.2 \leq X \leq 4.3)}{P(X > 1)}$$
Calculate numerator:
$$P(3.2 \leq X \leq 4.3) = \frac{4.3 - 3.2}{15 - 0} = \frac{1.1}{15} \approx 0.0733$$
Calculate denominator:
$$P(X > 1) = 1 - P(X \leq 1) = 1 - \frac{1 - 0}{15} = 1 - \frac{1}{15} = \frac{14}{15} \approx 0.9333$$
Therefore:
$$P(3.2 \leq X \leq 4.3 \mid X > 1) = \frac{0.0733}{0.9333} \approx 0.0785$$
### e. Find $t$ such that 57% of customers wait at least $t$ minutes
This means:
$$P(X \geq t) = 0.57 \implies P(X < t) = 1 - 0.57 = 0.43$$
Since $X$ is uniform:
$$P(X < t) = \frac{t - 0}{15 - 0} = \frac{t}{15} = 0.43$$
Solve for $t$:
$$t = 15 \times 0.43 = 6.45$$
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**Final answers:**
- a. Mean = $7.5$
- b. Standard deviation $\approx 4.3301$
- c. Probability wait more than 5 minutes $\approx 0.6667$
- d. Conditional probability $\approx 0.0785$
- e. 57% wait at least $6.45$ minutes