Your question seems to contain some typos and appears to be asking for the probability density function of waiting times at a station where a tram arrives at regular intervals of 9 minutes. Let's clarify and solve the problem step by step.
To find the probability density function of waiting times, here’s what we need to keep in mind:
1. Interval Details: The tram arrives every 9 minutes, so the waiting time for a tram is uniformly distributed between 0 and 9 minutes.
2. Uniform Distribution: For a uniform distribution over an interval, say [tex]\([a, b]\)[/tex], all outcomes are equally likely. The probability density function, [tex]\( f(x) \)[/tex], for a uniform distribution is given by:
[tex]\[ f(x) = \frac{1}{b - a} \text{ for } x \in [a, b] \][/tex]
[tex]\[ f(x) = 0 \text{ otherwise} \][/tex]
For our case:
- [tex]\( a = 0 \)[/tex]
- [tex]\( b = 9 \)[/tex]
So, the PDF [tex]\( f(x) \)[/tex] will be:
[tex]\[ f(x) = \frac{1}{9 - 0} = \frac{1}{9} \text{ for } x \in [0, 9] \][/tex]
[tex]\[ f(x) = 0 \text{ otherwise} \][/tex]
Given the provided options:
- The correct probability density function should be defined over the interval [tex]\( [0, 9] \)[/tex].
Assuming the options were:
A. [tex]\( f(\alpha) = \frac{7}{4} \)[/tex] on the interval [tex]\( 0 \leq \alpha \leq 2 \)[/tex].
B. [tex]\( f(x) = \frac{1}{4} \)[/tex] on the interval [tex]\( 0 \leq x \leq \infty \)[/tex].
Neither option is correctly describing the uniform distribution for waiting times between 0 and 9 minutes, given:
The correct PDF should be:
[tex]\[ f(x) = \frac{1}{9} \text{ for } 0 \leq x \leq 9 \][/tex]
So, let's propose the correct answer:
C. The probability density function is [tex]\( f(x) = \frac{1}{9} \)[/tex] on the interval [tex]\( 0 \leq x \leq 9 \)[/tex].
