The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 00 and 99 minutes. Find the probability that a randomly selected passenger has a waiting time less thanless than 1.751.75 minutes.
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Part 1
Find the probability that a randomly selected passenger has a waiting time less thanless than 1.751.75 minutes.
enter your response here ​(Simplify your answer. Round to three decimal places as​ needed.)

To solve this, we need to calculate the probability that a randomly selected passenger's waiting time is less than 1.75 minutes, given that the waiting times are uniformly distributed between 0 and 99 minutes.

For a uniform distribution, the probability density function is given by:
\[
f(x) = \frac{1}{b - a}
\]
where \( a \) and \( b \) are the lower and upper bounds of the distribution. Here, \( a = 0 \) and \( b = 99 \).

### Step 1: Calculate the total range of the distribution
The total range is:
\[
b - a = 99 - 0 = 99 \text{ minutes}
\]

### Step 2: Find the probability
Since the distribution is uniform, the probability of waiting less than 1.75 minutes is the length of the interval from 0 to 1.75 divided by the total range (99 minutes):
\[
P(X < 1.75) = \frac{1.75}{99}
\]

### Step 3: Perform the calculation
\[
P(X < 1.75) = \frac{1.75}{99} \approx 0.0177
\]

Thus, the probability that a randomly selected passenger has a waiting time of less than 1.75 minutes is approximately 0.018 (rounded to three decimal places).

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The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 00 and 99 minutes. Find the probability that a randomly selected passenger has a waiting time less than 1.75 minutes.

Find the probability that a passenger's waiting time is less than 1.75 minutes in a uniformly distributed time interval between 0 and 99 minutes. This problem explores concepts from probability theory, including uniform distribution, and is suitable for students in Grades 10-12.

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