The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between zero and six minutes find the probability that a randomly selected passenger has a waiting time greater than 4.25 minutes find the probability that a randomly selected passenger has a waiting time greater than 4.25 minutes please help

52.8%

Step-by-step explanation:

To solve this exercise, we must follow this formula:

Probability (P) = (Length of shaded region) * (Height of shaded region)

The best way to understand this exercise is to graph it. Attached, you can see the graphic representation of this uniform distribution (Graph 1).

1. Determine the total length and length of shaded area .

The length is 9, which is the maximum time that a passenger waits.

However, in the formula we are asked about the "length of the shaded area". According to the statement, the statistical event that is important to us is "waiting time greater than 4.25 minutes". Therefore, the length of the shaded area ranges from 4.25 to 9 (This can be seen in graph 2).

To calculate this length, we simply do this substraction:

9 - 4.25 = 4.75.

2. Determine the shaded region height:

To find this element we use the following formula :

Height = 1 over b-a

Where a is the minimum point of the total length, that is 0, and b is the maximum, that is 9.

When replacing we have:

Height = 1 over 9-0

You can see that the shaded region height equals the total height (Graphs 1 and Graphs 2).

3. Replace the data in the original formula:

P (time greater than 4.25) = (Shaded region length) * (Shaded region height)

P (time greater than 4.25) = (4.75) * (1/9)

P (time greater than 4.25) = 0.5277

We round and put in percentage terms:

(0.528)*100

= 52.8%

answer: 43/157

this is the answer and if you try to prove me wrong your a meanie

it has changed from the width and the height.

step-by-step explanation: