The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed between 0 and 6 minutes. find the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes.

Using the uniform distribution, it is found that there is a 0.625 = 62.5% probability that a randomly selected passenger has a waiting time greater than 2.25 minutes.

An uniform distribution has two bounds, a and b.  

The probability of finding a value above x is:

In this problem, uniformly distributed between 0 and 6 minutes, thus .

The probability of a time greater than 2.25 minutes is P(X > 2.25), thus:

0.625 = 62.5% probability that a randomly selected passenger has a waiting time greater than 2.25 minutes.

A similar problem is given at link

0.625

Step-by-step explanation:

Given : The waiting times between a subway departure schedule and the arrival of a passenger are uniformly distributed with interval= [0 minutes, 6 minutes]

The probability density function :-

The interval of  required event = [2.25 seconds, 6 seconds]

Now, the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes is given by :-

Hence, the probability that a randomly selected passenger has a waiting time greater than 2.25 minutes = 0.625

whatvare the choices

step-by-step explanation:

answer: 5 17/33 % (to the tenth)

step-by-step explanation:

he guessed: 34.5 pounds

actual weight is 32.7

there for he is off:

34.5 -32.7 = 1.8

32.7 is the 100%

1.8 represents %? of 32.7

therefore:

32.7   - 100%

1.8     -     x%

x= 1.8*100%/32.7% = 18*100%/327%

x=1800%/327%   = (5 165/327)%

to the tenth:

5 17/33 %