A manager of an apartment store reports that the time of a customer on the second floor must wait for the elevator has a uniform distribution ranging from 22 to 66 minutes. If it takes the elevator 3030 seconds to go from floor to floor, find the probability that a hurried customer can reach the first floor in less than 5.755.75 minutes after pushing the elevator button on the second floo

0.8125

Step-by-step explanation:

For a uniform distribution,

f(x) = 1/(b-a)

where b = upper limit of the uniform distribution = 6 minutes

a = lower limit of the uniform distribution = 2 minutes.

Then, probability of x falling between two intervals for a uniform distribution is simply given as

P(a < x < c) = ∫ᶜₐ f(x) dx = [1/(b-a)] ∫ᶜₐ dx

= (c - a)/(b - a)

For the question,

it takes the elevator 30 seconds to go from floor to floor, find the probability that a hurried customer can reach the first floor in less than 5.75 minutes after pushing the elevator button on the second floor.

The elevator spends 0.5 min to reach the first floor from the second floor, so the maximum waiting time for the hurried customer will be

5.75 - 0.5 = 5.25 minutes.

P(2 < x < 5.25) = (5.25-2)/(6-2) = (3.25/4) = 0.8125

Hope this Helps!!!

what i would say the best answer is a. 40.5