The probability that a randomly selected customer will wait more than 4 minutes at the deli is given below
- The probability that a randomly selected customer will wait more than 4 minutes =
![0.4286](https://tex.z-dn.net/?f=0.4286)
The waiting time for the customers is uniformly distributed between 0 - 7 minutes
Consider a random variabe X that represents the waiting time of customers and uniformly distributed between 0-7 minutes
Therefore, the density function of X is determined as
![f(x) = \frac{1}{b-a}\\\\f(x) = \frac{1}{7-0}\\\\f(x) = \frac{1}{7}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B1%7D%7Bb-a%7D%5C%5C%5C%5Cf%28x%29%20%3D%20%5Cfrac%7B1%7D%7B7-0%7D%5C%5C%5C%5Cf%28x%29%20%3D%20%5Cfrac%7B1%7D%7B7%7D)
The probability that the customer will wait more than 4minutes,
![P(X>4) = \int\limits^7_4 f{x} \,dx\\\\= \int\limits^7_4 \, \frac{1}{7}dx\\\\= \frac{1}{7} * [x]^7_4\\\\= \frac{7-4}{7} \\\\= 0.4286](https://tex.z-dn.net/?f=P%28X%3E4%29%20%3D%20%5Cint%5Climits%5E7_4%20f%7Bx%7D%20%5C%2Cdx%5C%5C%5C%5C%3D%20%5Cint%5Climits%5E7_4%20%20%5C%2C%20%20%20%5Cfrac%7B1%7D%7B7%7Ddx%5C%5C%5C%5C%3D%20%5Cfrac%7B1%7D%7B7%7D%20%2A%20%5Bx%5D%5E7_4%5C%5C%5C%5C%3D%20%5Cfrac%7B7-4%7D%7B7%7D%20%5C%5C%5C%5C%3D%200.4286)
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Answer:
i believe that the correct answer is true
Using the uniform distribution, it is found that the variance of x is of 2.0833.
<h3>What is the uniform probability distribution?</h3>
It is a distribution with two bounds, a and b, in which each outcome is equally as likely.
The variance is given by:
![V(X) = \frac{(b - a)^2}{12}](https://tex.z-dn.net/?f=V%28X%29%20%3D%20%5Cfrac%7B%28b%20-%20a%29%5E2%7D%7B12%7D)
In the die, all outcomes between 1 and 6 are equally as likely, hence the bounds are a = 1, b = 6, and the variance is given by:
![V(X) = \frac{(6 - 1)^2}{12} = 2.0833](https://tex.z-dn.net/?f=V%28X%29%20%3D%20%5Cfrac%7B%286%20-%201%29%5E2%7D%7B12%7D%20%3D%202.0833)
More can be learned about the uniform distribution at brainly.com/question/13889040