Question

6.3 The daily amount of coffee, in liters, dispensed by a machine located in an airport lobby is a random variable X having a continuous uniform distribution with A = 7 and B = 10. Find the probability that on a given day the amount of coffee dispensed by this machine will be (a) at most 8.8 liters; (b) more than 7.4 liters but less than 9.5 liters; (c) at least 8.5 liters.

Answer 1

a) 60% probability that on a given day the amount of coffee dispensed by this machine will be at most 8.8 liters.

b) 70% probability that on a given day the amount of coffee dispensed by this machine will be more than 7.4 liters but less than 9.5 liters

c) 50% probability that on a given day the amount of coffee dispensed by this machine will be at least 8.5 liters

Explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value X lower than x is given by the following formula.

`P(X \leq x) = (x - a)/(b-a)`

The probability of X being higher than x is:

`P(X > x) = 1 - (x - a)/(b-a)`

The probability of X being between c and d is:

`P(c \leq X \leq d) = (d - c)/(b - a)`

For this problem, we have that:

`a = 7, b = 10`

(a) at most 8.8 liters;

`P(X \leq 8.8) = (8.8 - 7)/(10 - 7) = 0.6`

60% probability that on a given day the amount of coffee dispensed by this machine will be at most 8.8 liters.

(b) more than 7.4 liters but less than 9.5 liters;

`P(7.4 \leq X \leq 9.5) = (9.5 - 7.4)/(10 - 7) = 0.7`

70% probability that on a given day the amount of coffee dispensed by this machine will be more than 7.4 liters but less than 9.5 liters

(c) at least 8.5 liters.

`P(X > 8.5) = 1 - (8.5 - 7)/(10 - 7) = 0.5`

50% probability that on a given day the amount of coffee dispensed by this machine will be at least 8.5 liters