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

Answer:

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

Step-by-step 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.

$P(X \leq x) = \frac{x - a}{b-a}$

The probability of X being higher than x is:

$P(X > x) = 1 - \frac{x - a}{b-a}$

The probability of X being between c and d is:

$P(c \leq X \leq d) = \frac{d - c}{b - a}$

For this problem, we have that:

$a = 7, b = 10$

(a) at most 8.8 liters;

$P(X \leq 8.8) = \frac{8.8 - 7}{10 - 7} = 0.6$

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;

$P(7.4 \leq X \leq 9.5) = \frac{9.5 - 7.4}{10 - 7} = 0.7$

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.

$P(X > 8.5) = 1 - \frac{8.5 - 7}{10 - 7} = 0.5$

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