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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.
Please show he solution on how you get it. Thanks.
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Each time the answer is the ratio of the lengths of the two corresponding measuring intervals.
(a) The condition means that the volume V is in the interval 7 <= V <= 8.8 liters.
The length of the possible outcome interval is 8.8 - 7 = 1.8 liters.
The ratio to the base interval of 10-7 = 3 liters is = 0.6,
so the probability is P = 0.6.
(b) The condition means that the volume V is in the interval 7.4 <= V <= 9.5 liters.
The length of the possible outcome interval is 9.5 - 7.4 = 2.1 liters.
The ratio to the base interval of 10-7 = 3 liters is = 0.7,
so the probability P = 0.7.
(c) The condition means that the volume V is in the interval 8.5 <= V <= 10 liters.
The length of the possible outcome interval is 10 - 8.5 = 1.5 liters.
The ratio to the base interval of 10-7 = 3 liters is = 0.5,
so the probability P = 0.5 in this case.
Solved, answered and completed.