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The time to fly between New York City and Atlanta is uniformly distributed with a minimum of 120 minutes and a maximum of 150 minutes. What is the probability that a flight takes more than 140 minutes?

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We model the situation with the continuous uniform distribution on the interval $[120,150]$.

So in principle the flight could last, for example, $142+(\pi\times 10^{-77})$ minutes. This may violate some law of physics, and measurements cannot be made with that kind of precision. But one should remember that a mathematical model is a model, it is not the reality. A model can be useful even if it fits reality only moderately well.

The interval $(140,150]$ has length $10$, while the whole interval $[120,150]$ has length $30$. Since the distribution is uniform, the required probability is $\dfrac{10}{30}$.

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  • $\begingroup$ Yes, judging by OPs other question on his profile I think this is the method that was intended to be used. $\endgroup$
    – user42538
    Feb 12, 2013 at 3:44
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\begin{align} P(X>140)&=\int_{140}^{150}\dfrac{1}{30}dx\\ &=\dfrac{10}{30}\\&=\dfrac{1}{3} \end{align} Alternately:

Probability is the area under the curve. enter image description here

You can think of the distribution to be a rectangle with a span from 120 to 150 and a height of $\dfrac{1}{30}$ to keep the area normalized to 1. Thus, the area from 140 to 150 is simply $10\times \dfrac{1}{30}$ which is the same as above.

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The probabilities being uniformly distributed means that the probability density function at any point between 120 and 150 is $\frac {1}{(150-120)} = \frac{1}{30}$, and the probability that the flight will last between 140 and 150 minutes is $(150-140)\frac{1} {30} = \frac{10}{30} = \frac{1}{3}$, so there's a 1 in 3 chance the flight will last between 140 and 150 minutes.

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