Question 1173096: Today, the waves are crashing onto the beach every 4.6 seconds. The times from when a person arrives at the shoreline until a crashing wave is observed follows a Uniform distribution from 0 to 4.6 seconds. Round to 4 decimal places where possible.
a. The mean of this distribution is
b. The standard deviation is
c. The probability that wave will crash onto the beach exactly 0.8 seconds after the person arrives is P(x = 0.8) =
d. The probability that the wave will crash onto the beach between 0.3 and 2.7 seconds after the person arrives is P(0.3 < x < 2.7) =
e. The probability that it will take longer than 2.02 seconds for the wave to crash onto the beach after the person arrives is P(x > 2.02) =
f. Suppose that the person has already been standing at the shoreline for 0.1 seconds without a wave crashing in. Find the probability that it will take between 1.4 and 1.7 seconds for the wave to crash onto the shoreline.
g. 89% of the time a person will wait at least how long before the wave crashes in? seconds.
h. Find the minimum for the upper quartile. seconds.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! the mean is midway between the limits or 2.3 sec.
variance is (b-a)^2/12 or 4.6^/12=1.766 sec^2
sd is sqrt (V)=1.33 sec.
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probability of an exact point is 0. This is a continuous function and only ranges have non-zero values.
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between 0.3 and 2.7 is 2.4 sec and the probability of that is 2.4/4.6 or 0.5217
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probability >2.02 sec is (4.6-2.02)/4.6=0.5609
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the uniform distribution has a memory, in that if one waits long enough, it will be certain at some point that the individual will experience whatever is being looked at. The starting point is when the person arrives at the shoreline, so the distribution is already operating at 0.1 sec.
So this means the the end point is now 4.5 sec. and the probability will be 0.3/4.5 or 0.0067
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89% of 4.6 sec is 4.094 sec.
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Upper quartile starts at 75% of the interval or 3.45 sec.
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