Observed data suggests that the depth (in centimeters) of the bioturbation layer in river sediment for a certain region is well
modeled by a uniform distribution on the interval [2.5,30]. (a) what are the mean and variance of depth? (b) what is the cumulative distribution function of depth? (c) what is the probability that a given observed depth is at most 15? between 15 and 20? (d) what is the probability that this observed depth is within one standard deviation of the mean value? within two standard deviations?
1 answer:
(a) No doubt your textbook tells you that for a uniform distribution on the interval [a, b]:
.. mean = (a+b)/2 = (2.5 +30)/2 = 16.25
.. variance = (b -a)^2/12 = 27.5^2/12 = 63 1/48
(b) Likewise, from your text
.. p(x) = (x-a)/(b-a) = (x -2.5)/27.5
.. cdf(x) = p(x)^2 = (x -2.5)^2/756.25 . . . cumulative distribution function
(c) cdf(15) ≈ 0.2066 . . . . . . . . . . . probability of being 15 or less
.. cdf(20) -cdf(15) ≈ 0.1983 . . . . . probability of being between 15 and 20
(d) cdf(μ+σ) -cdf(μ-σ) ≈ 0.5774 . . . probability of being withing 1σ of the mean
.. cdf(μ+2σ) -cdf(μ-2σ) = 1 . . . . . . . the entire distribution is within 2σ of the mean
.. mean = (a+b)/2 = (2.5 +30)/2 = 16.25
.. variance = (b -a)^2/12 = 27.5^2/12 = 63 1/48
(b) Likewise, from your text
.. p(x) = (x-a)/(b-a) = (x -2.5)/27.5
.. cdf(x) = p(x)^2 = (x -2.5)^2/756.25 . . . cumulative distribution function
(c) cdf(15) ≈ 0.2066 . . . . . . . . . . . probability of being 15 or less
.. cdf(20) -cdf(15) ≈ 0.1983 . . . . . probability of being between 15 and 20
(d) cdf(μ+σ) -cdf(μ-σ) ≈ 0.5774 . . . probability of being withing 1σ of the mean
.. cdf(μ+2σ) -cdf(μ-2σ) = 1 . . . . . . . the entire distribution is within 2σ of the mean
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