An article suggests the uniform distribution on the interval (7.5, 21) as a mode
ID: 3133284 • Letter: A
Question
An article suggests the uniform distribution on the interval (7.5, 21) as a model for depth (cm) of the bioturbation layer in sediment in a certain region.
(a) What are the mean and variance of depth? (Round your variance to two decimal places.)
(b) What is the cdf of depth?
F(x) =
(c) What is the probability that observed depth is at most 10? (Round your answer to four decimal places.)
What is the probability that observed depth is between 10 and 15? (Round your answer to four decimal places.)
(d) What is the probability that the observed depth is within 1 standard deviation of the mean value? (Round your answer to four decimal places.)
What is the probability that the observed depth is within 2 standard deviations of the mean value?
Explanation / Answer
A)
Note that here,
a = lower fence of the distribution = 7.5
b = upper fence of the distribution = 21
Thus, the mean, variance, and standard deviations are
u = mean = (b + a)/2 = 14.25 [ANSWER, MEAN]
s^2 = variance = (b -a)^2 / 12 = 15.1875 [ANSWER, VARIANCE]
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B)
Between 7.5 and 21, the cdf is
F(x) = x/(b-a) = x/(21-7.5) = 0.074074074x [ANSWER]
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c)
Hence,
P(x<10) = F(10) = 0.074074074*10 = 0.74074074 [ANSWER]
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P(10<x<15) = F(15) - F(10) = 0.074074074*15 - 0.074074074*10 = 0.37037037 [ANSWER]
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d)
As
s = standard deviation = sqrt(s^2) = 3.897114317
Then with 1 standard deviation is
P(within 1 standard deviation) = 2*s/(b-a) = 2*3.897114317/(21-7.5) = 0.577350269 [ANSWER]
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For uniform distributions, within 2 standard deviations is a sure event, as it is just 3.46 standard deviations long. Hence,
P(within 2 standard deviations) = 1 [ANSWER]
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