An article suggests the uniform distribution on the interval from 5.5 to 18 as a model for x = depth (in centimeters) of the bioturbation layer in sediment for a certain region.
(a) What is the height of the density curve?
(b) What is the probability that x is between 11 and 16? Between 8 and 13?
I have no idea how to do this problem
(a) What is the height of the density curve?
(b) What is the probability that x is between 11 and 16? Between 8 and 13?
I have no idea how to do this problem
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The uniform density function is of the form
f(x) = 1/(b - a), a < x < b and f(x) = 0 otherwise.
Then the uniform distribution function F(x) is given by
F(x) = 0, x < a; F(x) = (x - a)/(b - a), a < x < b; F(x) = 1, x > b
(a) The height is 1/(18 - 5.5) = 0.08
(b)
Between 11 and 16:
0.08(16 - 11) = 0.4
Between 8 and 13
0.08(13 - 8) = 0.4
f(x) = 1/(b - a), a < x < b and f(x) = 0 otherwise.
Then the uniform distribution function F(x) is given by
F(x) = 0, x < a; F(x) = (x - a)/(b - a), a < x < b; F(x) = 1, x > b
(a) The height is 1/(18 - 5.5) = 0.08
(b)
Between 11 and 16:
0.08(16 - 11) = 0.4
Between 8 and 13
0.08(13 - 8) = 0.4