An article suggests the uniform distribution on the interval (6.5, 19) as a mode

Question

An article suggests the uniform distribution on the interval (6.5, 19) 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? x 6.5 F(x) 6.5 x 19 19 s 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 decima 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.
PDF of Uniform Distribution f(x) = 1 / ( b - a ) for a < x < b
b = Maximum Value
a = Minimum Value
Mean = a + b / 2
Standard Deviation = Sqrt ( ( b - a ) ^ 2 / 12 )
f(x) = 1/(b-a) = 1 / (19-6.5) = 1 / 12.5 = 0.08

Mean = a + b / 2 = 12.75
Standard Deviation = Sqrt ( ( b - a ) ^ 2 / 12 ) = 3.608

c.
P(X < 10) = (10-6.5) * f(x)
= 3.5*0.08
= 0.28

d.
To find P(a < X < b) =( b - a ) * f(x)
P(10 < X < 15) = (15-10) * f(x)
= 5*0.08
= 0.4

e.
probability with one s.d of the mean value is [ 12.75 - 3.608, 12.75 + 3.608 ] = [ 9.142, 16.358 ]

f.
probability with two s.d of the mean value is [ 12.75 - 2*3.608, 12.75 + 2*3.608 ] = [ 5.534, 19.966]

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