The article "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants" (Water Research, $1984: 1169-1174$ ) suggests the uniform dis

Ayaana Buck

Ayaana Buck

Answered question

2021-01-02

The article "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants" (Water Research, $1984: 1169-1174$ ) suggests the uniform distribution on the interval (7.5,20) as a model for depth (cm) of the bioturbation layer in sediment in a certain region.
What are the mean and variance of depth?
b. What is the cdf of depth?
What is the probability that observed depth is at most 10? Between 10 and $15 ?$
What is the probability that the observed depth is within 1 standard deviation of the mean value? Within 2 standard deviations?

Answer & Explanation

Nathalie Redfern

Nathalie Redfern

Skilled2021-01-03Added 99 answers

Given:
*The interval of the uniform distribution is [a,b]=[7.5,20]
a) If the function has a uniform distribution on the interval, the mean depth is,
μ=b+a2
=7.5+202
=13.75 cm
The depth's variance is determined by:
σ2=112×(ba)2
=112×(207.5)2
=13.02 cm
b)
The cdf or cumulative distribution function of the depth is given by:
F(x)={(0x<7.5xaba=x7.5207.57.5x201x20) 
c)
Given is the probability that the observed depth is at most 10:
P((X<10))
=F(10)
=107.5207.5
=0.2
The probability that observed depth is in between 10 and 15 is given by:
P(10X15)=F(15)F(10)
Substitute the values in the above expression.
P=F(15)F(10)
=157.5207.5107.5207.5
=0.4
Therefore, the probability that observed depth is in between 10 and 15 is given by:
P(10X15)=0.4
d) The limit within one deviation from the mean is,
μσ=13.753.608=10.14
μ+σ=13.75+3.608=17.358
The probability that the observed depth is within 1 standard deviation of the mean value is given by:
P(10.14X17.358)=F(17.358)F(10.14)
Substitute the values in the above expression.
P=F(18.358)F(10.14)
=17.2587.5207.510.147.5207.5
=0.7890.211
=0.578
Hence, the probability that the observed depth is within 1 standard deviation of the mean value is,
P(10.14X17.358)=0.578
The limit within two deviation from the mean is,
μ2σ=13.752×3.608=6.534
μ+2σ=13.75+2×3.608=20.966
But the random variable is defined on the interval [7.5 and 20]
Hence, If the observed depth is within 2 standard deviations of the mean value, the probability is,
P(6.534X20.966)=P(6.534X7.5)+P(7.5X20)+P(20X20.966)
Substitute the values in the above expression.
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