The article modeling sediment and water column interactions for hydrophobic pollutants 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 stats.1. what is the mean and variance of depth?2. what is the cdf of depth?3. what is the probability that observed depth is at most 10? between 10 and 15?4.what is the probability that the observed depth is within one standard deviation of the mean value? within 2 standard deviations?

Answer:1)[tex]\mu=\frac{1}{2}(7.5+20) =13.75[/tex][tex]\sigma^2 = \frac{1}{12}(20-7.5)^2 =13.02[/tex]2) [tex]F(x)=\big\{0, x<a[/tex][tex]F(x) =\big\{ \frac{x-a}{b-a}=\frac{x-7.5}{20-7.5}, a\leq x<b[/tex][tex]F(x)=\big\{1, x\geq b[/tex]3) [tex]P(X<10)=F(10)=\frac{10-7.5}{20-7.5}=0.2[/tex][tex]P(10\leq X \leq 15)=F(15)-F(10)=\frac{15-7.5}{20-7.5} -\frac{10-7.5}{20-7.5}=0.6-0.2=0.4[/tex]4) [tex]P(10.142\leq X \leq 17.358)=F(17.358)-F(10.142)=\frac{17.358-7.5}{20-7.5} -\frac{10.142-7.5}{20-7.5}=0.789-0.211=0.578[/tex][tex]P(6.534\leqX\leq 20.966)=P(6.534\leq X<7.5)+P(7.5\leq X \leq 20)+P(20<X\leq 20.966)=0+1+0=1[/tex]Step-by-step explanation:A uniform distribution, sometimes also known as a rectangular distribution, is a distribution that has constant probability.Part 1If X is a random variable that follows an uniform distribution [tex]x\sim U(a,b)[/tex]. The mean for an uniform distribution is given by : [tex]\mu=\frac{1}{2}(a+b)[/tex]On this case a=7.5 and b=20 so if we replace we got:[tex]\mu=\frac{1}{2}(7.5+20) =13.75[/tex]The variance for the uniform distribution is given by this formula:[tex]\sigma^2 = \frac{1}{12}(b-a)^2 [/tex]And replacing we have:[tex]\sigma^2 = \frac{1}{12}(20-7.5)^2 =13.02[/tex]Part 2The cumulative distribution function is given by:[tex]F(x)=\big\{0, x<a[/tex][tex]F(x) =\big\{ \frac{x-a}{b-a}=\frac{x-7.5}{20-7.5}, a\leq x<b[/tex][tex]F(x)=\big\{1, x\geq b[/tex]Part 3What is the probability that observed depth is at most 10?We are interested on this probability:[tex]P(X<10)=F(10)=\frac{10-7.5}{20-7.5}=0.2[/tex]What is the probability that observed depth is between 10 and 15?On this case we want this probability:[tex]P(10\leq X \leq 15)=F(15)-F(10)=\frac{15-7.5}{20-7.5} -\frac{10-7.5}{20-7.5}=0.6-0.2=0.4[/tex]Part 4What is the probability that the observed depth is within one standard deviation of the mean value? within 2 standard deviations?First we find the limits within one deviation from the mean:[tex]\mu-\sigma= 13.75-3.608=10.142[/tex][tex]\mu-\sigma= 13.75+3.608=17.358[/tex]And we want this probability:[tex]P(10.142\leq X \leq 17.358)=F(17.358)-F(10.142)=\frac{17.358-7.5}{20-7.5} -\frac{10.142-7.5}{20-7.5}=0.789-0.211=0.578[/tex]Now we find the limits within two deviation's from the mean:[tex]\mu-2*\sigma= 13.75-2*3.608=6.534[/tex][tex]\mu-2*\sigma= 13.75+2*3.608=20.966[/tex]But since the random variable is defined just between (7.5 and 20) so we can find just the probability on these limits.[tex]P(6.534\leqX\leq 20.966)=P(6.534\leq X<7.5)+P(7.5\leq X \leq 20)+P(20<X\leq 20.966)=0+1+0=1[/tex]