The article "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants" (Water Research [1984]: 1169-1174) suggests the uniform distribution on the interval from 7.5 to 20 as a model for x = depth (in centimeters) of the bioturbation layer in sediment for a certain region.

a. Draw the density curve for x.

b. What is the height of the density curve?

c. What is the probability that x is at most 12?

d. What is the probability that x is between 10 and 15? Between 12 and 17? Why are these two probabilities equal?

#### Solution

The probability density curve is

` $$ \begin{aligned} f(x)=\left\{ \begin{array}{ll} \frac{1}{20 - 7.5} , & \hbox{$7.5 \leq x\leq 20$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{aligned} $$ `

a. The density curve of uniform distribution $U(7.5, 20)$ looks like

b. The height of the density curve is `$\frac{1}{20-7.5} =0.08$`

c. The probability that that x is at most 12 is

` $$ \begin{aligned} P(X\leq 12) &= \int_{7.5}^{12} f(x) \; dx\\ &= \frac{1}{20-7.5}\int_{7.5}^{12} \; dx\\ &= 0.08\big[x\big]_{7.5}^{12} \\ &= 0.08\big[ 12 -7.5]\\ &= 0.36 \end{aligned} $$ `

d. The probability that x is between 10 and 15 is

` $$ \begin{aligned} P(10< X< 15) &= \int_{10}^{15} f(x) \; dx\\ &= \frac{1}{20-7.5}\int_{10}^{15} \; dx\\ &= 0.08\big[x\big]_{10}^{15} \\ &= 0.08\big[ 15 -10]\\ &= 0.4 \end{aligned} $$ `

The probability that x is between 10 and 15 is

` $$ \begin{aligned} P(12< X< 17) &= \int_{12}^{17} f(x) \; dx\\ &= \frac{1}{20-7.5}\int_{12}^{17} \; dx\\ &= 0.08\big[x\big]_{12}^{17} \\ &= 0.08\big[ 17 -12]\\ &= 0.4 \end{aligned} $$ `

These two probabilities are same because the interval 10 to 15 and 12 to 17 are of equal length and the probabilities are uniformaly distributed.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators