Question

An article suggests the uniform distribution on the interval (7.5, 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.) mean variance


(b) What is the cdf of depth? F(x) = 0 x < 7.5 7.5 ≤ x < 19 1 19 ≤ x

Answer 1

Answer:

a) Mean: 13.25

Variance: 11.02

b)

0 if x below than 7.5

${x-a}{b-a}$ is x in between 7.5 and 19

1 if x higher than 19.

Step-by-step explanation:

This is a problem in which we use the uniform probability distribution.

(a) What are the mean and variance of depth?

The mean is the midpoint of 7.5 and 19. So

$M = \frac{7.5 + 19}{2} = 13.25$

The mean is 13.25

The variance is given by the following formula:

$V = \frac{(b-a)^{2}}{12}$

In which b and a are the limits of the interval. So $b = 19, a = 7.5$

So

$V = \frac{(b-a)^{2}}{12} = \frac{(19-7.5)^{2}}{12} = 11.02$

The variance is 11.02.

(b) What is the cdf of depth?

The uniform probability distribution has a cummulative distribution function, of 0 if is below a, $\frac{x-a}{b-a}$ for x in between a and b and 1 is x is greater than b.

So:

0 if x below than 7.5

${x-a}{b-a}$ is x in between 7.5 and 19

1 if x higher than 19.