Question

An article 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. what is the probability that x is at most 16

Answer 1

Answer:

a) Figure attached

b) For this case we want this probability:

$P(X>16)$

And we can use the complement rule:

$P(X>16) = 1-P(X<16)$

And we can use the cumulative distribution function given by:

$F(x)= \frac{x-a}{b-a}, a\leq X \leq b$

And replacing we got:

$P(X>16) =1- \frac{16-7.5}{20-7.5} = 1-0.68 = 0.32$

Step-by-step explanation:

For this case we define the random variable X= depth (in centimeters) of the bioturbation layer in sediment for a certain region, and we know the distribution for X, given by:

$X \sim Unif (a=7.5, b=20)$

Part a

For this case we can see the figure attached.

Part b

For this case we want this probability:

$P(X>16)$

And we can use the complement rule:

$P(X>16) = 1-P(X<16)$

And we can use the cumulative distribution function given by:

$F(x)= \frac{x-a}{b-a}, a\leq X \leq b$

And replacing we got:

$P(X>16) =1- \frac{16-7.5}{20-7.5} = 1-0.68 = 0.32$