Question

The thickness of a protective coating applied to a conductor designed to work in corrosive conditions follows a uniform distribution over the interval [20,40] microns. a. Find the mean and standard deviation of the thickness of the protective coating. b. Find also the probability that the coating is less than 35 microns thick

Answer 1

Answer:

a) For this case the mean is given by:

$E(X) = \frac{a+b}{2}= \frac{20+40}{2}= 35$

The variance is given by:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(40-20)^2}{12} =33.33$

And the deviation would be:

$Sd(X) = \sqrt{33.33}= 5.774$

b) For this case we want to find this probability:

$P(X$

And we can use the cumulative distribution function given by:

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

And using this we got:

$P(X$

Step-by-step explanation:

For this case we define the random variable X as the hickness of a protective coating applied to a conductor and we know that the distribution of X is given by:

$X \sim Unif (a= 20, b=40)$

Part a

For this case the mean is given by:

$E(X) = \frac{a+b}{2}= \frac{20+40}{2}= 35$

The variance is given by:

$Var(X) = \frac{(b-a)^2}{12}= \frac{(40-20)^2}{12} =33.33$

And the deviation would be:

$Sd(X) = \sqrt{33.33}= 5.774$

Part b

For this case we want to find this probability:

$P(X$

And we can use the cumulative distribution function given by:

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

And using this we got:

$P(X$