Respuesta :
Answer:
a) For this case the mean is given by:
[tex] E(X) = \frac{a+b}{2}= \frac{20+40}{2}= 35[/tex]
The variance is given by:
[tex] Var(X) = \frac{(b-a)^2}{12}= \frac{(40-20)^2}{12} =33.33[/tex]
And the deviation would be:
[tex] Sd(X) = \sqrt{33.33}= 5.774[/tex]
b) For this case we want to find this probability:
[tex]P(X<35) [/tex]
And we can use the cumulative distribution function given by:
[tex] F(X) = \frac{x-a}{b-a} , a \leq X \leq b[/tex]
And using this we got:
[tex] P(X<35) = F(35) = \frac{35-20}{40-20}= 0.75[/tex]
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:
[tex]X \sim Unif (a= 20, b=40)[/tex]
Part a
For this case the mean is given by:
[tex] E(X) = \frac{a+b}{2}= \frac{20+40}{2}= 35[/tex]
The variance is given by:
[tex] Var(X) = \frac{(b-a)^2}{12}= \frac{(40-20)^2}{12} =33.33[/tex]
And the deviation would be:
[tex] Sd(X) = \sqrt{33.33}= 5.774[/tex]
Part b
For this case we want to find this probability:
[tex]P(X<35) [/tex]
And we can use the cumulative distribution function given by:
[tex] F(X) = \frac{x-a}{b-a} , a \leq X \leq b[/tex]
And using this we got:
[tex] P(X<35) = F(35) = \frac{35-20}{40-20}= 0.75[/tex]
