Answer:
a) [tex] P(X>10) =1-P(X<10) = 1-F(10) = 1-\frac{10-6}{12-6} = 0.333[/tex]
b) [tex] P(X<11)= F(11) = \frac{11-6}{12-6}= 0.833[/tex]
c) [tex] P(6.5<X<8)= F(8) -F(6.5) = \frac{8-6}{12-6}-\frac{6.5-6}{12-6}= 0.333-0.0833 = 0.250[/tex]
In the figure attached we have the illustration for the probability for each part.
Step-by-step explanation:
For this case we define the random variable X as the voltages in a circuit and the distribution for X is given by:
[tex]X \sim Unif (a= 6, b=12) [/tex]
Part a
we want this probability:
[tex] P(X>10)[/tex]
We can use the cumulative distirbution function given by:
[tex] F(x) = \frac{x-a}{b-a}, a\leq X \leq b[/tex]
And using the complement rule we have this:
[tex] P(X>10) =1-P(X<10) = 1-F(10) = 1-\frac{10-6}{12-6} = 0.333[/tex]
Part b
We want this probability:
[tex] P(X<11)[/tex]
And using the cumulative distirbution function we got:
[tex] P(X<11)= F(11) = \frac{11-6}{12-6}= 0.833[/tex]
Part c
We want this probability:
[tex] P(6.5<X<8)[/tex]
And using the cumulative distirbution function we got:
[tex] P(6.5<X<8)= F(8) -F(6.5) = \frac{8-6}{12-6}-\frac{6.5-6}{12-6}= 0.333-0.0833 = 0.250[/tex]
In the figure attached we have the illustration for the probability for each part.
