Question

Assume that voltages in a circuit vary between 6 volts and 12 volts, and voltages are spread evenly over the range of possibilities, so that there is a uniform distribution. Find the probability of the given range of voltage levels. In each case, draw a sketch and leave answers as a fraction. a. Greater than 10 volts b. Less than 11 volts c. Between 6.5 volts and 8 volts

Answer 1

a)
`P(X>10) =1-P(X<10) = 1-F(10) = 1-(10-6)/(12-6) = 0.333`

b)
`P(X<11)= F(11) = (11-6)/(12-6)= 0.833`

c)
`P(6.5<X<8)= F(8) -F(6.5) = (8-6)/(12-6)-(6.5-6)/(12-6)= 0.333-0.0833 = 0.250`

In the figure attached we have the illustration for the probability for each part.

Explanation:

For this case we define the random variable X as the voltages in a circuit and the distribution for X is given by:

`X \sim Unif (a= 6, b=12)`

Part a

we want this probability:

`P(X>10)`

We can use the cumulative distirbution function given by:

`F(x) = (x-a)/(b-a), a\leq X \leq b`

And using the complement rule we have this:

`P(X>10) =1-P(X<10) = 1-F(10) = 1-(10-6)/(12-6) = 0.333`

Part b

We want this probability:

`P(X<11)`

And using the cumulative distirbution function we got:

`P(X<11)= F(11) = (11-6)/(12-6)= 0.833`

Part c

We want this probability:

`P(6.5<X<8)`

And using the cumulative distirbution function we got:

`P(6.5<X<8)= F(8) -F(6.5) = (8-6)/(12-6)-(6.5-6)/(12-6)= 0.333-0.0833 = 0.250`

In the figure attached we have the illustration for the probability for each part.