Question

4. The cycle time for trucks hauling concrete to a highway construction site is uniformly distributed over the interval 50 to 70 minutes. What is the probability that the cycle time exceeds 65 minutes?

Answer 1

The probability that the cycle time exceeds 65 minutes
`P(Y>65|Y>55)` is
`(1)/(3)`.

∴
`P(Y>65|Y>55)=(1)/(3)`.

Explanation:

Given that the cycle time for trucks to uniformly distributed over the interval (50,70)

Let Y be the random variable for cycle time

To find P(Y>65|Y>55) :

We know that The density function is inversely proportional to the length of a interval for a uniform distribution ( and 0 elsewhere)

`f(y)=\left\{\begin{array}{l}(1)/(70-50)=(1)/(20),50<y<70\\ 0, elsewhere\end{array}\right.`

The formula for Conditional probability is

`P(A|B)=(P(A\bigcap B))/(P(B))`

Now we have that
`P(Y>65|Y>55)=(P(Y>65\bigcap Y>55))/(P(Y>55))`

Now we have to first find P(Y>65|Y>55) :

We have the common interval between interval Y>55 is (55,70) and Y>65 is (65,70)

∴
`Y>55\bigcap Y>65=Y>65`

∴
`P(Y>55\bigcap Y>65)=P(Y>65)`

`=P(65<Y<70)`

∴
`P(Y>55\bigcap Y>65)=P(65<Y<70)`

The formula
`P(a<Y<b)=\int\limits_a^b f(y)dy`

So we can write as
`P(65<Y<70)=\int\limits_(65)^(70) f(y)dy`

Since
`f(y)=(1)/(20)`
`\forall` y∈(50,70)

`P(65<Y<70)=\int\limits_(65)^(70) ((1)/(20))dy`

`=(1)/(20)y]\limits_(65)^(70)`

`=(1)/(20)(70-65)`

`=(1)/(20)(5)`

`=0.25`

∴
`P(65<Y<70)=0.25`

Similarly we find P(Y>55)=P(55<Y<70)

`P(55<Y<70)=\int\limits_(55)^(70) ((1)/(20))dy`

`=(1)/(20)y]\limits_(55)^(70)`

`=(1)/(20)(70-55)`

`=(1)/(20)(25)`

`=0.75`

∴
`P(55<Y<70)=0.75`

`P(Y>65|Y>55)=(Y>55\bigcap Y>65)/(P(Y>55))`

`=(0.25)/(0.75)` ( substituting the values)

`=(1)/(3)`

∴
`P(Y>65|Y>55)=(1)/(3)`.

The probability that the cycle time exceeds 65 minutes
`P(Y>65|Y>55)` is
`(1)/(3)`.