Respuesta :
Answer:
The probability that the cycle time exceeds 65 minutes [tex]P(Y>65|Y>55)[/tex] is [tex]\frac{1}{3}[/tex].
∴ [tex]P(Y>65|Y>55)=\frac{1}{3}[/tex].
Step-by-step 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)
[tex]f(y)=\left\{\begin{array}{l}\frac{1}{70-50}=\frac{1}{20},50<y<70\\ 0, elsewhere\end{array}\right.[/tex]
The formula for Conditional probability is
[tex]P(A|B)=\frac{P(A\bigcap B)}{P(B)}[/tex]
Now we have that [tex]P(Y>65|Y>55)=\frac{P(Y>65\bigcap Y>55)}{P(Y>55)}[/tex]
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)
∴ [tex]Y>55\bigcap Y>65=Y>65[/tex]
∴ [tex]P(Y>55\bigcap Y>65)=P(Y>65)[/tex]
[tex]=P(65<Y<70)[/tex]
∴ [tex]P(Y>55\bigcap Y>65)=P(65<Y<70)[/tex]
The formula [tex]P(a<Y<b)=\int\limits_a^b f(y)dy[/tex]
So we can write as [tex]P(65<Y<70)=\int\limits_{65}^{70} f(y)dy[/tex]
Since [tex]f(y)=\frac{1}{20}[/tex] [tex]\forall[/tex] y∈(50,70)
[tex]P(65<Y<70)=\int\limits_{65}^{70} (\frac{1}{20})dy[/tex]
[tex]=\frac{1}{20}y]\limits_{65}^{70}[/tex]
[tex]=\frac{1}{20}(70-65)[/tex]
[tex]=\frac{1}{20}(5)[/tex]
[tex]=0.25[/tex]
∴ [tex]P(65<Y<70)=0.25[/tex]
Similarly we find P(Y>55)=P(55<Y<70)
[tex]P(55<Y<70)=\int\limits_{55}^{70} (\frac{1}{20})dy[/tex]
[tex]=\frac{1}{20}y]\limits_{55}^{70}[/tex]
[tex]=\frac{1}{20}(70-55)[/tex]
[tex]=\frac{1}{20}(25)[/tex]
[tex]=0.75[/tex]
∴ [tex]P(55<Y<70)=0.75[/tex]
[tex]P(Y>65|Y>55)=\frac{Y>55\bigcap Y>65}{P(Y>55)}[/tex]
[tex]=\frac{0.25}{0.75}[/tex] ( substituting the values)
[tex]=\frac{1}{3}[/tex]
∴ [tex]P(Y>65|Y>55)=\frac{1}{3}[/tex].
The probability that the cycle time exceeds 65 minutes [tex]P(Y>65|Y>55)[/tex] is [tex]\frac{1}{3}[/tex].
