Respuesta :
Answer:
The probability that the second pipe (with length Y) is more than 0.11 feet longer than the first pipe (with length X) is 0.2059
Step-by-step explanation:
Let X and Y represent the lengths of two different pipes produced by the process.
Density function of X and Y is :
[tex]f(X)=\frac{1}{b-a}=\frac{1}{10.57-10}=\frac{1}{0.57}\\f(Y)=\frac{1}{b-a}=\frac{1}{10.57-10}=\frac{1}{0.57}[/tex]
Expected length of 1 pipe is :
[tex]E(X)=\int_{x} x \cdot f(x) = \int_{10}^{10.57} x \frac{1}{0.57}=10.285 feet[/tex]
[tex]E(X^2)=\int_{x} x^2 \cdot f(x) = \int_{10}^{10.57} x^2 \frac{1}{0.57}=105.808 feet[/tex]
The variance of length of pipe :
[tex]V(X)=E(X^2)-(E(X))^2\\V(X)=105.808-(10.285)^2=0.0268[/tex]
To Find the probability that the second pipe (with length Y) is more than 0.11 feet longer than the first pipe (with length X)
[tex]P(Y>0.11+X)=P(Y-X>0.11)\\P(Y>0.11+X)=P(\frac{(Y-X)-E(Y-X)}{SD(Y-X)}\\P(Y>0.11+X)=P(\frac{0.11-[E(Y)-E(X)]}{\sqrt{V(Y)+V(X)}}\\P(Y>0.11+X)=P(Z>\frac{0.11-(10.285-10.285)}{\sqrt{0.0268+0.0268}}\\P(Y>0.11+X)=P(Z>0.8207)\\P(Y>0.11+X)=1-P(Z\leq 0.8207)\\P(Y>0.11+X)=1-0.7941\\P(Y>0.11+X)=0.2059[/tex]
Hence the probability that the second pipe (with length Y) is more than 0.11 feet longer than the first pipe (with length X) is 0.2059
