Question

A manufacturer has designed a process to produce pipes that are 10 feet long. The distribution...

A manufacturer has designed a process to produce pipes that are 10 feet long. The distribution of the pipe length, however, is actually Uniform on the interval 10 feet to 10.57 feet. Assume that the lengths of individual pipes produced by the process are independent. Let X and Y represent the lengths of two different pipes produced by the process.

a) What is the joint pdf for X and Y? Choose one of the following.

f(x,y) = xy 10 < x < 10.57, 10 < y < 10.57

f(x,y) = 1 10 < x < 10.57, 10 < y < 10.57  

f(x,y) = 1/(0.57)2 10 < x < 10.57, 10 < y < 10.57

f(x,y) = 1/(0.57)2   10 < x < 11, 10 < y < 11


b) What is the probability that a single pipe will be between 10.14 feet and 10.45 feet long? Give your answer to four decimal places.


c) What is the probability that both pieces of pipe are between 10.14 feet and 10.45 feet long? Give your answer to four decimal places. Hint: Try to avoid doing calculus to solve this problem.


d) What is the expected length of a single pipe? Give your answer to three decimal places.


e) What is the expected total length of the two pieces of pipe? Give your answer to three decimal places.


f) What is the variance of the length of a single pipe? Give your answer to four decimal places.


g) What is the variance of the total length of both pipes? Give your answer to four decimal places.


h)What is the probability that the second pipe (with length Y) is more than 0.27 feet longer than the first pipe (with length X)? Give your answer to four decimal places. Hint: Do not use calculus to get your answer.

Homework Answers

Answer #1

a)f(x,y) = 1/(0.57)2 10 < x < 10.57, 10 < y < 10.57

b) probability that a single pipe will be between 10.14 feet and 10.45 feet long=(10.45-10.14)/0.57 =0.5439

c)probability that both pieces of pipe are between 10.14 feet and 10.45 feet long =0.54392 =0.2958

d)expected length of a single pipe=(10+10.57)/2=10.285

e) expected total length of the two pieces of pipe=2*10.285 =20.57

f)variance of the length of a single pipe =0.572/12 =0.3037

g) variance of the total length of both pipes 0.3037+0.3037 =0.6074

h) probability that the second pipe (with length Y) is more than 0.27 feet longer than the first pipe

=(1/2)*(0.57-0.27)*(0.57-0.27)/(0.57*0.57)=0.1385

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