Question

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.

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 =0.5439^{2} =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 =0.57^{2}/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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