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

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
1. For a discrete distribution the pmf = x/10   for x = 1, 2, 3, 4....
1. For a discrete distribution the pmf = x/10   for x = 1, 2, 3, 4. What is the expected value for this distribution? Give your answer to four decimal places. 2. The pmf for a discrete distribution is given by p(x) = 1/3 for x = -1, 0, 1. What is the variance of this distribution? Give your answer to four decimal places. 3. You have invented a new procedure for which the outcome can be classified as either...
The heights of pecan trees are normally distributed with a mean of 10 feet and a...
The heights of pecan trees are normally distributed with a mean of 10 feet and a standard deviation of 2 feet. 13. Show all work. Just the answer, without supporting work, will receive no credit. (a) What is the probability that a randomly selected pecan tree is between 9 and 12 feet tall? (Round the answer to 4 decimal places) (b) Find the 75th percentile of the pecan tree height distribution. (Round the answer to 2 decimal places) (c) For...
a)  The defect rate for a wire making process has a Poisson distribution with 1.23 defects per...
a)  The defect rate for a wire making process has a Poisson distribution with 1.23 defects per 1000 feet. Let X be the number of defects widgets in a randomly-selected 1000-foot length of wire. What is the probability that there will be no defects in that length of wire? State your answer rounded to three decimal digits. b) Let X be the net weight of a randomly-selected cereal box. The distribution of net weight for this filling process is uniformly-distributed between...
The defect length of a corrosion defect in a pressurized steel pipe is normally distributed with...
The defect length of a corrosion defect in a pressurized steel pipe is normally distributed with mean value 28 mm and standard deviation 7.8 mm. (a) What is the probability that defect length is at most 20 mm? Less than 20 mm? (Round your answers to four decimal places.) at most 20mmless than 20mm (b) What is the 75th percentile of the defect length distribution—that is, the value that separates the smallest 75% of all lengths from the largest 25%?...
The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0...
The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min. If one such class is randomly selected, find the probability that the class length is less than 51.3 min. P(X < 51.3) = (Report answer accurate to 2 decimal places.) The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min. If one such class is randomly selected, find the probability that the class length is between...
An experiment was designed to compare the lengths of time that four different drugs provided pain...
An experiment was designed to compare the lengths of time that four different drugs provided pain relief after surgery. The results (in hours) follow. Is there enough evidence to reject the null hypothesis that there is no significant difference in the length of pain relief provided by the four drugs at α = .05? Drug A B C D 4 5 11 3 5 4 9 3 3 3 12 4 6 5 11 10 (a) Find the test statistic....
The accompanying data represent x = treadmill run time to exhaustion (min) and y = 20...
The accompanying data represent x = treadmill run time to exhaustion (min) and y = 20 km ski time (min). x 7.7 8.4 8.7 9.0 9.6 9.6 10 10.3 10.5 11 11.8 y 71.0 71.4 65.0 68.7 64.4 69.4 63 64.9 66.7 62.2 61.6 sum x = 106.6 sum x^2 = 1047.44 sum y = 728.3 sum xy = 7024.64 sum y^2 = 48340.67 (a) Determine the equation of the estimated regression line. (Give the answers to four decimal places.)...
Define the joint pmf of two random variables X and Y to be: f(0, 10) =...
Define the joint pmf of two random variables X and Y to be: f(0, 10) = 1/24 f(0, 20) = 1/24 f(1, 10) = 1/8 f(1, 30) = 1/8 f(1, 20) = 1/4 f(2, 30) = 5/12 Give your answer to three decimal places. a) E[Y | X = 2] = b) E[Y] =
An experiment was designed to compare the lengths of time that four different drugs provided pain...
An experiment was designed to compare the lengths of time that four different drugs provided pain relief after surgery. The results (in hours) follow. Is there enough evidence to reject the null hypothesis that there is no significant difference in the length of pain relief provided by the four drugs at α = .05? Drug A B C D 7 6 8 3 2 5 12 4 3 5 10 4 7 4 8 8 (a) Find the test statistic....
An article suggests the uniform distribution on the interval (7.5, 20) as a model for depth...
An article suggests the uniform distribution on the interval (7.5, 20) as a model for depth (cm) of the bioturbation layer in sediment in a certain region. (a) What are the mean and variance of depth? (Round your variance to two decimal places.) mean      variance      (b) What is the cdf of depth? F(x) =      0      x < 7.5 7.5 ≤ x < 20      1 20 ≤ x (c) What is the probability that observed depth is at...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT