To assemble a piece of furniture, a wood peg must be inserted into a predrilled hole. Suppose that the diameter of a randomly selected peg is a random variable with mean 0.33 in. and standard deviation 0.008 in. and that the diameter of a randomly selected hole is a random variable with mean 0.334 in. and standard deviation 0.003 in. Let x₁ denote peg diameter, and let x₂ denote hole diameter. (a) Why would the random variable y, defined as y = X₂X₁, be of interest to the furniture manufacturer? O If y is negative, the peg will not fit in the hole. O If y is positive, the peg will not fit in the hole. O y must be zero for the peg to be inserted in the hole. (b) What is the mean value (in inches) of the random variable y? in (c) Assuming that x₁ and x₂ are independent, what is the standard deviation (in inches) of y? (Round the answer to three decimal places.) in (d) Based on your answers to parts (b) and (c), do you think that finding a peg that is too big to fit in the predrilled hole would be a relatively common or a relatively rare occurrence? Explain. Since 0 is ---Select--- standard deviation(s) from the mean in the distribution of y, this will be a relatively --Select--- occurrence.
To assemble a piece of furniture, a wood peg must be inserted into a predrilled hole. Suppose that the diameter of a randomly selected peg is a random variable with mean 0.33 in. and standard deviation 0.008 in. and that the diameter of a randomly selected hole is a random variable with mean 0.334 in. and standard deviation 0.003 in. Let x₁ denote peg diameter, and let x₂ denote hole diameter. (a) Why would the random variable y, defined as y = X₂X₁, be of interest to the furniture manufacturer? O If y is negative, the peg will not fit in the hole. O If y is positive, the peg will not fit in the hole. O y must be zero for the peg to be inserted in the hole. (b) What is the mean value (in inches) of the random variable y? in (c) Assuming that x₁ and x₂ are independent, what is the standard deviation (in inches) of y? (Round the answer to three decimal places.) in (d) Based on your answers to parts (b) and (c), do you think that finding a peg that is too big to fit in the predrilled hole would be a relatively common or a relatively rare occurrence? Explain. Since 0 is ---Select--- standard deviation(s) from the mean in the distribution of y, this will be a relatively --Select--- occurrence.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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PLEASE ANSWER ALL THE PARTS OF THE QUESTION( THIS IS NOT A GRADED ASSIGNMENT)
Transcribed Image Text:To assemble a piece of furniture, a wood peg must be inserted into a predrilled hole. Suppose that the diameter of a randomly selected peg is a random variable
with mean 0.33 in. and standard deviation 0.008 in. and that the diameter of a randomly selected hole is a random variable with mean 0.334 in. and standard
deviation 0.003 in. Let x₁ denote peg diameter, and let x₂ denote hole diameter.
(a) Why would the random variable y, defined as y = x₂ - X₁, be of interest to the furniture manufacturer?
O If y is negative, the peg will not fit in the hole.
O If y is positive, the peg will not fit in the hole.
O y must be zero for the peg to be inserted in the hole.
(b) What is the mean value (in inches) of the random variable y?
in
(c) Assuming that x₁ and x₂ are independent, what is the standard deviation (in inches) of y? (Round the answer to three decimal places.)
in
(d) Based on your answers to parts (b) and (c), do you think that finding a peg that is too big to fit in the predrilled hole would be a relatively common or a
relatively rare occurrence? Explain.
Since 0 is ---Select---
standard deviation(s) from the mean in the distribution of y, this will be a relatively ---Select--- ✓ occurrence.
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