Question

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 \(0.25\) in. and standard deviation \(0.006\) in. and that the diameter of a randomly selected hole is a random variable with mean \(0.253\) in. and standard deviation \(0.002\) in. Let \(x_{1}=\) peg diameter, and let \(x_{2}=\) denote hole diameter. a. Why would the random variable \(y\), defined as \(y=\) \(x_{2}-x_{1}\), be of interest to the furniture manufacturer? b. What is the mean value of the random variable \(y\) ? c. Assuming that \(x_{1}\) and \(x_{2}\) are independent, what is the standard deviation of \(y\) ? d. Is it reasonable to think that \(x_{1}\) and \(x_{2}\) are independent? Explain. e. 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.

Short answer

The mean value of the random variable y is 0.003 in. and its standard deviation is 0.00632 in. Whether the variables peg diameter and hole diameter are independent depends on the specifics of the manufacturing process. However, given the calculated mean and standard deviation, it would be a relatively rare occurrence to find a peg that is too big for its hole, assuming a normal distribution model.

Step-by-step solution

Understanding the Variable y

The variable \(y = x_{2} - x_{1}\) represents the discrepancy between the diameter of the predrilled hole and the diameter of the wood peg. This would be of interest to the furniture manufacturer because if \(y\) is negative (i.e. the hole is smaller than the peg), the peg will not fit into the hole and the piece of furniture cannot be assembled.

Calculating the Mean of y

The mean value of random variable \(y\) (peg's diameter subtracted from hole's diameter), can be computed as the difference of the means of \(x_{1}\) and \(x_{2}\). Hence, the mean value of \(y\) is \(0.253 - 0.25 = 0.003\) in.

Computing the Standard Deviation of y

Given that \(x_{1}\) and \(x_{2}\) are stated to be independent, the variance of \(y\) would be the sum of the variances of \(x_{1}\) and \(x_{2}\). The standard deviation of \(y\) would thereby be the square root of this sum. Hence, standard deviation of \(y\) = \(\sqrt{(0.006^{2} + 0.002^{2})} = 0.00632\) in.

Assessing the Independence of Variables

While we have so far assumed \(x_{1}\) and \(x_{2}\) to be independent for computing purpose, in reality it depends on the way the manufacturing process is designed. If the size of the pegs and holes are determined separately without any mutual influence, they would indeed be independent. However, if one's size affects the other's (as in processes designed with such interdependencies), they may not be independent.

Estimating the Probability of a Peg not Fitting into the Hole

With known mean and standard deviation, if it can be assumed that \(y\) is normally distributed, it is relatively rare for it to have a value more than three standard deviations away from the mean. This means that it would be unlikely to find a peg that is too big for its hole since pegs that are larger than the hole to a significant degree (0.00632*3 = 0.01896 in. greater than the mean diameter difference) would be a rare occurrence in a normal distribution model.