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 10.85 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.
What is the probability that a single pipe will be between 10.07 feet and 10.39 feet long?
You need a double integral.
integral from 10 to 10.85
integral from 10.07 to 10.39 ( only one changed because it is asking for a single pipe.)
integrate the function 1/(10.85-10)^2
To find the probability that a single pipe will be between 10.07 feet and 10.39 feet long, we need to calculate the area under the probability density function (PDF) curve between these two values.
Given that the distribution of pipe lengths follows a Uniform distribution from 10 feet to 10.85 feet, we can determine the probability density function as follows:
- The PDF is constant over the range of the distribution, which means that the height of the PDF curve is constant.
- The width of the range is 0.85 feet, so the width of the PDF curve is 0.85 feet.
- To find the height of the PDF curve, we divide 1 by the width to ensure that the area under the curve is equal to 1.
Therefore, the PDF of the pipe length distribution is:
f(x) = 1 / 0.85
To calculate the probability that a single pipe will have a length between 10.07 feet and 10.39 feet, we need to find the area under the PDF curve between these two values.
P(10.07 ≤ X ≤ 10.39) = ∫[10.07, 10.39] f(x) dx
Integrating the PDF function between 10.07 and 10.39 gives us:
P(10.07 ≤ X ≤ 10.39) = (10.39 - 10.07) * f(x)
P(10.07 ≤ X ≤ 10.39) = (10.39 - 10.07) / 0.85
Calculating this value will give us the probability that a single pipe will be between 10.07 feet and 10.39 feet long.