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Chapter 4: Problem 2

US The thickness of a flange on an aircraft component is uniformly distributed between 0.95 and 1.05 millimeters. Determine the following: a. Cumulative distribution function of flange thickness b. Proportion of flanges that exceeds 1.02 millimeters c. Thickness exceeded by \(90 \%\) of the flanges d. Mean and variance of flange thickness

Short Answer

Expert verified
a. CDF: \( F(x) = \frac{x - 0.95}{0.1} \); b. 30% exceed 1.02mm; c. 0.96mm exceeded by 90%; d. Mean = 1.00mm, Variance = 0.000833.

Step by step solution

01

Understand Uniform Distribution

A uniform distribution has a constant probability over a defined interval. For this problem, the flange thickness is uniformly distributed between 0.95 and 1.05 mm, making the lower limit \(a = 0.95\) and the upper limit \(b = 1.05\).
02

Cumulative Distribution Function (CDF)

The CDF \(F(x)\) for a uniform distribution is given by \( F(x) = \frac{x - a}{b - a} \). Substitute \(a = 0.95\) and \(b = 1.05\) to get \( F(x) = \frac{x - 0.95}{0.1} \) for \(0.95 \leq x \leq 1.05\).
03

Proportion Exceeding 1.02 mm

To find the proportion exceeding 1.02 mm, calculate \(1 - F(1.02)\). Substituting into the CDF, \( F(1.02) = \frac{1.02 - 0.95}{0.1} = 0.7 \). Thus, the proportion exceeding \(1.02\) mm is \(1 - 0.7 = 0.3\) or 30\%.
04

Thickness Exceeded by 90% of Flanges

To find the thickness exceeded by 90% of flanges, set \( F(x) = 0.1 \). Solve \( \frac{x - 0.95}{0.1} = 0.1 \) to get \(x = 0.96\) mm.
05

Mean and Variance

The mean \(\mu\) of a uniform distribution is \( \mu = \frac{a+b}{2} = \frac{0.95+1.05}{2} = 1.00\) mm. The variance \(\sigma^2\) is given by \( \sigma^2 = \frac{(b-a)^2}{12} = \frac{(1.05-0.95)^2}{12} = 0.000833\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cumulative Distribution Function
In a uniform distribution, every outcome in a certain range has an equal chance of occurring. To understand how likely a flange thickness is within or at a given point in this range, we use the Cumulative Distribution Function (CDF). The CDF, denoted as \( F(x) \), helps us calculate the probability that a random variable \( X \) is less than or equal to \( x \). The formula for the CDF in a uniform distribution is:

\[ F(x) = \frac{x - a}{b - a} \]
Where:
  • \(x\) is the point up to which we want to calculate the cumulative probability.
  • \(a\) is the start of the interval.
  • \(b\) is the end of the interval.
For the uniform distribution of flange thickness, the CDF would provide probabilities for thickness values from 0.95 to 1.05 mm.
Proportion Exceeding a Value
To determine the proportion of flange thicknesses that exceed a specific value, you can leverage the CDF. In this case, we are asked to find out how many flanges are thicker than 1.02 mm. You simply calculate \(1 - F(x)\), where \(F(x)\) is the cumulative distribution function evaluated at the specific thickness:

\[ F(1.02) = \frac{1.02 - 0.95}{0.1} = 0.7 \]
This means 70% of the flanges have a thickness of 1.02 mm or less, and thus:
- The proportion exceeding 1.02 mm is \(1 - 0.7 = 0.3\) or 30%
This method helps to understand how much of the distribution surpasses a particular threshold.
Mean and Variance
For any probability distribution, the mean and variance are essential statistics that describe its center and spread. In a uniform distribution:
  • The mean \( \mu \) is calculated as the midpoint of the interval, given by: \( \mu = \frac{a+b}{2} \)
  • The variance \( \sigma^2 \) measures how much the values differ from the mean, calculated as: \( \sigma^2 = \frac{(b-a)^2}{12} \)
For our flange thickness case:
- The mean thickness is \( \mu = \frac{0.95 + 1.05}{2} = 1.00 \) mm
- The variance is \( \sigma^2 = \frac{(1.05 - 0.95)^2}{12} = 0.000833 \) mm
These values highlight that the average flange thickness is 1.00 mm, with a very small spread or variability around this mean.
Probability Distribution
A probability distribution provides a mathematical function that describes the likelihood of various outcomes. For the flange thickness, it's described by a uniform distribution which tells us that every thickness between 0.95 and 1.05 mm is equally likely. Understanding this distribution involves:
  • Defining the interval \([a, b]\) where outcomes are possible.
  • Knowing that the probability density function (PDF) for a uniform distribution is \( f(x) = \frac{1}{b-a} \) for \(a \leq x \leq b\).
Thus, each possible thickness has an equal probability, creating a flat or 'uniform' appearance in the PDF. By recognizing this distribution, one can predict the behavior and characteristics of the flange thickness reliably.

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