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Chapter 5: Problem 66

A U-shaped component is to be formed from the three parts \(A, B,\) and \(C .\) The picture is shown in Fig. \(5-20 .\) The length of \(A\) is normally distributed with a mean of 10 millimeters and a standard deviation of 0.1 millimeter. The thickness of parts \(B\) and \(C\) is normally distributed with a mean of 2 millimeters and a standard deviation of 0.05 millimeter. Assume all dimensions are independent. (a) Determine the mean and standard deviation of the length of the gap \(D\). (b) What is the probability that the gap \(D\) is less than 5.9 millimeters?

Short Answer

Expert verified
(a) Mean: 6 mm, SD: 0.1118 mm. (b) Probability: 0.1859.

Step by step solution

01

Define the Length of the Gap D

The gap \(D\) is given by the expression \(D = A - (B + C)\), where \(A\) is the length of the middle part, \(B\) and \(C\) are the thicknesses of the side parts.
02

Calculate Mean of D

Using the linearity of expectation, the mean of \(D\) is \(\mu_D = \mu_A - (\mu_B + \mu_C)\). Given that \(\mu_A = 10\) mm, \(\mu_B = 2\) mm, and \(\mu_C = 2\) mm, we find \(\mu_D = 10 - (2 + 2) = 6\) mm.
03

Determine Standard Deviations for Each Part

The standard deviations are already provided: \(\sigma_A = 0.1\) mm, \(\sigma_B = 0.05\) mm, and \(\sigma_C = 0.05\) mm.
04

Calculate Standard Deviation of D

Since \(A\), \(B\), and \(C\) are independent, the variance of \(D\) is \(\sigma_D^2 = \sigma_A^2 + \sigma_B^2 + \sigma_C^2\). Thus, \(\sigma_D^2 = 0.1^2 + 0.05^2 + 0.05^2 = 0.0125\). So, \(\sigma_D = \sqrt{0.0125} \approx 0.1118\) mm.
05

Calculate Probability D < 5.9

Convert the problem to a standard normal distribution problem by using the Z-score: \(Z = \frac{D - \mu_D}{\sigma_D}\). Thus, \(Z = \frac{5.9 - 6}{0.1118} \approx -0.894\). Using Z-tables or a calculator, the probability \(P(Z < -0.894)\) gives approximately 0.1859.

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Key Concepts

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

Normal Distribution
In a normal distribution, data is symmetrically spread around a mean value, depicting the famous bell curve shape. This natural pattern is common in real-world phenomena like height or test scores. For our U-shaped component problem, the lengths of parts are normally distributed, meaning:
  • The majority of lengths cluster around the mean.
  • As lengths deviate further from the mean, they become less common.
The normal distribution is defined by two key parameters: mean (\( \mu \)) and standard deviation (\( \sigma \)). For example, the length of part A has a mean of 10 mm and a standard deviation of 0.1 mm. This tells us that most lengths will hover near 10 mm, with a higher likelihood of observing values within one standard deviation (9.9 to 10.1 mm) of the mean.
Standard Deviation
Standard deviation is a measure of dispersion within a dataset. It indicates how spread out numbers are around the mean. A lower standard deviation implies data points are closely clustered, while a higher means they are more spread out.For our exercise:
  • Part A has \( \sigma_A = 0.1 \) mm, indicating tight clustering around the mean of 10 mm.
  • Parts B and C have \( \sigma_B = \sigma_C = 0.05 \) mm, meaning their values are similarly tightly packed around the mean of 2 mm.
Calculating the standard deviation of a sum or difference of independent variables involves summing their variances, \( \sigma^2 \). When these values are combined to find the gap D鈥檚 standard deviation, the result is approximately 0.1118 mm, reflecting the combined variability of all parts involved.
Expectation and Variance
Expectation, or expected value, is the theoretical mean of a random variable, indicating where values tend to cluster in the long run. Variance describes the spread of these values around the mean.To determine these for the gap D, we use the following principles:
  • Expectation is linear: To find \( \mu_D \), calculate: \( \mu_D = \mu_A - (\mu_B + \mu_C) \). Having \( \mu_A = 10 \) mm, \( \mu_B = 2 \) mm, and \( \mu_C = 2 \) mm, results in \( \mu_D = 6 \) mm.
  • Variance for independent variables adds up: \( \sigma_D^2 = \sigma_A^2 + \sigma_B^2 + \sigma_C^2 = 0.0125 \). Taking the square root yields the standard deviation, \( \sigma_D = 0.1118 \) mm.
Knowing these values allows us to further analyze probabilities or potential outcomes involving D.
Independent Random Variables
Variables are considered independent when the occurrence of one does not affect the probability of another. This is key in many statistical calculations and probability assessments. In our context:
  • The lengths of A, B, and C are all assumed to be independent. This independence makes it possible to combine their variances to find the variance of any resulting sum or difference, like gap D.
  • Independence simplifies the computation of probabilities and other statistical measures, as there is no need to account for dependencies between variables.
By understanding and assuming independence, we confidently deduce the spread (variance and standard deviation) and mean of D, leading to easier calculations of probabilities, such as D being less than a specific threshold.

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Most popular questions from this chapter

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