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Chapter 5: Problem 30

A company that manufactures rivets believes the shear strength (in pounds) is modeled by \(N(800,50)\) a) Draw and label the Normal model. b) Would it be safe to use these rivets in a situation requiring a shear strength of 750 pounds? Explain. c) About what percent of these rivets would you expect to fall below 900 pounds? d) Rivets are used in a variety of applications with varying shear strength requirements. What is the maximum shear strength for which you would feel comfortable approving this company's rivets? Explain your reasoning.

Short Answer

Expert verified
15.87% of rivets are below 750 pounds. 97.72% are below 900 pounds. Maximum safe is about 717.75 pounds.

Step by step solution

01

Understanding the Normal Model

The shear strength of the rivets is given by a normal distribution with a mean (\(\mu\)) of 800 pounds and a standard deviation (\(\sigma\)) of 50 pounds. To draw and label the normal model, we plot a bell-shaped curve centered at 800 with points marked at \(\mu \pm \sigma, \mu \pm 2\sigma,\) and \(\mu \pm 3\sigma\). This gives us points at 750, 800, 850 for \(\pm \sigma\), 700, 900 for \(\pm 2\sigma\), and 650, 950 for \(\pm 3\sigma\).
02

Analyzing Rivet Safety for 750 Pounds

To determine if a rivet with 750 pounds of shear strength is safe, calculate how often these rivets fall below this threshold using the Z-score: \(Z = \frac{750 - 800}{50} = -1\). Consulting the standard normal distribution table, a Z-score of -1 corresponds to the 15.87th percentile. Thus, about 15.87% of rivets have a shear strength less than 750 pounds, meaning most exceed this threshold.
03

Calculating Rivet Strength Below 900 Pounds

To find what percent of rivets are below 900 pounds, calculate the Z-score: \(Z = \frac{900 - 800}{50} = 2\). A Z-score of 2 corresponds to the 97.72nd percentile. Therefore, about 97.72% of rivets fall below 900 pounds of shear strength.
04

Determining Maximum Safe Shear Strength

To find a maximum safe shear strength, we assume it keeps at least 95% of rivets above this value, aiming for a comfort margin similar to common engineering practices. Calculate the Z-score at the 5th percentile, which is -1.645. Then find the shear strength: \(800 + (-1.645 \times 50) = 717.75\). Thus, the maximum shear strength requirement for assurance is about 717.75 pounds.

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Key Concepts

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

Shear Strength
Shear strength refers to the ability of a material to resist forces that can cause its internal structure to slide against itself. In engineering applications, this measurement is crucial as it determines how well materials can withstand loads without failing. The given exercise involves rivets, where their shear strength is assumed to follow a normal distribution. The normal distribution helps us understand how shear strength values are spread around a mean, indicating the typical performance of the rivets.
Rivets with a shear strength modeled by a normal distribution centered around 800 pounds with a standard deviation of 50 pounds allow engineers to predict product behavior and performance under different loads.
Standard Deviation
Standard deviation is a measure of how much variation or dispersion exists from the average or mean value. In the context of this exercise, the standard deviation of 50 pounds indicates how much the shear strengths of the rivets deviate from the average of 800 pounds.
  • A smaller standard deviation would mean the rivets' shear strength values are closely clustered around the mean, indicating more consistent quality.
  • A larger standard deviation signifies more variability, implying a wider range of shear strength values which could affect reliability.
Understanding standard deviation helps engineers assess the consistency of the product, ensuring it meets specific safety standards.
Z-Score
A Z-score represents the number of standard deviations a data point is from the mean. In the exercise, the Z-score is used to determine how unusual a given shear strength value is within the normal distribution.
For example, a Z-score of -1, calculated for a shear strength of 750 pounds, indicates that this value is one standard deviation below the mean of 800 pounds. By computing the Z-score, engineers can use statistical tables to find out the percentage of rivets expected to fall below a specific strength, aiding in quality control and safety assessments.
Knowing the Z-score helps engineers and manufacturers decide if a product meets the required strength criteria safely.
Percentile
Percentiles are values below which a certain percentage of data in a data set falls. For instance, the 15.87th percentile in our exercise signifies that 15.87% of the rivets have a shear strength below 750 pounds, while the 97.72nd percentile indicates that 97.72% of rivets fall below 900 pounds.
  • Percentiles are useful for understanding how a specific shear strength value compares to the rest of the data.
  • They help engineers evaluate the proportion of rivets meeting or surpassing specific performance criteria.
By considering percentiles, engineers can ensure that they select materials or components that meet the desired reliability levels in their applications.
Engineering Applications
In engineering applications, understanding the normal distribution of shear strength is vital for ensuring safety and reliability. Rivets are commonly used in structures to hold parts together, so knowing their shear strength can be a matter of public safety.
For instance, in civil engineering, bridges or buildings must use rivets that meet certain shear strength standards to prevent structural failure.
  • Engineers assess if the majority of the rivets exceed necessary safety thresholds, opting for designs that ensure at least 95% of components can withstand the required loads.
  • By understanding statistical concepts such as the normal distribution, standard deviation, and percentiles, engineers can make informed decisions about materials, ensuring they meet rigorous industry standards for safety and performance.
Through such analyses, engineers can contribute to safer constructions and machinery that resist everyday stresses effectively.

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Most popular questions from this chapter

Assume the cholesterol levels of adult American women can be described by a Normal model with a mean of \(188 \mathrm{mg} / \mathrm{dL}\) and a standard deviation of 24 a) Draw and label the Normal model. b) What percent of adult women do you expect to have cholesterol levels over \(200 \mathrm{mg} / \mathrm{dL} ?\) c) What percent of adult women do you expect to have cholesterol levels between 150 and \(170 \mathrm{mg} / \mathrm{dL} ?\) d) Estimate the IQR of the cholesterol levels. e) Above what value are the highest \(15 \%\) of women's cholesterol levels?

Consider the Angus weights model \(N(1152,84)\) one last time. a) What weight represents the 40th percentile? b) What weight represents the 99th percentile? c) What's the IQR of the weights of these Angus steers?

The first Stat exam had a mean of 80 and a standard deviation of 4 points; the second had a mean of 70 and a standard deviation of 15 points. Reginald scored an 80 on the first test and an 85 on the second. Sara scored an 88 on the first but only a 65 on the second. Although Reginald's total score is higher, Sara feels she should get the higher grade. Explain her point of view.

Environmental Protection Agency (EPA) fuel economy estimates for automobile models tested recently predicted a mean of \(24.8 \mathrm{mpg}\) and a standard deviation of 6.2 mpg for highway driving. Assume that a Normal model can be applied. a) Draw the model for auto fuel economy. Clearly label it, showing what the \(68-95-99.7\) Rule predicts. b) In what interval would you expect the central \(68 \%\) of autos to be found? c) About what percent of autos should get more than \(31 \mathrm{mpg} ?\) d) About what percent of cars should get between 31 and \(37.2 \mathrm{mpg} ?\) e) Describe the gas mileage of the worst \(2.5 \%\) of all cars.

For the car speed data of Exercise 18 recall that the mean speed recorded was \(23.84 \mathrm{mph}\) with a standard deviation of \(3.56 \mathrm{mph}\). To see how many cars are speeding, John subtracts \(20 \mathrm{mph}\) from all speeds. a) What is the mean speed now? What is the new standard deviation? b) His friend in Berlin wants to study the speeds, so John converts all the original miles-per-hour readings to kilometers per hour by multiplying all speeds by 1.609 (km per mile). What is the mean now? What is the new standard deviation?
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