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Chapter 4: Q122E (page 269)

Shear strength of rock fractures. Understanding the characteristics

of rock masses, especially the nature of the fracturesis essential when building dams and power plants.The shear strength of rock fractures was investigated inEngineering Geology(May 12, 2010). The Joint RoughnessCoefficient (JRC) was used to measure shear strength.Civil engineers collected JRC data for over 750 rock fractures.The results (simulated from information provided in the article) are summarized in the accompanying SPSShistogram. Should the engineers use the normal probabilitydistribution to model the behavior of shear strength forrock fractures? Explain

Short Answer

Expert verified

Engineers should use the normal probability distribution to model the behavior of shear strength for rock fractures

Step by step solution

01

Given Information

The histograms for 750 rock fractures is given,

02

Explanation

From the above histogram, it is seen that the histogram looks like a normal probability curve. So, the engineers can use normal distribution to model the behavior of shear strength for rock fractures.

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Most popular questions from this chapter

Shopping vehicle and judgment. Refer to the Journal of Marketing Research (December 2011) study of whether you are more likely to choose a vice product (e.g., a candy bar) when your arm is flexed (as when carrying a shopping basket) than when your arm is extended (as when pushing a shopping cart), Exercise 2.85 (p. 112). The study measured choice scores (on a scale of 0 to 100, where higher scores indicate a greater preference for vice options) for consumers shopping under each of the two conditions. Recall that the average choice score for consumers with a flexed arm was 59, while the average for consumers with an extended arm was 43. For both conditions, assume that the standard deviation of the choice scores is 5. Also, assume that both distributions are approximately normally distributed.

a. In the flexed arm condition, what is the probability that a consumer has a choice score of 60 or greater?

b. In the extended arm condition, what is the probability that a consumer has a choice score of 60 or greater?

4.110 Manufacturing hourly pay rate. Government data indicate that the mean hourly wage for manufacturing workers in the United States is \(20.10 (Bureau of Labor Statistics, January 2016). Suppose the distribution of manufacturing wage rates nationwide can be approximated by a normal distribution with a standard deviation \)1.25 per hour. The first manufacturing firm contacted by a particular worker seeking a new job pays \(21.40 per hour.

a. If the worker were to undertake a nationwide job search, approximately what proportion of the wage rates would be greater than \)21.40 per hour?

b. If the worker were to randomly select a U.S. manufacturing firm, what is the probability the firm would pay more than $21.40 per hour?

c. The population median, call it η, of a continuous random

variable xis the value such that P(x≥η)=P(x≤η)=0.5that is, the median is the value such that half the area under the probability distribution lies above and half lies below it. Find the median of the random variable corresponding to the wage rate and compare it with the mean wage rate.

Compute the following:

a.7!3!(7−3)!

b.(94)

c. (50)

d.(44)

e.(54)

NHTSA crash tests. Refer to the NHTSA crash tests of new car models, Exercise 4.3 (p. 217). A summary of the driver-side star ratings for the 98 cars in the file is reproduced in the accompanying Minitab printout. Assume that one of the 98 cars is selected randomly and let x equal the number of stars in the car’s driver-side star rating.

  1. Use the information in the printout to find the probability distribution for x.
  2. FindP(x=5).
  3. FindP(x≤2).
  4. Find μ=E(x)and practically interpret the result.

Testing for spoiled wine. Suppose that you are purchasing cases of wine (12 bottles per case) and that, periodically, you select a test case to determine the adequacy of the bottles’ seals. To do this, you randomly select and test 3 bottles in the case. If a case contains 1 spoiled bottle of wine, what is the probability that this bottle will turn up in your sample?

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