William Navidi received a B. A in mathematics from Michigan State University, and a Ph. Navidi is a professor of applied mathematics and statistics at the Colorado School of Mines in Golden, Colorado. He has taught mathematics and statistics at all levels, from developmental through the graduate level.
This is a very large number. Solutions for Exercises in Chapter 2 19 2. Solutions for Exercises in Chapter 2 21 2. Solutions for Exercises in Chapter 2 23 2. Denote by H the event that you picked door A and the host opened door B, while there is no prize behind the door B. So, more likely engineer 1 did the job. Chapter 3 Random Variables and Probability Distributions 3. Let S and N stand for a spade and not a spade, respectively.
Let D and N stand for a dime and nickel, respectively. Let G and B stand for the colors of green and black, respectively. It is not extremely rare. So, this is a density function. Solutions for Exercises in Chapter 3 33 y 3. The probability for one combination of such a situation is 5! Since there are y! So, the conjecture is false. A random selection of 4 pieces of fruit can be made in 4 ways. Solutions for Exercises in Chapter 3 35 b From the row totals of Exercise 3.
Let W, Z represent a typical outcome of the experiment. The particular outcome 1, 0 indicating a total of 1 head and no heads on the first toss corresponds to the event T H. Similar calcu- lations for the outcomes 0, 0 , 1, 1 , and 2, 1 lead to the following joint probability distribution: w f w, z 0 1 2 z 0 0.
Solutions for Exercises in Chapter 3 37 1 1 3. Solutions for Exercises in Chapter 3 39 1. This is a continuous uniform distribution.
So, fX1 x1 is a density function. Chapter 4 Mathematical Expectation 4. This should not be surprised due to the symmetry of the density at Solutions for Exercises in Chapter 4 45 4. So, in the actual profit, 1 the variance is 18 2. Solutions for Exercises in Chapter 4 47 4. Using the approximation formula, we have e 2 7. One reason is that the first order approxi- mation may not always be good enough. Since 0. Since there will be 70 positions, the applicant will have the job.
Solving 0. So, we need to calculate the average of this quantity. The marginal densities of X and Y are, respectively, x 0 1 2 y 0 1 2 3 4 5 g x 0. Chapter 5 Some Discrete Probability Distributions 5. This probability is not very small so this is not a rare event.
Solutions for Exercises in Chapter 5 57 5. Solutions for Exercises in Chapter 5 59 5. Solutions for Exercises in Chapter 5 61 5. However, the 1st one can be either bad or good. So, there is a small prospects for bankruptcy. Perhaps more items should be sampled. Solutions for Exercises in Chapter 5 65 22 30 5. Therefore, the claim does not seem right.
Chapter 6 Some Continuous Probability Distributions 1 6. Therefore, from Table A. From Table A. Therefore, the total area to the left of k is 0. Therefore, 0. Therefore, Therefore, 1. He is late P Fraction of poodles weighing over 9. Fraction of poodles weighing at most 8.
Fraction of poodles weighing between 7. Proportion of components exceeding That is from 0 to Let Y be the number of days a person is served in less than 3 minutes. To compute median, notice the c. Hence a product is undesirable is 2. However, for smaller values such as 10, the normal population will give you smaller probabilities. Since the average time between two calls in 6. Therefore, the mean and variance of the number of calls per hour should all be 6. The negative number in reaction time is not reasonable.
So, it means that the normal model may not be accurate enough. Thus the drill bit of problem 6. A drill bit is a mechanical part that certainly will have significant wear over time. Hence the exponential distribution would not apply.
Chapter 7 Functions of Random Variables 7. Solutions for Exercises in Chapter 7 81 7. Solutions for Exercises in Chapter 7 83 7. This is a uniform 0,1 distribution. The mean should not be used on account of the extreme value 95, and the mode is not desirable because the sample size is too small. Therefore, the variance of the sample mean is reduced from 0.
Therefore, the variance of the sample mean is increased from 0. So, 8. So, P Therefore, the number of sample means between Therefore, about 0. There- fore, the mean amount to be 0. So, P 3. This means that the assumption of the equality of the population means are not reasonable. Solutions for Exercises in Chapter 8 89 8. No, this is not very strong evidence that the population mean of the process exceeds the government limit.
Conclusion values are not valid. Since the value Solutions for Exercises in Chapter 8 91 8. So, the result is inconclusive.
Since, from Table A. Hence the variances may not be equal. Hence, by solving 2. Hence, by equating 2. Note that Table A. However, one can deduce the conclusion based on the values in the last line of the table. Also, computer software gives the value of 0. Also, z0. Solutions for Exercises in Chapter 9 99 9. The tolerance interval is So, The tolerance interval is 1.
So, the upper limit is 3. Since t0. So, the interval is So, the tolerance interval is So, the tolerance limit is Since 62 exceeds the lower bound of the interval, yes, this is a cause of concern. Since 6. The lower bound of the one-sided tolerance interval is Their claim is not necessarily correct. Hence, the tolerance limits are 1. Hence the claim is valid. Solutions for Exercises in Chapter 9 9. It is known that z0. The treatment appears to reduce the mean amount of metal removed.
So, 4. Solutions for Exercises in Chapter 9 80 40 9. From this study we conclude that there is a sig- nificantly higher proportion of women in electrical engineering than there is in chemical engineering. So, 4 0. Hence, 19 6. Hence, 8 0. Hence, 11 2.
So, 1. This is called the largest order statistic of the sample. Then solve them to obtain the maximum likelihood estimates. Solutions for Exercises in Chapter 9 28 9. Since z0. Since the interval contains 0.
So, 2. Since 0 is not in the interval, the claim appears valid. Hence, polishing does increase the average endurance limit. Solutions for Exercises in Chapter 9 b Since the sample sizes are large enough, it is not necessary to assume the normality due to the Central Limit Theorem. It is known that f0. Hence, the sample sizes in Review Exercise 9. Using Table A. So, the limit of the one-sided tolerance interval is 6. Since this interval contains 10, the claim by the union leaders appears valid.
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