Probability and statistics for engineers and scientists solutions ninth edition

The variability for decreased also as seen by looking at the picture in a. The gap represents an increase of over ppm. It appears from the data that hydrocarbon emissions decreased considerably between and and that the extreme large emission over ppm were no longer in evidence. Relative Frequency 1. Solutions for Exercises in Chapter 1 7 c Use the double-stem-and-leaf plot, we have the following.

This trimmed mean is in the middle of the mean and median using the full amount of data. Due to the skewness of the data to the right see plot in c , it is common to use trimmed data to have a more robust result. Actually if a scatter plot of these two data sets is made, it is easy to see that some outlier would influence the trend.

It does show certain relationship. One reason is that there is an extreme value which influence the mean value at the load High Low Also, the variation of the shrinkage is a little larger for the low injection velocity than that for the high injection velocity.

Solutions for Exercises in Chapter 1 9 1. Also, the variation for the former group is much higher as well. Chapter 2 Probability 2. America, S. Solutions for Exercises in Chapter 2 13 2. Sample Space P S F 2. Solutions for Exercises in Chapter 2 15 2. The other 3 persons can then be placed in line in 3! By Theorem 2. Therefore, there are 9! Any of the remaining 5 nonzero digits can be chosen for the hundreds position, leaving a choice of 5 digits for the tens position. If a 3 is used in the hundreds position, then a 4, 5, or 6 must be used in the tens position leaving 5 choices for the units position.

Then using Theorem 2. Solutions for Exercises in Chapter 2 17 2. There are total days in a year. 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. Apparently, the mean operating costs of type A engines are higher than those of type B engines.

We assumed normality in the calculation. Still, we need to assume normality in the distribution. So, the upper bound is 3. We then obtain Also, from Exercise Comparing results from Exercises Solutions for Exercises in Chapter 10 0. OC curve 0. Decision: reject H0.

The White Cheddar Popcorn, on average, weighs less than 5. Decision: Reject H0 and conclude that men who use TM, on average, mediate more than 8 hours per week. Decision: Fail to reject H0. Solutions for Exercises in Chapter 10 Decision: Reject H0 and conclude that it takes less than 35 minutes, on the average, to take the test. So, fail to reject H0 and conclude that the 6. Do not 1.

Using Decision: Do not reject H0. Degrees of freedom is calculated as Degrees of freedom is calculated as 0. Decision: Reject H0.

Decision: Reject H0 ; length of storage influences sorbic acid residual concentrations. Decision: Reject H0 when a significance level is above 0. The sample size needed is 6. The sample size would be 5. Hence, fail to reject H0. So, 0. Reject H0 ; breathing frequency significantly higher in the presence of CO. Denote by X for those who choose lasagna. Decision: Reject H0 at level 0. The proportion of urban residents who favor the nuclear plant is larger than the proportion of suburban residents who favor the nuclear plant.

The proportion of couples married less than 2 years and planning to have children is significantly higher than that of couples married 5 years and planning to have children. It cannot be shown that breast cancer is more prevalent in the urban community. Using the table, 0. Since f0. So, the variability of the time to assemble the product is not significantly greater for men.

Solutions for Exercises in Chapter 10 A goodness-of-fit test with 6 degrees of freedom is based on the following data: oi 8 4 5 11 14 14 4 ei 6. Computation: A goodness-of-fit test with 5 degrees of freedom is based on the following data: oi 5 4 13 8 5 5 ei 6. Reject H0 ; the speed is increased by using the facilitation tools.

There is no significant change in WBC leukograms. First we do the f -test s2 to test equality of the variances. Since in this experimental data, those two variables can be controlled each at two levels, the interaction can be inves-. America, S. The other 3 persons can then be placed in line in 3! By Theorem 2. Therefore, there are 9! Any of the remaining 5 nonzero digits can be chosen for the hundreds position, leaving a choice of 5 digits for the tens position. If a 3 is used in the hundreds position, then a 4, 5, or 6 must be used in the tens position leaving 5 choices for the units position.

Then using. There are total days in a year. This is a very large number. N Therefore, n. Then the probability that he receives a speeding ticket: 4. 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. Millions discover their favorite reads on issuu every month. Give your content the digital home it deserves.

Get it to any device in seconds. Solutions manual for probability and statistics for engineers and scientists 9th edition by walpole.