Basic Business Statistics, Student Value Edition (13th Edition)
13th Edition
ISBN: 9780321946393
Author: Mark L. Berenson, David M. Levine, Kathryn A. Szabat
Publisher: PEARSON
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Chapter 5, Problem 34PS
a.
To determine
Find the
b.
To determine
Find the probability that at least one bag will be mishandled.
c.
To determine
Find the probability that at least two bags will be mishandled.
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Ch 6 Sec 4 Homework
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Stress at work: In a poll conducted by the General Social Survey, 81% of respondents said that their jobs were sometimes or always stressful. Two hundred
workers are chosen at random. Use the TI-84 Plus calculator as needed. Round your answer to at least four decimal places.
(a) Approximate the probability that 155 or fewer workers find their jobs stressful.
(b) Approximate the probability that more than 145 workers find their jobs stressful.
(c) Approximate the probability that the number of workers who find their jobs stressful is between 154 and 172 inclusive.
Part 1 of 3
The probability that 155 or fewer workers find their jobs stressful is 0.1207
Part 2 of 3
bility that more than 145 workers find their jobs…
Chapter 5 Solutions
Basic Business Statistics, Student Value Edition (13th Edition)
Ch. 5 - Given the following probability distribution: a....Ch. 5 - Prob. 2PSCh. 5 - Recently , a regional automobile dealership sent...Ch. 5 - In the carnival game Under-or -Over-Seven, a pair...Ch. 5 - Prob. 5PSCh. 5 - Prob. 6PSCh. 5 - Prob. 7PSCh. 5 - Given the following probability distribution for...Ch. 5 - Prob. 9PSCh. 5 - The process of being served at a bank consists of...
Ch. 5 - In the portfolio example in this section (see page...Ch. 5 - Prob. 12PSCh. 5 - Prob. 13PSCh. 5 - Prob. 14PSCh. 5 - Suppose that in Example 5.1 on page 193, you...Ch. 5 - Prob. 16PSCh. 5 - Suppose that in Problem 5.16 you wanted to create...Ch. 5 - Determine the following:...Ch. 5 - Prob. 19PSCh. 5 - Determine the mean and standard deviation of the...Ch. 5 - The increase or decrease in the price of a stock...Ch. 5 - Prob. 22PSCh. 5 - Prob. 23PSCh. 5 - A manufacturing company regularly conducts quality...Ch. 5 - When a customer places an order with Rudy’s...Ch. 5 - Prob. 26PSCh. 5 - In Example 5.5 on page 200, you and two friends...Ch. 5 - Assume a Poisson distribution....Ch. 5 - Assume a Poisson distribution....Ch. 5 - Assume a Poisson distribution with =5.0. What is...Ch. 5 - Prob. 31PSCh. 5 - The quality control manager of Marilyn’s Cookies...Ch. 5 - Refer to Problem 5.22. How many cookies in a batch...Ch. 5 - Prob. 34PSCh. 5 - Prob. 35PSCh. 5 - The Consumer Financial Protection Bureau’s...Ch. 5 - J.D. Power and Associates calculates and publishes...Ch. 5 - Prob. 38PSCh. 5 - Prob. 39PSCh. 5 - Refer to Problem 5.27. if you purchased a Toyota...Ch. 5 - A toll-free phone number is available from 9 A.M....Ch. 5 - Prob. 42PSCh. 5 - Prob. 43PSCh. 5 - An auditor for the Internal Revenue Service is...Ch. 5 - Prob. 45PSCh. 5 - Prob. 46PSCh. 5 - Prob. 47PSCh. 5 - Prob. 48PSCh. 5 - Prob. 49PSCh. 5 - What are the four properties that must be present...Ch. 5 - Prob. 51PSCh. 5 - Prob. 52PSCh. 5 - Darwin Head, a 35-year-old sawmill worker, won 1...Ch. 5 - Between 1896-when the Dow Jones index was...Ch. 5 - Smartphone adoption among American teens has...Ch. 5 - One theory concerning the Dow jones industrial...Ch. 5 - Prob. 57PSCh. 5 - Prob. 58PSCh. 5 - Social log-ins involve recommending of sharing an...Ch. 5 - The Consumer Financial Protection Bureau’s...Ch. 5 - Prob. 61PSCh. 5 - One theory concerning the...Ch. 5 - Spurious correlation refers to the apparent...Ch. 5 - Prob. 64PSCh. 5 - Prob. 65PS
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- A case-control (or retrospective) study was conducted to investigate a relationship between the colors of helmets worn by motorcycle drivers and whether they are injured or killed in a crash. Results are given in the accompanying table. Using a 0.01 significance level, test the claim that injuries are independent of helmet color. Color of Helmet Black White Yellow Red Blue Controls (not injured) 499 373 32 159 79 Cases (injured 221 108 8 66 38 or killed) Click here to view the chi-square distribution table. Chi-square distribution table Area to the Right of the Critical Value Degrees of Freedom 0.995 0.99 0.975 0.95 0.90 0.10 0.05 0.025 0.01 0.005 C. Ho: Injuries and neimet color are dependent H₁: Injuries and helmet color are independent D. Ho: Whether a crash occurs and helmet color are dependent 1 0.001 0.004 0.016 2.706 3.841 5.024 6.635 7.879 2 0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 10.597 3 0.072 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.345 12.838 4 0.207 0.297…arrow_forwardConduct the hypothesis test and provide the test statistic and the critical value, and state the conclusion. A person drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed frequencies for the outcomes of 1, 2, 3, 4, 5, and 6, respectively: 28, 32, 46, 39, 29, 26. Use a 0.025 significance level to test the claim that the outcomes are not equally likely. Does it appear that the loaded die behaves differently than a fair die? Click here to view the chi-square distribution table. The test statistic is (Round to three decimal places as needed.) Chi-square distribution table Area to the Right of the Critical Value Degrees of Freedom 0.995 0.99 0.975 0.95 0.90 0.10 0.05 0.025 0.01 0.005 1 0.001 0.004 0.016 2.706 3.841 5.024 6.635 2 0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 7.879 10.597 3 0.072 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.345 12.838 4 0.207 0.297 0.484 0.711 1.064 7.779 9.488 11.143 13.277 14.860 5…arrow_forwardThe online clothing retailer e-Parel is conducting a study to estimate the average size of the orders placed by visitors to its website. The project manager desires a $60 bound on the error of estimation at 90% confidence. The population standard deviation is unknown, and a “best guess” of $175 is used as the planning value for σ. Use the Distributions tool to help you answer the questions that follow. 0123 Select a Distribution The z-value for a 90% confidence interval of the population mean is . In order to satisfy the requirement of a $60 bound on the error of estimation, a sample size no smaller than is needed.arrow_forward
- A local electronics store just received a shipment of 620 HDMI cables. The manager wants to estimate the number of defective HDMI cables in the shipment. Rather than checking every HDMI cable, the manager plans to take a simple random sample of size 62 in order to estimate the proportion of defective HDMI cables in the shipment. If the sample proportion of defective HDMI cables, p̂p̂, is greater than 0.0323 (there are more than two defective HDMI cables in the sample), the manager will file a complaint and request a new shipment. Suppose that the true proportion of defective HDMI cables in the shipment is approximately p = 0.02. What is the expected value of the sample proportion? E(Pˆ)E(P^)= Since the sample is to be drawn from a finite population, and since the sample is 5% of the population size, the finite population correction factor needed when you calculate the standard deviation of the sampling distribution. What is the standard deviation of the…arrow_forwardAn automobile battery manufacturer offers a 39/50 warranty on its batteries. The first number in the warranty code is the free-replacement period; the second number is the prorated-credit period. Under this warranty, if a battery fails within 39 months of purchase, the manufacturer replaces the battery at no charge to the consumer. If the battery fails after 39 months but within 50 months, the manufacturer provides a prorated credit toward the purchase of a new battery. The manufacturer assumes that X, the lifetime of its auto batteries, is normally distributed with a mean of 44 months and a standard deviation of 3.6 months. Use the following Distributions tool to help you answer the questions that follow. (Hint: When you adjust the parameters of a distribution, you must reposition the vertical line (or lines) for the correct areas to be displayed.) 0123 Select a Distribution If the manufacturer’s assumptions are correct, it would need to replace of its…arrow_forwardIn regards to conducting a linear contrast after a one-way ANOVA, can you explain how seemingly arbitrary weights that "emphasize or de-emphasize" certain variables in a linear combination and sum to zero are able to provide information about how certain groups differ from each other? For example, if we havethree groups A, B, and C, and we want tocompare the mean of group A with theaverage of groups B and C, the weights inthis case are 1 for group A, and -0.5 for groupsB and C, which sum to zero. But how do these numbers model the relationship of comparing one group to the average of the other two? Does it have to do with how the math is carried out, such as how the test statistic is created?arrow_forward
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