Each day a machine is put in operation for five hours. Suppose that the number of hours of operation until failure of the machine is Exponential with mean of 4 hours. If the machine operates independently each day, the expected number of days each week that the machine would operate without failing is
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of hours of operation until failure of the machine is Exponential with mean of 4
hours. If the machine operates independently each day, the expected number of
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- A five years Business Informatics student studying at TU Wien does some temporary substitute work on the banking sector every 31 days on average, and this time follows an exponential distribution. After acquiring one of these works, its duration is also aleatoric, exponentially distributed and its mean duration is 60 daysThe money earned depends on the total number of days worked in each contract, paid at $100 daily. For unemployment periods, the student has taken out an insurance for which $50 is paid for each day on the dole. Build a queuing model to establish the mean annual income (1 year = 365 days) that the student earns. (Hint: single server, finite customer population of 1.)The time (in hour) to failure of a component after starting a new machine is exponentially distributed with mean 3 hours. If 2 hours have passed with no component failure after starting the machine, what is the probability that the component will fail in the next hour? Choose the correct equation to aswer the question. Оа. Р(X 2) — Р(X 3|X 3|X > 2) = P(X > 1)According to scientists, the cockroach has had 300 million years to develop a resistance to destruction. In a study conducted by researchers, 5,000 roaches (the expected number in a roach-infested house) were released in the test kitchen. One week later, the kitchen was fumigated and 18,857 dead roaches were counted, a gain of 13,857 roaches for the 1-week period. Assume that none of the original roaches died during the 1-week period and that the standard deviation of x, the number of roaches produced per roach in a 1-week period, is 1.4. Use the number of roaches produced by the sample of 5,000 roaches to find a 90% confidence interval for the mean number of roaches produced per week for each roach in a typical roach-infested house. Find a 90% confidence interval for the mean number of roaches produced per week for each roach in a typical roach-infested house. (Round to three decimal places as needed.)
- One of the factors that determines the degree of risk a pesticide poses to human health is the rate at which a pesticide is absorbed into skin after contact. An important question is whether the amount in the skin continues to increase with the length of the contact, or whether it increases for only a short time before leveling off. To investigate this, measured amounts of a certain pesticide were applied to 20 samples of rat skin. Four skins were analyzed at each of the time intervals 1 , 2 , 4 , 10 , and 24 hours. The amounts of the chemical (in micrograms) that were in the skin are given in the following table. Duration Amounts Absorbed 1 1.8 1.9 1.8 2.2 2 1.5 2.4 1.5 2.5 4 1.5 2.7 1.3 1.9 10 1.3 1.5 1.4 2.6 24 1.8 1.3 2.3 1.8 Construct an ANOVA table. Round your answers to four decimal places. One-way ANOVA: Pesticide Amount…Researchers have created every possible "knockout" line in yeast. Each line has exactly one gene deleted and all the other genes present (Steinmetz et al. 2002). The growth rate—how fast the number of cells increases per hour—of each of these yeast lines has also been measured, expressed as a multiple of the growth rate of the wild type that has all the genes present. In other words, a growth rate greater than 11 means that a given knockout line grows faster than the wild type, whereas a growth rate less than 11 means it grows more slowly. The growth rate of a random sample of knockout lines is 0.86, 1.02, 1.02, 1.01, 1.02, 1, 0.99, 1.01, 0.91, 0.83, 1.01 What is the standard deviation and variance of growth rate for this sample?Many runners believe that listening to music while running enhances their performance. The authors of a paper wondered if this is true for experienced runners. They recorded time to exhaustion for 11 triathletes while running on a treadmill at a speed determined to be near their peak running velocity. The time to exhaustion was recorded for each participant on two different days. On one day, each participant ran while listening to music that the runner selected as motivational. On a different day, each participant ran with no music playing. You can assume that it is reasonable to regard these 11 triathletes as representative of the population of experienced triathletes. Only summary quantities were given in the paper, but the data in the table below are consistent with the means and standard deviations given in the paper. Time to exhaustion (in seconds) Motivational music Runner 1 3 4 5 6 7 8 9 10 11 535 534 527 521 432 498 555 399 536 543 524 No music 467 449 483 574 563 593 474 499…
- The sum of two positive quantities is equal to 2n. Find the probability that their product is not less than 3/4 times their greatest product.For 30 recent Academy Award ceremonies, ages of Best Supporting Actors (x) and ages of Best Supporting Actresses (y) are recorded. The 30 paired ages yield x = 52.1 years, y = 37.3 years, r = 0.076, P-value = 0.691, and yn = 34.4 + 0.0547x. Find the best predicted value of yn (age of Best Supporting Actress) in 1982, when the age of the Best Supporting Actor (x) was 46 years.
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