Organizers of a fishing tournament believe that the lake holds a sizable population of largemouth bass....

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Organizers of a fishing tournament believe that the lake holds a sizable population of largemouth bass. They assume that the weights of these fish have a model that is skewed to the right with a mean of 5.4 pounds and a standard deviation of 1.46 pounds. Use this information to answer parts (a) through (e). (a) Why does a skewed model make sense here? A. A skewed to the right model makes sense because it would be expected that there are many small fish and only a few large ones in the lake. B. A skewed to the right model makes sense because only the large fish are catchable. C. A skewed to the right model does not make sense. (b) Is it possible to determine the probability that a bass randomly selected ("caught") from the lake weighs more than 5 pounds. If it is possible, determine the probability. A. The probability is . (Round to four decimal places as needed.) B. Since the distribution is not normal and the exact distribution is not given, it is not possible to determine the probability. C. The number of largemouth bass caught in the contest is not given so it is not possible to determine the probability. D. Since catching a fish is not perfectly random, the probability cannot be calculated. (c) Each contestant catches 5 fish each day. Is it possible to determine the probability that someone's catch averages more than 5 pounds? If it is possible, determine the probability. A. The probability is . (Round to four decimal places as needed.) B. Since the distribution is skewed to the right and the sample size of 5 is not large enough to use the Central Limit Theorem, it is not possible to determine the probability. (d) The 10 contestants competing each caught the limit of 5 fish. What is the standard deviation of the mean weight of the 50 fish caught? SD(y) = (Round to two decimal places as needed.) (e) Would it be surprising if the mean weight of the 50 fish caught in the competition was more than 6.4 pounds? Use the 68-95-99.7 Rule. A. It would not be surprising since 6.4 pounds is fewer than 1 pound away from the mean. B. It would be surprising since 6.4 pounds is more than 2 standard deviations away from the mean. C. It would not be surprising since 6.4 pounds is fewer than 2 standard deviations away from the mean. OD. It would be surprising since 6.4 pounds is more than 1 pound away from the mean. Organizers of a fishing tournament believe that the lake holds a sizable population of largemouth bass. They assume that the weights of these fish have a model that is skewed to the right with a mean of 5.4 pounds and a standard deviation of 1.46 pounds. Use this information to answer parts (a) through (e). (a) Why does a skewed model make sense here? A. A skewed to the right model makes sense because it would be expected that there are many small fish and only a few large ones in the lake. B. A skewed to the right model makes sense because only the large fish are catchable. C. A skewed to the right model does not make sense. (b) Is it possible to determine the probability that a bass randomly selected ("caught") from the lake weighs more than 5 pounds. If it is possible, determine the probability. A. The probability is . (Round to four decimal places as needed.) B. Since the distribution is not normal and the exact distribution is not given, it is not possible to determine the probability. C. The number of largemouth bass caught in the contest is not given so it is not possible to determine the probability. D. Since catching a fish is not perfectly random, the probability cannot be calculated. (c) Each contestant catches 5 fish each day. Is it possible to determine the probability that someone's catch averages more than 5 pounds? If it is possible, determine the probability. A. The probability is . (Round to four decimal places as needed.) B. Since the distribution is skewed to the right and the sample size of 5 is not large enough to use the Central Limit Theorem, it is not possible to determine the probability. (d) The 10 contestants competing each caught the limit of 5 fish. What is the standard deviation of the mean weight of the 50 fish caught? SD(y) = (Round to two decimal places as needed.) (e) Would it be surprising if the mean weight of the 50 fish caught in the competition was more than 6.4 pounds? Use the 68-95-99.7 Rule. A. It would not be surprising since 6.4 pounds is fewer than 1 pound away from the mean. B. It would be surprising since 6.4 pounds is more than 2 standard deviations away from the mean. C. It would not be surprising since 6.4 pounds is fewer than 2 standard deviations away from the mean. OD. It would be surprising since 6.4 pounds is more than 1 pound away from the mean.