Question: Please explain how these answers were obtains, more specific processes would be better and I will be very appreciated! 5. In a city with over

Please explain how these answers were obtains, more specific processes would be better and I will be very appreciated!

Please explain how these answers were obtains, more specific processes would bebetter and I will be very appreciated! 5. In a city with

5. In a city with over 500,000 registered vehicles, the costs of auto insurance of all regis- tered vehicles have a mean of $1423 and a standard deviation of $533.2. Sixty-five percent of all registered vehicles have auto insurance that costs over $1423. a) True or false? The distribution of auto insurance costs of all registered vehicles is approximately Normal. (3 marks] Circle your answer: True False Justify your answer: The distribution is not symmetric - more than 50% of Insurance costs one greater than the mean. Hence the distribution b) Sixteen registered vehicles are randomly chosen. What is the standard deviation cannot be Normal. of the total auto insurance costs of the 16 vehicles? Circle your answer. [2 marks] A. $33.325 B. $133.3 C $2132.8 D. $8531.2c) Sixteen registered vehicles are randomly chosen. The number of vehicles (out of the 16 vehicles) that have auto insurance costing over $1423 will follow (circle only one answer) [2.5 marks) A. approximately the Normal distribution with standard deviation $132 V16 B. approximately the Normal distribution with standard deviation v 16(0.65) (1 - 0.65). C. approximately the Normal distribution with standard deviation (0.65) 1-0.65 16 (D the Binomial distribution with standard deviation ( 16(0.65)(1 - 0.65). E. both (B) and (D). F. both (C) and (D). d) Consider repeated random samples of 100 registered vehicles. The average auto insurance cost of the 100 registered vehicles over these repeated samples will follow (circle only one answer) (2.5 marks] A. a non-Normal distribution with standard deviation 533.2. .)approximately the Normal distribution with standard deviation 533? V100 C. approximately the Normal distribution with standard deviation 100(0.65)(1 - 0.65). D. approximately the Normal distribution with standard deviation (0.651 1-0.43 100 E. the Binomial distribution with standard deviation 100(0.65) (1 - 0.65). F. Both (C) and (E)

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