If it is appropriate to do so, use the normal approximation to the p^ �^ -distribution to calculate the indicated probability: n=80,p=0.715= P( p̂ > 0.75) =
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P( p̂ > 0.75) =
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- The average number of miles (in thousands) that a car's tire will function before needing replacement is 66 and the standard deviation is 17. Suppose that 42 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of ¯xx¯? ¯xx¯ ~ N(,) If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 63.5 and 66.1. For the 42 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 63.5 and 66.1.If it is appropriate to do so, use the normal approximation to the p^ distribution to calculate the indicated probability: Standard Normal Distribution Table n=80,p=0.715P( p̂ > 0.75)=The average number of miles (in thousands) that a car's tire will function before needing replacement is 66 and the standard deviation is 20. Suppose that 16 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of X? X ~ N(,) What is the distribution of ¯x¯? ¯x¯ ~ N(,) If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 70.3 and 77.6. For the 16 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 70.3 and 77.6. For part d), is the assumption that the distribution is normal necessary? YesNo
- The average number of miles (in thousands) that a car's tire will function before needing replacement is 65 and the standard deviation is 17. Suppose that 50 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution. b. What is the distribution of ¯xx¯? ¯xx¯~ N( ? ), (?) c. If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 67.4 and 69.6? d. For the 50 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 67.4 and 69.6 ?The average number of miles (in thousands) that a car's tire will function before needing replacement is 66 and the standard deviation is 19. Suppose that 44 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution. If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 65.8 and 68.1. Incorrect For the 44 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 65.8 and 68.1. Incorrect For part d), is the assumption that the distribution is normal necessary? Yes or NoThe average number of miles (in thousands) that a car's tire will function before needing replacement is 66 and the standard deviation is 19. Suppose that 45 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of X? X ~ N( , ) What is the distribution of ¯x? x¯ ~ N( , ) If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 64.4 and 67. For the 45 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 64.4 and 67. For part d), is the assumption that the distribution is normal necessary? NoYes
- The average number of miles (in thousands) that a car's tire will function before needing replacement is 74 and the standard deviation is 19. Suppose that 45 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of XX? XX ~ N(,) What is the distribution of ¯xx¯? ¯xx¯ ~ N(,) If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 74.3 and 78.1. For the 45 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 74.3 and 78.1. For part d), is the assumption that the distribution is normal necessary? NoYesThe average number of miles (in thousands) that a car's tire will function before needing replacement is 72 and the standard deviation is 19. Suppose that 47 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of XX? XX ~ N(,) What is the distribution of ¯xx¯? ¯xx¯ ~ N(,) If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 69 and 71.6.The average number of miles (in thousands) that a car's tire will function before needing replacement is 72 and the standard deviation is 14. Suppose that 12 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of XX? XX ~ N(,) What is the distribution of ¯xx¯? ¯xx¯ ~ N(,) If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 68.9 and 74.4. For the 12 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 68.9 and 74.4. For part d), is the assumption that the distribution is normal necessary? YesNo
- The life expectancy of a fluorescent tube is normally distributed with a mean of 6,050 hours and a standard deviation of 800 hours. Find the probability that a tube lasts for more than 6,850 hours. Use the Empirical Rule. Empirical Rule for a Normal Distribution In a normal distribution, approximately - 68% of the data lie within 1 standard deviation of the mean. • 95% of the data lie within 2 standard deviations of the mean. - 99.7% of the data lie within 3 standard deviations of the mean. A normal distribution 0.15 2.35% 13.5% 34% 34% 13.5% 2.35% O.15 H- 30 u - 20 H +20 u +30 68% of the data 95% of the data 99.7% of the dataThe average number of miles (in thousands) that a car's tire will function before needing replacement is 65 and the standard deviation is 18. Suppose that 48 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of XX? XX ~ N(,) What is the distribution of ¯xx¯? ¯xx¯ ~ N(,) If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 66.6 and 69.5. For the 48 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 66.6 and 69.5. For part d), is the assumption that the distribution is normal necessary?A simple random sample of 100 commute times is collected from the entire population of commuters in the United States. The mean of these 100 commute times is 45 minutes, with a standard deviation of about 30 minutes. mean = 45, standard deviation = 30, n = 100 Using the appropriate probability distribution, determine the critical value. Enter this value below to three decimal places. Calculate the margin of error using the critical value you determined in part (a). Enter this value below to three decimal places. Determine the confidence interval for the population mean μ\muμ. Express your answer in interval form. Interpret this confidence interval.
