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 weight of adult Atlantic salmons is 10.02 pounds with a standard deviation of 2.89 pounds. It is also assumed that their weights are normally distributed. What is the probability to 2 decimal places that a randomly selected Atlantic salmon weighs less than 13 pounds?
- Given a normal distribution with m = 83 and s^2 = 16, find the probability that X ... assumes a value less than 80 = more than 91 = between 75 and 85 =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 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? NoYesAn average tire used by a transportation company lasts 20,000 km with a standard deviation of 100 km. Assuming that the distance traveled by a tire is normally distributed, what is the probability that a tire used by the company wil| last at most 30,000 km?
- 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? NoYesIt costs $6.75 to play a very simple game, in which a dealer gives you one card from a deck of 52 cards. If the card is a heart, spade, or diamond, you lose. If the card is a club other than the queen of clubs, you win $10.50. If the card is the queen of clubs, you win $49.00. The random variable x represents your net gain from playing this game once, or your winnings minus the cost to play. What is the mean of x, rounded to the nearest penny? $iThe 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.
