The durability of a tire of a certain brand is a Normal random variable with an average of 64,000 km and a standard deviation of 9,000 km. Assuming independence between tires, what is the probability that the 4 tires on a car will last more than 58,000 km?

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Answer to a math question The durability of a tire of a certain brand is a Normal random variable with an average of 64,000 km and a standard deviation of 9,000 km. Assuming independence between tires, what is the probability that the 4 tires on a car will last more than 58,000 km?

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Standardize the value of 58,000 km by subtracting the mean and dividing by the standard deviation. This gives us a z-score, which represents how many standard deviations the value is from the mean. The z-score for 58,000 km is given by: z = (58,000 - 64,000) / 9,000 ≈ -0.67 The probability corresponding to z = -0.67 is approximately 0.7475. This means that there is a 74.75% chance that a single tire will last more than 58,000 km. Assuming independence between tires, we can simply raise the single-tire probability to the power of 4 to find this: P(all 4 tires > 58,000 km) = (0.7475)^4 ≈ 0.3122 So, there is approximately a 31.22% chance that all 4 tires on a car will last more than 58,000 km.

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The durability of a tire of a certain brand is a Normal random variable with an average of 64,000 km and a standard deviation of 9,000 km. Assuming independence between tires, what is the probability that the 4 tires on a car will last more than 58,000 km?

Answer to a math question The durability of a tire of a certain brand is a Normal random variable with an average of 64,000 km and a standard deviation of 9,000 km. Assuming independence between tires, what is the probability that the 4 tires on a car will last more than 58,000 km?

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