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a lawyer commutes daily from his suburban home to his midtown office. the average time for a one-way trip is 24 minutes, with a standard deviation of 3.8 minutes. assume the distribution of trip times to be normally distributed. (a) what is the probability that a trip will take at least 1/2 hour? (b) if the office opens at 9:00 a.m. and the lawyer leaves his house at 8:45 a.m. daily, what percentage of the time is he late for work

User Akametta
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Answer:

(a) To find the probability that a trip will take at least 1/2 hour (30 minutes), we need to find the area under the normal distribution curve to the right of 30 minutes. We can standardize the distribution using the formula z = (x - μ) / σ, where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.

z = (30 - 24) / 3.8 = 1.58

Using a standard normal distribution table or a calculator with a normal distribution function, we can find the probability that a trip will take at least 30 minutes is approximately 0.0571 or 5.71%.

(b) If the office opens at 9:00 a.m. and the lawyer leaves his house at 8:45 a.m. daily, he needs to arrive at the office before 9:00 a.m. to be on time. We can find the percentage of the time he is late for work by finding the area under the normal distribution curve to the right of 15 minutes (the difference between 8:45 a.m. and 9:00 a.m.), and then subtracting that value from 1 to get the percentage of the time he is on time or early.

z = (15 - 24) / 3.8 = -2.37

Using a standard normal distribution table or a calculator with a normal distribution function, we can find the probability that he is late for work is approximately 0.008 or 0.8%. Therefore, he is on time or early approximately 99.2% of the time.

User Dtmland
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