標題:
發問:
Each weekday a man goes to work by bus .His arrival time at the bus stop is normally distributed with standard deviation 3 minutes.Hid mean arrival time is 8:30 am .Buses leave promptly every 5 minutes at 8:21 am,8:26 am ,etc.Find the probability that he catches the buuses at 8:36
最佳解答:
For the man to catch the bus at 8:36 am, he has to arrive the bus stop after 8:31 am but before 8:36 am. They are deviated from the mean arrival time by 1 min to 6 mins. The bound of the test statistics, (difference between arrival time and mean) / standard deviation So, P[(1/3 < z < 6/3)] = P(1/3 < z < 2) = P(0 < z < 2) - P(0 < z < 1/3) From table Prob = 0.4772 - 0.130567 Prob = 0.3466
其他解答:
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al APPLIED發問:
Each weekday a man goes to work by bus .His arrival time at the bus stop is normally distributed with standard deviation 3 minutes.Hid mean arrival time is 8:30 am .Buses leave promptly every 5 minutes at 8:21 am,8:26 am ,etc.Find the probability that he catches the buuses at 8:36
最佳解答:
For the man to catch the bus at 8:36 am, he has to arrive the bus stop after 8:31 am but before 8:36 am. They are deviated from the mean arrival time by 1 min to 6 mins. The bound of the test statistics, (difference between arrival time and mean) / standard deviation So, P[(1/3 < z < 6/3)] = P(1/3 < z < 2) = P(0 < z < 2) - P(0 < z < 1/3) From table Prob = 0.4772 - 0.130567 Prob = 0.3466
其他解答:
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