excans:9c543a08dc: Difference between revisions
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(Created page with "'''Solution: B''' The sum of the breaking strengths is approximately normally distributed with mean 100*10 = 1000 and variance 100. Hence the approximate probability that the rope will support a weight of 985 pounds is approximately the probability that a standard normal exceeds -15/10 = -1.5, which equals 0.0668.") |
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'''Solution: B''' | '''Solution: B''' | ||
The sum of the breaking strengths is approximately normally distributed with mean 100*10 = 1000 and variance 100. Hence the approximate probability that the rope will support a weight of 985 pounds is approximately the probability that a standard normal exceeds -15/10 = -1.5, which equals 0. | The sum of the breaking strengths is approximately normally distributed with mean 100*10 = 1000 and variance 100. Hence the approximate probability that the rope will support a weight of 985 pounds is approximately the probability that a standard normal exceeds -15/10 = -1.5, which equals 0.9332 | ||
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Latest revision as of 17:34, 28 June 2024
Solution: B
The sum of the breaking strengths is approximately normally distributed with mean 100*10 = 1000 and variance 100. Hence the approximate probability that the rope will support a weight of 985 pounds is approximately the probability that a standard normal exceeds -15/10 = -1.5, which equals 0.9332 .