Normal distribution problem??!?!
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Normal distribution problem??!?!

[From: ] [author: ] [Date: 11-04-23] [Hit: ]
875 is 0.The area under the standard normal curveleft to z = + 1.875 represents the probability.Therefore,Required probability = 0.5000 + 0.......
Suppose that the breaking strength of a rope (in pounds) is normally distributed, with a mean of 100 pounds and a standard deviation of 16.
What is the probability that a certain rope will break when subjected to a force of 130 pounds or less?

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z = (X-Mean)/S.D
z value corresponding to X=130 is
z = (130-100)/16 = 30/16 = 1.875
The area under the standard normal curve corresponding to z = 1.875 is 0.4696
The area under the standard normal curve left to z = + 1.875 represents the probability. Therefore,
Required probability = 0.5000 + 0.4696 = 0.9696 or 96.96%

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(130 - 100) / 16 = z
1.875 = z

P(z = 1.875) = 0.969603638
96.96%
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