SOLUTION: I have a final today and there will be similar questions as this one. Can someone please help me understand this problem. The fracture strength of a certain type of manufactured

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Question 257493: I have a final today and there will be similar questions as this one. Can someone please help me understand this problem.
The fracture strength of a certain type of manufactured glass is normally distributed with a mean of 579 MPa with a standard deviation of 14 MPa. (a) What is the probability that a randomly chosen sample of glass will break at less than 579 MPa? (b) More than 590 MPa? (c) Less than 600 MPa?
Thank you
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
The fracture strength of a certain type of manufactured glass is normally distributed with a mean of 579 MPa with a standard deviation of 14 MPa. (a) What is the probability that a randomly chosen sample of glass will break at less than 579 MPa? (b) More than 590 MPa? (c) Less than 600 MPa?
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You have to convert to z-scores.
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(a) What is the probability that a randomly chosen sample of glass will break at less than 579 MPa?
z(579)= (579-579)/14 = 0
P(x < 579) = P(z < 0) = 0.5000
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(b) More than 590 MPa?
z(590) = (590-579)/14 = 11/14= 0.7857...
P(x > 590) = P(z > 0.7857..) = 0.2160...
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(c) Less than 600 MPa?
z(600) = (600-579)/14 = 21/14 = 3/2
P(x < 600) = P(z< 3/2) = 0.9332
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Cheers,
Stan H.