SOLUTION: The fracture strength of a certain type of manufactured glass is normally distributed with a mean of 581 MPa with a standard deviation of 13 MPa. What is the probability that a

Question 321228: The fracture strength of a certain type of manufactured glass is normally distributed with a mean of 581 MPa with a standard deviation of 13 MPa.
What is the probability that a randomly chosen sample of glass will break at, (Round your answers to 4 decimal places.)

(a) less than 581 MPa?
(b) more than 592 Mpa?
(c) less than 602 MPa?
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 581 MPa with a standard deviation of 13 MPa.
What is the probability that a randomly chosen sample of glass will break at, (Round your answers to 4 decimal places.)
(a) less than 581 MPa?
z(581) = (581-581)/13 = 0
P(x< 581) = P(z< 0)= 0.5000
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(b) more than 592 Mpa
z(592) = (592-581)/13 = 11/13
--
P(x > 592) = P(z > 11/13) = normalcdf(11/13,1000) = 0.1987
---------------------------------

(c) less than 602 MPa?
= P(z < z(602)) = normalcdf(0,602,581,13) = 0.9469
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Comment: I have used several methods to find the 3 answers.
Let me know if this is confusing.
The normalcdf is a function on a TI-84 calculator.
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Cheers,
Stan H.
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