The fracture strength of a certain type of manufactured glas

The fracture strength of a certain type of manufactured glass is normally distributed with a mean of 560 MPa with a standard deviation of 12 MPa.

(Round to 4 decimal places)

a)  

What is the probability that a randomly chosen sample of glass will break at less than 560 MPa?

(b)

What is the probability that a randomly chosen sample of glass will break at more than 582 Mpa?

(c)What is the probability that a randomly chosen sample of glass will break at less than 593 MPa?

Solution

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    560      
u = mean =    560      
          
s = standard deviation =    12      
          
Thus,          
          
z = (x - u) / s =    0      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   0   ) =    0.5 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    582      
u = mean =    560      
          
s = standard deviation =    12      
          
Thus,          
          
z = (x - u) / s =    1.833333333      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.833333333   ) =    0.033376508 [ANSWER]

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c)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    593      
u = mean =    560      
          
s = standard deviation =    12      
          
Thus,          
          
z = (x - u) / s =    2.75      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   2.75   ) =    0.997020237 [ANSWER]