Question 13 and 14 refer to the following setup The fracture

Question 13 and 14 refer to the following setup: The fracture toughness of a sheet of shatter-proof glass is known to be normally distributed, with a mean of 1.59 MPa/root m, and a standard deviation is 0.2 MPa/root m. 13. What is the probability that a randomly selected sheet of shatter-proof glass will have a fracture toughness greater than 1.62 MPa/root m? Answer: 14. Find the value k such that 98.5% of sheets of shatter-proof glass have a fracture toughness greater than k.

Solution

13.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    1.62      
u = mean =    1.59      
n = sample size =    1      
s = standard deviation =    0.2      
          
Thus,          
          
z =    0.15      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.15   ) =    0.440382308 [ANSWER]

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First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.015      
          
Then, using table or technology,          
          
z =    -2.170090378      
          
As x = u + z * s          
          
where          
          
u = mean =    1.59      
z = the critical z score =    -2.170090378      
s = standard deviation =    0.2      
          
Then          
          
x = critical value =    1.155981924   [ANSWER]