An article suggested that yield strength ksi for A36 grade s

An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with mu = 42 and sigma = 4.5. What is the probability that yield strength is at most 38? Greater than 58? (Round your answers to four decimal places.) at most 38 greater than 58 What yield strength value separates the strongest 75% from the others? (Round your answer to three decimal places.) ksi

Solution

a)

at most 38:

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    38      
u = mean =    42      
          
s = standard deviation =    4.5      
          
Thus,          
          
z = (x - u) / s =    -0.888888889      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   -0.888888889   ) =    0.187031399 [answer]

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greater than 58:

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    58      
u = mean =    42      
          
s = standard deviation =    4.5      
          
Thus,          
          
z = (x - u) / s =    3.555555556      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   3.555555556   ) =    0.000188591 [answer]

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b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.25      
          
Then, using table or technology,          
          
z =    -0.67448975      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    42      
z = the critical z score =    -0.67448975      
s = standard deviation =    4.5      
          
Then          
          
x = critical value =    38.96479612   [ANSWER]