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 = 42 and = 5.5.

(a) What is the probability that yield strength is at most 38? Greater than 63? (Round your answers to four decimal places.)

(b) What yield strength value separates the strongest 75% from the others? (Round your answer to three decimal places.)
ksi

Solution

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    38      
u = mean =    42      
          
s = standard deviation =    5.5      
          
Thus,          
          
z = (x - u) / s =    -0.727272727      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.727272727   ) =    0.233529451 [ANSWER]

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b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area = 1 - 0.75 =   0.25      
          
Then, using table or technology,          
          
z =    -0.67448975      
          
As x = u + z * s,          
          
where          
          
u = mean =    42      
z = the critical z score =    -0.67448975      
s = standard deviation =    5.5      
          
Then          
          
x = critical value =    38.29030637   [ANSWER]