Question

An article suggested that yield strength (ksi) for A36 grade steel is normally distributed with μ = 44 and σ = 5.0.

(a) What is the probability that yield strength is at most 40? Greater than 64? (Round your answers to four decimal places.)

at most 40
greater than 64

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

Answer 1

a)

µ =    44          
σ =    5          
              
P( X ≤    40   ) = P( (X-µ)/σ ≤ (40-44) /5)      
=P(Z ≤   -0.800   ) =   0.2119   (answer)
------------------

µ =    44                  
σ =    5                  
                      
P ( X ≥   64   ) = P( (X-µ)/σ ≥ (64-44) / 5)              
= P(Z ≥   4.000   ) = P( Z <   -4.000   ) =    0.0000   (answer)

at most 40	0.2119
greater than 64	0.0000

b)

µ=   44                  
σ =    5                  
proportion= 1-0.75 = 0.25                  
                      
Z value at    0.25   =   -0.674   (excel formula =NORMSINV(   0.25   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -0.674   *   5   +   44  
X   =   40.628   (answer)