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 = 45 and = 5.5.

(a) What is the probability that yield strength is at most 39? Greater than 67? (Round your answers to four decimal places.)

You may need to use the appropriate table in the Appendix of Tables to answer this question.

Solution

Mean ( u ) =45
Standard Deviation ( sd )=5.5
Number ( n ) = 36
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a)
P(X > 39) = (39-45)/5.5
= -6/5.5 = -1.0909
= P ( Z >-1.091) From Standard Normal Table
= 0.8623                  
P(X < = 39) = (1 - P(X > 39)
= 1 - 0.8623 = 0.1377                  
b)
P(X > 67) = (67-45)/5.5
= 22/5.5 = 4
= P ( Z >4) From Standard Normal Table
= 0