Rockwell hardness of pins of a certain type is known to have

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.2. (Round your answers to four decimal places.) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 11 pins is at least 51? What is the (approximate) probability that the sample mean hardness for a random sample of 43 pins is at least 51? You may need to use the appropriate table in the Appendix of Tables to answer this question.

Solution

Mean ( u ) =50
Standard Deviation ( sd )=1.2/Sqrt(11) = 0.3618
Number ( n ) = 11
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a)
P(X < 51) = (51-50)/1.2/ Sqrt ( 11 )
= 1/0.3618= 2.7639
= P ( Z <2.7639) From Standard NOrmal Table
= 0.9971                  
P(X > = 51) = 1 - P(X < 51)
= 1 - 0.9971 = 0.0029                  
b)
P(X < 51) = (51-50)/1.2/ Sqrt ( 43 )
= 1/0.183= 5.4645
= P ( Z <5.4645) From Standard NOrmal Table
= 1                  
P(X > = 51) = 1 - P(X < 51)
= 1 - 1 = 0