The inside diameter of a randomly selected piston ring is a

The inside diameter of a randomly selected piston ring is a random variable with mean value 8 cm and standard deviation 0.05 cm. Suppose the distribution of the diameter is normal. (Round your answers to four decimal places.) Calculate P(7.99 LE X LE 8.01) when n = 16. P(7.99 LE X LE 8.01) = How likely is it that the sample mean diameter exceeds 8.01 when n = 25? P(X GE 8.01) = You may need to use the appropriate table in the Appendix of Tables to answer this question.

Solution

Mean ( u ) =8
Standard Deviation ( sd )=0.05
Number ( n ) = 16
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
a)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 7.99) = (7.99-8)/0.05/ Sqrt ( 16 )
= -0.01/0.0125
= -0.8
= P ( Z <-0.8) From Standard Normal Table
= 0.21186
P(X < 8.01) = (8.01-8)/0.05/ Sqrt ( 16 )
= 0.01/0.0125 = 0.8
= P ( Z <0.8) From Standard Normal Table
= 0.78814
P(7.99 < X < 8.01) = 0.78814-0.21186 = 0.5763                  

b)
P(X < 8.01) = (8.01-8)/0.05/ Sqrt ( 25 )
= 0.01/0.01= 1
= P ( Z <1) From Standard NOrmal Table
= 0.8413                  
P(X > = 8.01) = 1 - P(X < 8.01)
= 1 - 0.8413 = 0.1587