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 15 cm and standard deviation 0.08 cm. Suppose the distribution of the diameter is normal. (Round your answers to four decimal places.)

(a) Calculate

P(14.99 X 15.01)

when n = 16.

P(14.99 X 15.01)

=  

(b) How likely is it that the sample mean diameter exceeds 15.01 when n = 25?

P(X 15.01)

=

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

Solution

Normal Distribution
Mean ( u ) =15
Standard Deviation ( sd )=0.08
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 14.99) = (14.99-15)/0.08
= -0.01/0.08 = -0.125
= P ( Z <-0.125) From Standard Normal Table
= 0.45026
P(X < 15.01) = (15.01-15)/0.08
= 0.01/0.08 = 0.125
= P ( Z <0.125) From Standard Normal Table
= 0.54974
P(14.99 < X < 15.01) = 0.54974-0.45026 = 0.0995                  

WHEN n=16

To find P(a <= Z <=b) = F(b) - F(a)
P(X < 14.99) = (14.99-15)/0.08/ Sqrt ( 16 )
= -0.01/0.02
= -0.5
= P ( Z <-0.5) From Standard Normal Table
= 0.30854
P(X < 15.01) = (15.01-15)/0.08/ Sqrt ( 16 )
= 0.01/0.02 = 0.5
= P ( Z <0.5) From Standard Normal Table
= 0.69146
P(14.99 < X < 15.01) = 0.69146-0.30854 = 0.3829      
          
b)
P(X < 15.01) = (15.01-15)/0.08/ Sqrt ( 25 )
= 0.01/0.016= 0.625
= P ( Z <0.625) From Standard NOrmal Table
= 0.734                  
P(X > = 15.01) = 1 - P(X < 15.01)
= 1 - 0.734 = 0.266