A manufacturer of heat exchangers requires that the plate sp

A manufacturer of heat exchangers requires that the plate spacings of its exchange be between .240 and .260 inches. A quality control engineer sampled 20 exchangers and measured the spacing of the plates on each exchanger. If the sample mean and sample standard deviation of these 20 measurements are .254 and .005, estimate the fraction of all exchangers whose plate spacings fall outside the specified region. Assume that the plate spacings have a normal distribution.

Solution

Normal Distribution
Mean ( u ) =0.254
Standard Deviation ( sd )=0.005
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < 0.24) = (0.24-0.254)/0.005
= -0.014/0.005= -2.8
= P ( Z <-2.8) From Standard Normal Table
= 0.0026
P(X > 0.26) = (0.26-0.254)/0.005
= 0.006/0.005 = 1.2
= P ( Z >1.2) From Standard Normal Table
= 0.1151
P( X < 0.24 OR X > 0.26) = 0.0026+0.1151 = 0.1176