In: Statistics and Probability
The length of an injected-molded plastic case that holds tape is normally distributed with a mean length=90.2 millimeters and a standard deviation of 0.1 millimeter.
What is the proportion of the molded plastic cases that fall between 90.3 millimeters and 89.7 millimeters?
Solution:
Given: The length of an injected-molded plastic case that holds tape is normally distributed with a mean length=90.2 millimeters and a standard deviation of 0.1 millimeter.
We have to find:
P(89.7 < X < 90.3 ) = .............?
Find z score for x = 89.7 and 90.3
and
Thus
P(89.7 < X < 90.3 ) = P( -5.00 < Z < 1.00 )
P(89.7 < X < 90.3 ) = P( Z < 1.00 ) - P( Z < -5.00 )
Look in z table for z = 1.0 and 0.00 and find corresponding area.
P( Z< 1.00) = 0.8413
Since z = -5.00 is less than 3 standard deviations from mean, area below -5.00 is approximately 0.
that is:
P( Z < -5.00) = 0.0000
Thus
P(89.7 < X < 90.3 ) = P( Z < 1.00 ) - P( Z < -5.00 )
P(89.7 < X < 90.3 ) = 0.8413 - 0.0000
P(89.7 < X < 90.3 ) = 0.8413
Thus the proportion of the molded plastic cases that fall between 90.3 millimeters and 89.7 millimeters is 0.8413