Question

In: Statistics and Probability

The length of an injected-molded plastic case that holds tape is normally distributed with a mean...

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?

Solutions

Expert Solution

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


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