Question

The lengths of 3-inch nails manufactured on a machine are normally distributed with a mean of 3.0 inches and a standard deviation of 0.009 inch. The nails that are either shorter than 2.981 inches or longer than 3.019 inches are unusable. What percentage of all the nails produced by this machine are unusable?

The answer I was getting was 1.9652 and that is wrong.

Answer #1

Solution:

Given that,

mean = = 3.0 inches

standard deviation = = 0.009 inches

P( x < 2.981 )

P ( x - / ) < ( 2.981 - 3.0 / 0.009)

P ( z < - 0.019 / 0.009 )

P ( z < - 2.11 )

Using z table

= 0.0174

P( x > 3.019 )

= 1 - P( x <3.019 )

= 1 - P ( x - / ) < ( 3.019 - 3.0 / 0.009)

= 1 - P ( z < 0.019 / 0.009 )

= 1 - P ( z < 2.11 )

Using z table

= 1 - 0.9826

= 0.0174

= 0.0174 += 0.0174 =0.0348

Probability =0.0348 = 3.48%

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