The lengths of nails produced in a factory are normally distributed with a mean of 5.13 centimeters and a standard deviation of 0.04 0.04 centimeters. Find the two lengths that separate the top 8% 8 % and the bottom 8% 8 % . These lengths could serve as limits used to identify which nails should be rejected. Round your answer to the nearest hundredth, if necessary.
The lengths 'a' and 'b' will be symmetric with the center as probability on each side of the center = 0.5.
So, P(X < a) = P(X > b) = 0.08
So, using the standardized normal score, we can write :
Using Z-table we get the value of Z for which cumulative probability is 0.06 as -
P(Z < -1.41) = 0.08
So a-5.13/0.04=-1.41
a=5.13-1.41*0.04
=5.07
As the value 'b' is symmetric with 'a' about center, so -
5.13 - a = b - 5.13
So, b = 10.26 - 5.07
Hence, b = 5.19
So, the two lengths that separate the top 8% and the bottom 8% of the lengths are 5.07 cm and 5.19 cm respectively.
Hope this will be Helpful. Thanks and God Bless You :)
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