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Chapter 6: Problem 34

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 \(.009\) inch. The nails that are either shorter than \(2.98\) inches or longer than \(3.02\) inches are unusable. What percentage of all the nails produced by this machine are unusable?

Short Answer

Expert verified
Approximately 2.62% of the nails produced by this machine are expected to be unusable.

Step by step solution

01

Calculate the Z-Score for 2.98

The z-score is calculated by using the formula \( Z = (X - 渭) / 蟽 \) , where X is the value, 渭 is the mean and 蟽 is the standard deviation. For the first bound, we have X = 2.98, 渭 = 3 and 蟽 = .009. Plugging these values into the formula, we get \( Z1 = (2.98 - 3) / .009 = -2.22 \)
02

Calculate the Z-Score for 3.02

For the other bound, we have X = 3.02, with the mean and standard deviation still the same. By similar computations, we get \( Z2 = (3.02 - 3) / .009 = 2.22 \)
03

Computation of Probability

The Z-score represents the number of standard deviations an element is from the mean. A negative Z-score indicates the value lies below the mean, while a positive indicates it's above the mean. We need to find out % of nails outside the range of -2.22 < Z < 2.22. Checking these values in the standard normal distribution table (or by using a Z-table calculator or appropriate function in a statistics software), we find that the combined area is 0.9738 (or 97.38%) which represents the % of nails in our desired range. Since we need those outside this range, we subtract this from 100%. We get % unusable nails as \(100% - 97.38% = 2.62% \).
04

Answer Verification

To validate the answer, check if the calculated probability lies between 0 and 1 (or 0% and 100%) and the Z-scores computed earlier makes sense in context of the problem. Since all these checks pass, we have the final solution.

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