What percentage?

Question

The lengths of 5-inch nails manufactured on a machine are normally distributed with a mean of 5.0 inches and a standard deviation of 0.009 inch. The nails that are either shorter than 4.98 inches or longer that 5.02 inches are unusable. What percentage of all the nails produced by this machine are unusable?

Answer 1

Since we are dealing with normal distribution, we can use the z-score to evaluate the percentage area in the normal curve.

Z= (x-μ)/σ

Where x = score = 4.98 and 5.02

μ = mean = 5

σ = standard deviation. = 0.009

Z₁= (4.98-5)/0.009

Z₁= -2.2222

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Z₂= (5.02-5)/0.009

Z₂= 2.2222

According to the z-score calculator:

P(x<4.98) = P(z<-2.2222) ≈ 0.013134

P(x>5.02) = P(z > 2.2222) ≈ 0.013134

0.013134 + 0.013134 = 0.026268

The percentage of all the nails produced by the machine that are unusable is 2.6268%