Question

Consider again the company making tires for bikes is concerned about the exact width of their cyclocross tires. The company has a lower specification limit of 22.6 mm and an upper specification limit of 23 mm. The standard deviation is 0.13 mm and the mean is 22.8 mm.

(Round your answer to 4 decimal places.)
a. What is the probability that a tire will be too narrow?
(Round your answer to 4 decimal places.)
b. What is the probability that a tire will be too wide?
(Round your answer to 3 decimal places.)
c. What is the probability that a tire will be defective?

Answer 1

a. 0.0620

b. 0.0620

c. 0.124

Explanation:

You want the probability that a tire will have widths less than 22.6 mm, greater than 23.0 mm, or either when the distribution of widths is normal with a mean of 22.8 mm and a standard deviation of 0.13 mm.

Too narrow

The tire will be too narrow if the width is less than 22.6 mm, or 0.20/0.13 ≈ 1.5384615 standard deviations below the mean. The attached calculator display shows that probability to be about 0.0620.

P(width < 22.6 mm) ≈ 0.0620

Too wide

The tire will be too wide if the width is greater than 23.0 mm, or 0.20/0.13 ≈ 1.5384615 standard deviations above the mean. Since this Z-score is the same as for "too narrow," the probability is the same:

P(width > 23.0 mm) ≈ 0.0620

Defective

If the only defects considered are "too narrow" and "too wide", then the probability of a defective tire is the sum of the probabilities above:

P(defective) = 0.0620 +0.0620 = 0.124

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Additional comment

The Z-score needs to be accurate to better than 4 significant figures to get a probability with sufficient accuracy.

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