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Chapter 6: Problem 86

Major League Baseball rules require that the balls used in baseball games must have circumferences between 9 and \(9.25\) inches. Suppose the balls produced by the factory that supplies balls to Major League Baseball have circumferences normally distributed with a mean of \(9.125\) inches and a standard deviation of \(.06\) inch. What percentage of these baseballs fail to meet the circumference requirement?

Short Answer

Expert verified
Therefore, the percentage of baseballs that fail to meet the circumference requirement is \( 98.15% + 1.85% = 100% \). However, considering only the balls produced by the factory, the expected percentage of balls outside the accepted circumference is \( 1.85% + 1.85% = 3.7% \).

Step by step solution

01

Calculate the Z-scores

Z-scores are calculated by taking the difference between a particular value and the mean of a set of values, then dividing by the standard deviation. Let's calculate the Z-scores of the two circumference limits, 9 and 9.25 inches, using the formula \( Z = (X - 渭) / 蟽 \) where \( X \) is the value, \( 渭 \) is the mean, and \( 蟽 \) is the standard deviation.
02

Identify the Z-scores

In case of the lower limit (9 inches), \( Z = (9 - 9.125) / .06 = -2.083 \). Similarly, for the upper limit (9.25 inches), \( Z = (9.25 - 9.125) / .06 = 2.083 \).
03

Find the corresponding percentages associated with the Z-scores

Using a Z-table, we can find the area under a standard normal curve to the left of any given Z-score. In this case, -2.083 corresponds to 1.85% and 2.083 corresponds to 98.15%.
04

Calculate the percentages failing to meet the requirement

Subtract the percentages found in step 3 from 100. For the balls with less than 9-inch circumference, the percentage is \( 100 - 1.85 = 98.15% \). For those with more than 9.25-inch circumference, the percentage is \( 100 - 98.15 = 1.85% \).

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