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Answer :
Final answer:
Using the z-score formula, the weight ranges for 132 lb and 208 lb correspond approximately to the z-scores -4.45 and 2.92 respectively. Based on standard normal distribution values, very few women fall within this weight range, indicating that the majority of women might be excluded from these specifications.
Explanation:
Assuming that the weights of women are normally distributed with a mean of 177.9 lb and a standard deviation of 10.3 lb (as the original 1779 lb was a typo), we can determine the percentage of women that fall within the specified range using the z-score formula (Z = (X - μ) / σ). This allows us to adjust the weights such that they have a mean of 0 and a standard deviation of 1, which represents a standard normal distribution.
The z-score for 132 lb would be (132 - 177.9) / 10.3 which equals approximately -4.45. The z-score for 208 lb is (208 - 177.9) / 10.3 which equals approximately 2.92.
In a standard normal distribution graph, 95% of the data falls within -2 and 2, meaning the z-scores for the specified range hold a very small percentage of the total data. Based on the standard normal distribution values, the percentage of women that fall within the 132 lb - 208 lb weight range are unlikely to be more than a few percent. Hence, a large proportion of women might be excluded.
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