Assume that military aircraft te ejection seals designed for men weighing between 1332 lb and 208 lb. If women's weights are normally distributed with a mean of 1779ib anda standard deviation of 10.3 lb, what percentage of women have weights that are within those limits? Are many women excluded with those specifications? The percentage of women that have weights between those limits is 1% (Round to two decimal places as needed)

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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