Assume that military aircraft uso djection seats designed for men weighing between 1488 lb and 208 lb. If women's weights are normally distributed with a mean of 1652 lb and a standard deviation of 428 lb, what percentage of women have weights that are within those limits? Are many women excluded with those specifications? . BE The percentage of women that have weights between those limits is 0% (Round to two decimal places as needed.)

To find the percentage of women with weights within the limits of 1488 lb and 208 lb, we need to calculate the z-scores for these limits using the given mean and standard deviation.

For the lower limit of 208 lb:

z = (208 - 1652) / 428 = -1.08

For the upper limit of 1488 lb:

z = (1488 - 1652) / 428 = -0.38

Now, we can refer to a standard normal distribution table or use a calculator to find the percentage of data within these z-scores.

According to the standard normal distribution table, the percentage of data within a z-score of -1.08 is approximately 0.1401, and the percentage of data within a z-score of -0.38 is approximately 0.3520.

Therefore, the percentage of women with weights between 1488 lb and 208 lb is:

0.3520 - 0.1401 = 0.2119, or 21.19% (rounded to two decimal places).

Based on these specifications, no women are excluded as the percentage is greater than 0%.

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