Assume that military arcraft se ejection seats designed for men weighing between 133 2 lb and 208 ib. It women's weights are normally distributed with a man of 172 9 banda standard deviation of 40 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 concepts from normal distribution, z-scores, and z-table, we calculate that approximately 64.43% of women fall within the weight range required for the ejection seats. Therefore, quite a significant percentage of women are excluded as they don't meet the weight requirement.

Explanation:

To solve this problem, we would need to use normal distribution. A normal distribution is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. In graph form, it appears as a bell curve.

The first step is to transform the weights into z-scores. The z-score for a weight x is given by the formula (x - mean)/standard deviation. Hence, z1 = (133.2 - 172.9) / 40.3 = -0.98 and z2 = (208 - 172.9) / 40.3 = 0.87.

Next, we consult a z-table, also known as the standard normal table, to find the probability or area under the curve for these z-scores. For z1 = -0.98, the corresponding value in the table is 0.1635 which means 16.35% of women weigh less than 133.2 lbs. For z2 = 0.87, the corresponding value is 0.8078 meaning about 80.78% of women weigh less than 208 lbs.

To get the percentage of women who fall within these weight range, we subtract the lower percentage from the higher percentage. Hence, 80.78% - 16.35% = 64.43%. So, approximately 64.43% of women meet the weight requirements for the ejection seat.

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