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------------------------------------------------ Assume that military aircraft use ejection seats designed for individuals weighing between 147 lb and 205 lb. If women's weights are normally distributed with a mean of 173.7 lb and a standard deviation of 43.2 lb, what percentage of women have weights that are within those limits?

Round to two decimal places as needed.

Are many women excluded with those specifications?

Answer :

The percentage of women who have weights that are within those limits are 49.97%.

Yes, many women were excluded with those specifications.

How to determine the required percentage?

In Mathematics and Statistics, the z-score of a given sample size or data set can be calculated by using the following formula:

Z-score, z = (x - μ)/σ

Where:

  • σ represents the standard deviation.
  • x represents the sample score.
  • μ represents the mean score.

For a weight of 147 lb, the z-score is given by:

Z-score, z = (147 - 173.7)/43.2

Z-score, z = -26.7/43.2

Z-score, z = -0.62.

For a weight of 205 lb, the z-score is given by:

Z-score, z = (205 - 173.7)/43.2

Z-score, z = 31.3/43.2

Z-score, z = 0.73.

By using the standardized normal distribution table, the percentage of women who have weights between 147 lb and 205 lb can be calculated as follows:

P(-0.62 ≤ z ≤ 0.73) = P(z ≤ 0.73) - P(z ≤ -0.62)

P(z ≤ 0.73) - P(z ≤ -0.62) = (0.7673 - 0.2676) × 100

P(z ≤ 0.73) - P(z ≤ -0.62) = 0.4997 × 100

Percentage = 49.97%.

For the percentage of women who are excluded from this limit, we have:

The percentage excluded = 100% - 49.97%.

The percentage excluded = 50.03%.

In this context, we can conclude that many women were excluded with those specifications.

Read more on z-score here: brainly.com/question/26714379

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