Save \( \begin{array}{l}\text { Assume that military aircraft use ejection seats designed for men weighing between } 140.2 \mathrm{lb} \text { and } 213 \mathrm{lb} \text {. } \\ \text { If women's weights are normally distributed with a mean of } 174.2 \mathrm{lb} \text { and a standard deviation of } 46.4 \mathrm{lb}, \\ \text { what percentage of women have weights that are within those limits? Are many women excluded with } \\ \text { those specifications? }\end{array} \) \( \begin{array}{l}\text { The percentage of women that have weights between those limits is } \square \% . \\ \text { (Round to two decimal places as needed.) }\end{array} \)
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To find the percentage of women whose weights fall between 140.2 lb and 213 lb, given that weights are normally distributed with a mean (μ) of 174.2 lb and a standard deviation (σ) of 46.4 lb, we can use the Z-score formula. First, we calculate the Z-scores for the weight limits: 1. For 140.2 lb: \[ Z = \frac{X - \mu}{\sigma} = \frac{140.2 - 174.2}{46.4} \approx -0.734 \] 2. For 213 lb: \[ Z = \frac{X - \mu}{\sigma} = \frac{213 - 174.2}{46.4} \approx 0.837 \] Next, we can look up these Z-scores in a standard normal distribution table or use a calculator: - The percentage of women weighing less than 140.2 lb (Z = -0.734) is about 23.5%. - The percentage of women weighing less than 213 lb (Z = 0.837) is about 79.8%. Now, we find the percentage of women within the specified weight limits: \[ \text{Percentage within limits} = P(Z < 0.837) - P(Z < -0.734) = 79.8\% - 23.5\% = 56.3\% \] Thus, the percentage of women that have weights between 140.2 lb and 213 lb is approximately 56.30%. This means a significant number of women are excluded from using those ejection seats, leaving nearly 43.7% of women unable to meet the weight requirements specified for military aircraft ejection seats.
