Question

Assume that military aircraft use ejection seats designed for men weighing between 135.4 lb and 216 lb. If women's

weights are normally distributed with a mean of 167.9 lb and a standard deviation of 48.1 lb, what percentage of
women have weights that are within those limits? Are many women excluded with those specifications?
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The percentage of women that have weights between those limits is%.
(Round to two decimal places as needed.)

Answer 1

To find the percentage of women with weights between 135.4 lb and 216 lb, you can use the z-score formula:

\[ Z = \frac{{X - \mu}}{{\sigma}} \]

where:

- \( X \) is the weight limit,

- \( \mu \) is the mean weight,

- \( \sigma \) is the standard deviation.

For the lower limit (135.4 lb):

\[ Z_{\text{lower}} = \frac{{135.4 - 167.9}}{{48.1}} \]

For the upper limit (216 lb):

\[ Z_{\text{upper}} = \frac{{216 - 167.9}}{{48.1}} \]

Once you have the z-scores, you can use a standard normal distribution table or calculator to find the percentage of women within those limits. Subtract the cumulative percentage corresponding to the lower z-score from the cumulative percentage corresponding to the upper z-score.

Calculate these values to find the percentage and let me know if you need further assistance.