Assume that military aircraft use ejection seats designed for men weighing between 141.4 lb and 210 lb.

Question

If​ women's weights are normally distributed with a mean of 179.6 lb and a standard deviation of

47.64 ​lb, what percentage of women have weights that are within those​ limits? Are many women excluded with those​ specifications?

Answer 1

To find the percentage of women that have weights between 141 and 210 pounds, we need to standardize these values and then compare them to a normal distribution table. (It can also be called a z-score table, and can be found here: https://byjus.com/maths/z-score-table/).

To standardize the values, subtract the mean of the distribution of women's weights and divide by the standard deviation of the same distribution. In this case, we calculate (141-179.6)/47.64 = -.802. This is the z-score for the value of 141 within the distribution of women's weights. It means that 141 is 80.2% of a standard deviation below the mean. We do the same for 210 pounds, and find it's z-score is equal to .638.

Next, we find these values on the z-score table. First, the value of .638 is associated with the cumulative probability of roughly .7389. This means that roughly 73.89% of observations will be below 210 pounds in the distribution of women's weights. The probability associated with the z-score of -.802 is .2119, meaning that 21.19% of observations are below this value in the distribution of women's weights.

The final step is to subtract the second probability from the first: .7389 - .2119 = .5270. We can interpret this final answer as indicating that 52.7% of the distribution of women's weights lies between the values of 141.4 and 210 pounds. In other words, 52.7% of women will be covered by the ejection seats, meaning that the seats are not suitable for almost half of women.