Question Help Assume that military aircraft use ejection seats designed for men weighing between 143.6143.6 lb and 208208 lb. If women's weights are normally distributed with a mean of 178.6178.6 lb and a standard deviation of 48.148.1 lb, what percentage of women have weights that are within those limits? Are many women excluded with those specifications

49.95% of women have weights that are within those limits. More than half of women are excluded with those specifications, so yes, there are many women excluded.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Between 143.6 lb and 208lb.

pvalue of Z when X = 208 subtracted by the pvalue of Z when X = 143.6. So

X = 208

has a pvalue of 0.7291

X = 143.6

has a pvalue of 0.2296

0.7291 - 0.2296 = 0.4995

49.95% of women have weights that are within those limits. More than half of women are excluded with those specifications, so yes, there are many women excluded.

all the angles opposite to the angle 150 would be 150 and all the angles that are next to 150 are 30 because they equal 180 it you add it up.

step-by-step explanation: