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.

Answer is B

The equation is correct since(log2^10)(log4^8)(log10^4)=log10/log2*log8/log4*log4/log10

=log8/log2

=3

Step-by-step explanation: