The heights of adult men in America are normally distributed, with a mean of 69.8 inches and a standard deviation of 2.67 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.8 inches and a standard deviation of 2.54 inches. What percentage of men are SHORTER than 67 inches

14.69% of men are SHORTER than 67 inches

Step-by-step explanation:

When the distribution is normal, we use 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.

The heights of adult men in America are normally distributed, with a mean of 69.8 inches and a standard deviation of 2.67 inches.

This means that

What percentage of men are SHORTER than 67 inches

This is the pvalue of Z when X = 67. So

has a pvalue of 0.1469

14.69% of men are SHORTER than 67 inches

answer: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq   and the angle ∠orq   are opposite, therefore:

then, can be used to d to calculate the measure of ∠opq