Women have pulse rates that are normally distributed with a mean of 75.6 beats per minute and a standard deviation of 11.6 beats per minute. complete parts a through c below. a. find the percentiles upper p 1 and upper p 99. upper p 1equals nothing beats per minute

The upper 1% have a heart rate of at least 102.57 beats per minute.

The upper 99% have a heart rate of at least 48.63 beats per minute.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

Upper P1

The value of X when Z has a pvalue of 0.99.

So we use

The upper 1% have a heart rate of at least 102.57 beats per minute.

Upper P99.

The value of X when Z has a pvalue of 0.01.

So we use

The upper 99% have a heart rate of at least 48.63 beats per minute.

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.