Question

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

Answer 1

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.

Explanation:

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

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

`Z = (X - \mu)/(\sigma)`

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:

`\mu = 75.6, \sigma = 11.6`

Upper P1

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

So we use
`Z = 2.325`

`Z = (X - \mu)/(\sigma)`

`2.325 = (X - 75.6)/(11.6)`

`X - 75.6 = 11.6*2.325`

`X = 102.57`

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
`Z = -2.325`

`Z = (X - \mu)/(\sigma)`

`-2.325 = (X - 75.6)/(11.6)`

`X - 75.6 = 11.6*(-2.325)`

`X = 48.63`

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