Question

Assume that females have pulse rates that are normally distributed with a mean of u = 75.0 beats per minute and a standard deviation of sigma = 12.5 beats per minute. If 1 adult female is randomly​ selected, find the probability that her pulse rate is between 69 beats per minute and 81 beats per minute.

Answer 1

Answer:

36.88% probability that her pulse rate is between 69 beats per minute and 81 beats per minute.

Step-by-step explanation:

Problems of normally distributed samples are 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:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 75, \sigma = 12.5$

Find the probability that her pulse rate is between 69 beats per minute and 81 beats per minute.

This is the pvalue of Z when X = 81 subtracted by the pvalue of Z when X = 69.

X = 81

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{81 - 75}{12.5}$

$Z = 0.48$

$Z = 0.48$ has a pvalue of 0.6844

X = 69

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{69 - 75}{12.5}$

$Z = -0.48$

$Z = -0.48$ has a pvalue of 0.3156

0.6844 - 0.3156 = 0.3688

36.88% probability that her pulse rate is between 69 beats per minute and 81 beats per minute.

Answer 2

Answer:

$P(69<X<81)=P(\frac{69-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{81-\mu}{\sigma})=P(\frac{69-75}{12.5}<Z<\frac{81-75}{12.5})=P(-0.48<z<0.48)$

And we can find this probability with this difference:

$P(-0.48<z<0.48)=P(z<0.48)-P(z<-0.48)=0.684-0.316= 0.368$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the pulse rates of a population, and for this case we know the distribution for X is given by:

$X \sim N(75,12.5)$  

Where $\mu=75$ and $\sigma=12.5$

We are interested on this probability

$P(69<X<81)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(69<X<81)=P(\frac{69-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{81-\mu}{\sigma})=P(\frac{69-75}{12.5}<Z<\frac{81-75}{12.5})=P(-0.48<z<0.48)$

And we can find this probability with this difference:

$P(-0.48<z<0.48)=P(z<0.48)-P(z<-0.48)=0.684-0.316= 0.368$