Question

Suppose that the distribution of heart rates for medium-sized dogs is normally distributed with mean 115 beats per minute and standard deviation 18 beats per minute. Stacy takes her medium-sized dog, Scooter, to the veterinarian for a wellness check and learns that Scooter's heart rate is 126 beats per minute. Approximately what percentage of medium-sized dogs have a heart rate that is lower than Scooter's?

Answer 1

Answer:

72.91%, that is, approximately 73% of medium-sized dogs have a heart rate that is lower than Scooter's

Step-by-step 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:

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

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:

$\mu = 115, \sigma = 18$

Lower than 126 beats per minute

pvalue of Z when X = 126. So

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

$Z = \frac{126 - 115}{18}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of 0.7291

72.91%, that is, approximately 73% of medium-sized dogs have a heart rate that is lower than Scooter's