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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 :

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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:

[tex]\mu = 115, \sigma = 18[/tex]

Lower than 126 beats per minute

pvalue of Z when X = 126. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{126 - 115}{18}[/tex]

[tex]Z = 0.61[/tex]

[tex]Z = 0.61[/tex] 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

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