Respuesta :
Answer:
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 [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 = 75.6, \sigma = 11.6[/tex]
Upper P1
The value of X when Z has a pvalue of 0.99.
So we use [tex]Z = 2.325[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]2.325 = \frac{X - 75.6}{11.6}[/tex]
[tex]X - 75.6 = 11.6*2.325[/tex]
[tex]X = 102.57[/tex]
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 [tex]Z = -2.325[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-2.325 = \frac{X - 75.6}{11.6}[/tex]
[tex]X - 75.6 = 11.6*(-2.325)[/tex]
[tex]X = 48.63[/tex]
The upper 99% have a heart rate of at least 48.63 beats per minute.
