Answer:
49.87% of the population has a heart rate between 68 and 77.
Step-by-step explanation:
We are given that the mean of the data for the resting heart of adults is 68 beats per minute and the standard deviation is 3 beats per minute.
Let X = the data for the resting heart of adults
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 68 beats per minute
[tex]\sigma[/tex] = standard deviation = 3 beats per minute
Now, the percentage of the population that has a heart rate between 68 and 77 is given by = P(68 < X < 77)
P(68 < X < 77) = P(X < 77) - P(X [tex]\leq[/tex] 68)
P(X < 77) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{77-68}{3}[/tex] ) = P(Z < 3) = 0.9987
P(X [tex]\leq[/tex] 68) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{68-68}{3}[/tex] ) = P(Z [tex]\leq[/tex] 0) = 0.50
The above probability is calculated by looking at the value of x = 3 and x = 0 in the z table which has an area of 0.9987 and 0.50 respectively.
Therefore, P(68 < X < 77) = 0.9987 - 0.50 = 0.4987 or 49.87%
