1 answer:
You might be interested in
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 = <u><em>the data for the resting heart of adults</em></u>
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = population mean = 68 beats per minute
= 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 68)
P(X < 77) = P( <
) = P(Z < 3) = 0.9987
P(X 68) = P(
) = P(Z
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%