Answer:
a) The problem says that this represent the values within 3 deviations from the mean and using the empirical rule we know that on this case we have 68% of the data on this interval.
b) For this case we can use the z score formula again:
![z_1= \frac{98.73-98.38}{0.48}=0.729](https://tex.z-dn.net/?f=%20z_1%3D%20%5Cfrac%7B98.73-98.38%7D%7B0.48%7D%3D0.729)
![z_2= \frac{97.89-98.38}{0.48}=-1.020](https://tex.z-dn.net/?f=%20z_2%3D%20%5Cfrac%7B97.89-98.38%7D%7B0.48%7D%3D-1.020)
For this case we want this probability:
![P(97.89](https://tex.z-dn.net/?f=%20P%2897.89%3CX%3C98.73%29%20%3DP%28-1.02%3CZ%3C0.729%29%3D%20P%28Z%3C0.729%29-P%28Z%3C-1.02%29%3D0.767-0.154%3D%200.613)
So the approximate percentage of temperatures between 97.89F and 98.73F is 61.3%
Step-by-step explanation:
The empirical rule, also referred to as "the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)". The empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).
For this case we know that the body temperatures for a group of heatlhy adults represented with the random variable X follows this distribution:
![X \sim N(\mu =98.38F, \sigma=0.48 F)](https://tex.z-dn.net/?f=%20X%20%5Csim%20N%28%5Cmu%20%3D98.38F%2C%20%5Csigma%3D0.48%20F%29)
Part a
For this case we can use the z score formula to measure how many deviations we are within the mean, given by:
![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=%20z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
If we find the z score for the values given we got:
![z_1= \frac{99.57-98.38}{0.48}=2.479](https://tex.z-dn.net/?f=%20z_1%3D%20%5Cfrac%7B99.57-98.38%7D%7B0.48%7D%3D2.479)
![z_2= \frac{97.05-98.38}{0.48}=-2.771](https://tex.z-dn.net/?f=%20z_2%3D%20%5Cfrac%7B97.05-98.38%7D%7B0.48%7D%3D-2.771)
The problem says that this represent the values within 3 deviations from the mean and using the empirical rule we know that on this case we have 68% of the data on this interval.
Part b
For this case we can use the z score formula again:
![z_1= \frac{98.73-98.38}{0.48}=0.729](https://tex.z-dn.net/?f=%20z_1%3D%20%5Cfrac%7B98.73-98.38%7D%7B0.48%7D%3D0.729)
![z_2= \frac{97.89-98.38}{0.48}=-1.020](https://tex.z-dn.net/?f=%20z_2%3D%20%5Cfrac%7B97.89-98.38%7D%7B0.48%7D%3D-1.020)
For this case we want this probability:
![P(97.89](https://tex.z-dn.net/?f=%20P%2897.89%3CX%3C98.73%29%20%3DP%28-1.02%3CZ%3C0.729%29%3D%20P%28Z%3C0.729%29-P%28Z%3C-1.02%29%3D0.767-0.154%3D%200.613)
So the approximate percentage of temperatures between 97.89F and 98.73F is 61.3%