Answer :
To determine the standard deviation for each patient's body temperatures, we will first calculate the mean (average) of their temperatures, then find the variance, and finally the standard deviation.
Patient 1:
- Temperatures: 36.6, 36.8, 37.9, 37.3, 38.2, 36.7
Calculate the mean:
[tex]\text{Mean} = \frac{36.6 + 36.8 + 37.9 + 37.3 + 38.2 + 36.7}{6} = \frac{223.5}{6} = 37.25\,\degree C[/tex]Find the squared differences from the mean, then average them (variance):
[tex]\text{Variance} = \frac{(36.6-37.25)^2 + (36.8-37.25)^2 + (37.9-37.25)^2 + (37.3-37.25)^2 + (38.2-37.25)^2 + (36.7-37.25)^2}{6}[/tex][tex]= \frac{0.4225 + 0.2025 + 0.4225 + 0.0025 + 0.9025 + 0.3025}{6}
= \frac{2.255}{6} \approx 0.3758[/tex]Calculate the standard deviation (the square root of the variance):
[tex]\text{Standard Deviation} = \sqrt{0.3758} \approx 0.61\,\degree C[/tex]
Patient 2:
- Temperatures: 37.2, 37.8, 36.4, 38.3, 36.2
Calculate the mean:
[tex]\text{Mean} = \frac{37.2 + 37.8 + 36.4 + 38.3 + 36.2}{5} = \frac{185.9}{5} = 37.18\,\degree C[/tex]Find the squared differences from the mean, then average them (variance):
[tex]\text{Variance} = \frac{(37.2-37.18)^2 + (37.8-37.18)^2 + (36.4-37.18)^2 + (38.3-37.18)^2 + (36.2-37.18)^2}{5}[/tex][tex]= \frac{0.0004 + 0.3844 + 0.6084 + 1.2544 + 0.9604}{5}
= \frac{3.208}{5} \approx 0.6416[/tex]Calculate the standard deviation (the square root of the variance):
[tex]\text{Standard Deviation} = \sqrt{0.6416} \approx 0.80\,\degree C[/tex]
Conclusion:
From the standard deviations calculated:
- Patient 1's standard deviation is approximately [tex]0.61\,\degree C[/tex].
- Patient 2's standard deviation is approximately [tex]0.80\,\degree C[/tex].
The smaller the standard deviation, the more consistent the data. Therefore, Patient 1's body temperatures are more consistent than Patient 2's.
