Answer :
Answer:
a) The sample range 25.9 [tex]ml\slash kg\slash \min[/tex]
b) The sample variance is 49.344 [tex]ml^2 \slash kg^2 \slash min^2[/tex]
c) The sample standard deviation 7.0245 [tex]ml\slash kg\slash \min[/tex]
Step-by-step explanation:
We are given the following data on oxygen consumption (mL/kg/min):
28.6, 49.4, 30.3, 28.2, 28.9, 26.4, 33.8, 29.9, 23.5, 30.2
a) The sample range
Range = Maximum - Minimum
[tex]\text{Range} = 49.4 - 23.5 = 25.9[/tex]
The sample range 25.9 [tex]ml\slash kg\slash \min[/tex]
b) The sample variance
[tex]\text{Variance} = \displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{309.2}{10} = 30.92[/tex]
Sum of squares of differences =
5.3824 + 341.5104 + 0.3844 + 7.3984 + 4.0804 + 20.4304 + 8.2944 + 1.0404 + 55.0564 + 0.5184 = 444.096
[tex]s^2 = \dfrac{444.096}{9} = 49.344[/tex]
The sample variance is 49.344 [tex]ml^2 \slash kg^2 \slash min^2[/tex]
c) The sample standard deviation
It is the square root of sample variance.
[tex]s = \sqrt{s^2} = \sqrt{49.344} = 7.0245[/tex]
The sample standard deviation 7.0245 [tex]ml\slash kg\slash \min[/tex]