Respuesta :
Answer:
[tex] Range = 1.48-0.63=0.850 W/kg[/tex]
[tex] s= 0.320 W/Kg[/tex]
[tex] s^2 = 0.320^2= 0.103 W^2 /kg^2[/tex]
D. No, because some models of cell phones will have a larger market share than others. Measures from different models should be weighted according to their size in the population.
Step-by-step explanation:
For this case we have the following data values:
0.95,0.73,0.63,0.91,1.32,1.48,0.63,1.23,0.91,1.41,0.67
The first step on this case is order the datase on increasing way and we got:
0.63 0.63 0.67 0.73 0.91 0.91 0.95 1.23 1.32 1.41 1.48
The range is defined as [tex] Range = Max-Min[/tex]
And if we replace we got:
[tex] Range = 1.48-0.63=0.850 W/kg[/tex]
The sample standard deviation is given by this formula:
[tex] s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
And if we replace we got [tex] s= 0.320 W/Kg[/tex]
And the sample variance is just the standard deviation squared so we got:
[tex] s^2 = 0.320^2= 0.103 W^2 /kg^2[/tex]
And for the last question about : If one of each model is measured for radiation and the results are used to find the measures of variation, are the results typical of the population of cell phones that are in use?
We can conclude this:
D. No, because some models of cell phones will have a larger market share than others. Measures from different models should be weighted according to their size in the population.
