Respuesta :
Answer:
[tex] Range = 149-131=18[/tex]
[tex] s^2 =\frac{(131-139.5)^2 +(137-139.5)^2 +(138-139.5)^2 +(141-139.5)^2 +(141-139.5)^2 +(149-139.5)^2}{6-1}=35.1[/tex][tex] s =\sqrt{35.1}=5.9[/tex]
B. ​Ideally, the standard deviation would be zero because all the measurements should be the same.
Step-by-step explanation:
For this case we have the following data:
131 137 138 141 141 149
For this case the range is defined as [tex] Range = Max-Min[/tex]
And for our case we have [tex] Range = 149-131=18[/tex]
First we need to calculate the average given by this formula:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}=\frac{837}{6}=139.5[/tex]
We can calculate the sample variance with the following formula:
[tex] s^2 = \frac{\sum_{i=1}^n (X_i -\bar x)^2}{n-1}[/tex]
And if we replace we got:
[tex] s^2 =\frac{(131-139.5)^2 +(137-139.5)^2 +(138-139.5)^2 +(141-139.5)^2 +(141-139.5)^2 +(149-139.5)^2}{6-1}=35.1[/tex]
And the standard deviation is just the square root of the variance so then we got:
[tex] s =\sqrt{35.1}=5.9[/tex]
If the​ subject's blood pressure remains constant and the medical students correctly apply the same measurement​ technique, what should be the value of the standard​ deviation?
For this case the variance and deviation should be 0 since we not evidence change then we not have variation. And for this case the best answer is:
B. ​Ideally, the standard deviation would be zero because all the measurements should be the same.
