Respuesta :
Answer:
The data set marked as B has the largest standard deviation
Explanation:
Standard Deviation
It's a number used to show how a set of measurements is spread out from the average value. A low standard deviation means that most of the values are close to the average. A high standard deviation means that the numbers are more spread out.
The formula for the standard deviation is
[tex]\displaystyle \sigma=\sqrt{\frac{\sum (x_i-\mu)^2}{n}}[/tex]
Where [tex]x_i[/tex] is the value of each measurement, n is the number of elements in the set, and [tex]\mu[/tex] is the average or media of the values, defined as
[tex]\displaystyle \mu=\frac{\sum x_i}{n}[/tex]
Let's analyze each set of data:
A.3,4,3,4,3,4,3
The average is
[tex]\displaystyle \mu=\frac{3+4+3+4+3+4+3}{7}=3.43[/tex]
Computing the stardard deviation:
[tex]\sigma=\sqrt{\frac{(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2}{7}}[/tex]
[tex]\sigma=0.5[/tex]
B.1,6,3,15,4,12,8
The average is
[tex]\displaystyle \mu=\frac{1+6+3+15+4+12+8}{7}=7[/tex]
Computing the stardard deviation:
[tex]\sigma=\sqrt{\frac{(1-7)^2+(6-7)^2+(3-7)^2+(15-7)^2+(4-7)^2+(12-7)^2+(8-7)^2}{7}}[/tex]
[tex]\sigma=4.7[/tex]
C. 20, 21,23,19,19,20,20
The average is
[tex]\displaystyle \mu=\frac{20+21+23+19+19+20+20}{7}=20.29[/tex]
Computing the stardard deviation:
[tex]\sigma=\sqrt{\frac{(20-20.29)^2+(21-20.29)^2+(23-20.29)^2+(19-20.29)^2+(19-20.29)^2+(20-20.29)^2+(20-20.29)^2}{7}}[/tex]
[tex]\sigma=1.3[/tex]
D.12,14,13,14,12,13,12
The average is
[tex]\displaystyle \mu=\frac{12+14+13+14+12+13+12}{7}=12.86[/tex]
Computing the stardard deviation:
[tex]\sigma=\sqrt{\frac{(12-12.86)^2+(14-12.86)^2+(13-12.86)^2+(14-12.86)^2+(12-12.86)^2+(13-12.86)^2+(12-12.86)^2}{7}}[/tex]
[tex]\sigma=0.8[/tex]
We can see the data set marked as B has the largest standard deviation
