Respuesta :
Answer:
[tex] \bar X = \frac{\sum_{i=1}^n x_i f_i}{N}= \frac{1385.5}{81}= 17.1[/tex]
Now  in order to calculate the variance we can use the following formula:
[tex] s^2 = \frac{\sum fx^2 -[\frac{(\sum fx)^2}{N}]}{N-1}[/tex]
and replacing we got:
[tex] s^2 = \frac{32790.25 -\frac{(1385.5)^2}{80}}{79}= 111.3[/tex]
And the deviation would be the square root of the variance:
[tex] s = \sqrt{111.33} = 10.6[/tex]
Step-by-step explanation:
We assume the following dataset
Age range (years) Â Â Â Â Â Â Â 1-10 Â Â Â Â 11-20 Â Â Â 21-30 Â Â Â >31
Number of individuals     30       18       23      10
Solution to the problem
We can solve the problem creating the following table:
Class    Midpoint(xi)  fi     xi*fi     xi^2 *fi
1-10 Â Â Â Â Â Â 5.5 Â Â Â Â Â Â 30 Â Â Â 165 Â Â Â Â 907.5
11-20 Â Â Â Â Â 15.5 Â Â Â Â Â 18 Â Â Â 279 Â Â Â 4324.5
21-30 Â Â Â Â Â 25.5 Â Â Â Â Â 23 Â Â Â 586.5 Â 14955.75
>31 Â Â Â Â Â Â Â 35.5 Â Â Â Â Â 10 Â Â Â Â 355 Â Â Â 12602.5
___________________________________
Total               81    1385.5   32790.25
The midpoint is calculated as the average between the lower and the upper interval values.
We can calculate the mean with the following formula:
[tex] \bar X = \frac{\sum_{i=1}^n x_i f_i}{N}= \frac{1385.5}{81}= 17.1[/tex]
Now  in order to calculate the variance we can use the following formula:
[tex] s^2 = \frac{\sum fx^2 -[\frac{(\sum fx)^2}{N}]}{N-1}[/tex]
and replacing we got:
[tex] s^2 = \frac{32790.25 -\frac{(1385.5)^2}{80}}{79}= 111.3[/tex]
And the deviation would be the square root of the variance:
[tex] s = \sqrt{111.33} = 10.6[/tex]
