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) 1-10 11-20 21-30 >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 5.5 30 165 907.5
11-20 15.5 18 279 4324.5
21-30 25.5 23 586.5 14955.75
>31 35.5 10 355 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]
