Respuesta :
Answer:
[tex]\bar X = \frac{\sum_{i=1}^n x_i f_i}{\sum_{i=1}^n f_i}= \frac{23716.8}{182}= 130.312 \approx 130.3[/tex]
And the variance can be calculated with this formula:
[tex]s^2 = \frac{\sum fx^2 - \frac{(\sum x f)^2}{n}}{n-1}[/tex]
And replacing we got:
[tex] s^2 = \frac{3094539.88 -\frac{(23716.8)^2}{182}}{181}=21.85 \approx 21.9[/tex]
Step-by-step explanation:
For this case we can calculate the mean with the following table:
Class Midpoint (xi) fi xi *fi xi^2 *fi
121.6- 125.2 123.4 31 3825.4 472054.36
125.3-128.9 127.1 46 5846.6 743102.86
129.0-132.6 130.8 40 5232 684345.6
132.7-136.3 134.5 46 6187 832151.5
136.4-140.0 138.2 19 2625.8 362885.56
____________________________________________
Total 182 23716.8 3094539.88
And we can calculate the mean with the following formula
[tex]\bar X = \frac{\sum_{i=1}^n x_i f_i}{\sum_{i=1}^n f_i}= \frac{23716.8}{182}= 130.312 \approx 130.3[/tex]
And the variance can be calculated with this formula:
[tex]s^2 = \frac{\sum fx^2 - \frac{(\sum x f)^2}{n}}{n-1}[/tex]
And replacing we got:
[tex] s^2 = \frac{3094539.88 -\frac{(23716.8)^2}{182}}{181}=21.85 \approx 21.9[/tex]
