Respuesta :
Answer:
[tex] \bar X = \frac{827.5}{50}= 16.55 \approx 16.6[/tex]
[tex] s^2 = \frac{15431.13 -\frac{(827.5)^2}{50}}{49}= 35.429 \approx 35.4[/tex]
[tex] s = \sqrt{35.429}= 5.952 \approx 6.0[/tex]
Step-by-step explanation:
For this case is useful construct the following table:
Class Midpoint (xi) fi xi* fi xi^2* fi
_________________________________________
8.6-12.5 10.55 15 158.25 1669.538
12.6-16.5 14.55 18 261.9 3810.645
16.6-20.5 18.55 4 74.2 1376.41
20.6-24.5 22.55 3 67.65 1525/508
24.6-28.5 26.55 10 265.5 7049.025
_________________________________________
Total 50 827.5 15431.13
For this case the sample mean can be calculated with this formula:
[tex] \bar X=\frac{\sum_{i=1}^n x_i f_i}{n}[/tex]
And if we replace we got:
[tex] \bar X = \frac{827.5}{50}= 16.55 \approx 16.6[/tex]
For the sample variance we can ise the following formula:
[tex] s^2 = \frac{\sum x^2_i f_i -\frac{(\sum x_i f_i)^2}{n}}{n-1}[/tex]
And replacing we got:
[tex] s^2 = \frac{15431.13 -\frac{(827.5)^2}{50}}{49}= 35.429 \approx 35.4[/tex]
And for the deviation we just take the square root of the variance and we got:
[tex] s = \sqrt{35.429}= 5.952 \approx 6.0[/tex]
