Respuesta :
Answer:
[tex]\bar X = \frac{\sum_{i=1}^n X_i f_i}{n}= \frac{22026.1}{171}=128.81[/tex]
[tex] s^2 = \frac{2839948.85- \frac{(22026.1)^2}{171}}{171-1}= 16.59[/tex]
[tex] s= \sqrt{16.587}=4.07[/tex]
Step-by-step explanation:
For this case we can calculate the sample variance and deviation with the following table
Class Midpoint (Xi) fi Xi*fi Xi^2 *fi
120.6-123.6 122.1 17 2075.7 253443
123.7-126.7 125.2 49 6134.8 768077
126.8-129.8 128.3 29 3720.7 477365.8
129.9-132.9 131.4 41 5387.4 707904.4
133.0-136.0 134.5 35 4007.5 633158.8
___________________________________________
Total 171 22026.1 2839948.85
For this case we can calculate the mean or expected value with the following formula:
[tex]\bar X = \frac{\sum_{i=1}^n X_i f_i}{n}= \frac{22026.1}{171}=128.81[/tex]
Now we can calculate the sample variance with the following formula:
[tex]s^2 =\frac{\sum f_i X^2_i -[\frac{(\sum X_i f_i)}{n}]^2}{n-1}[/tex]
And if we replace we got:
[tex] s^2 = \frac{2839948.85- \frac{(22026.1)^2}{171}}{171-1}= 16.59[/tex]
And the standard deviation would be the square root of the variance and we got:
[tex] s= \sqrt{16.59}=4.07[/tex]
