Answer:
![\bar X = \frac{\sum_{i=1}^n X_i f_i}{n}= \frac{22026.1}{171}=128.81](https://tex.z-dn.net/?f=%5Cbar%20X%20%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20X_i%20f_i%7D%7Bn%7D%3D%20%5Cfrac%7B22026.1%7D%7B171%7D%3D128.81)
![s^2 = \frac{2839948.85- \frac{(22026.1)^2}{171}}{171-1}= 16.59](https://tex.z-dn.net/?f=%20s%5E2%20%3D%20%5Cfrac%7B2839948.85-%20%5Cfrac%7B%2822026.1%29%5E2%7D%7B171%7D%7D%7B171-1%7D%3D%2016.59)
![s= \sqrt{16.587}=4.07](https://tex.z-dn.net/?f=%20s%3D%20%5Csqrt%7B16.587%7D%3D4.07)
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:
![\bar X = \frac{\sum_{i=1}^n X_i f_i}{n}= \frac{22026.1}{171}=128.81](https://tex.z-dn.net/?f=%5Cbar%20X%20%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20X_i%20f_i%7D%7Bn%7D%3D%20%5Cfrac%7B22026.1%7D%7B171%7D%3D128.81)
Now we can calculate the sample variance with the following formula:
![s^2 =\frac{\sum f_i X^2_i -[\frac{(\sum X_i f_i)}{n}]^2}{n-1}](https://tex.z-dn.net/?f=s%5E2%20%3D%5Cfrac%7B%5Csum%20f_i%20X%5E2_i%20-%5B%5Cfrac%7B%28%5Csum%20X_i%20f_i%29%7D%7Bn%7D%5D%5E2%7D%7Bn-1%7D)
And if we replace we got:
![s^2 = \frac{2839948.85- \frac{(22026.1)^2}{171}}{171-1}= 16.59](https://tex.z-dn.net/?f=%20s%5E2%20%3D%20%5Cfrac%7B2839948.85-%20%5Cfrac%7B%2822026.1%29%5E2%7D%7B171%7D%7D%7B171-1%7D%3D%2016.59)
And the standard deviation would be the square root of the variance and we got:
![s= \sqrt{16.59}=4.07](https://tex.z-dn.net/?f=%20s%3D%20%5Csqrt%7B16.59%7D%3D4.07)