Answer:
12.1, 12.3,12.4,12.5,12.3,12.1,12.2
![\bar X= \frac{12.1+12.3+12.4+12.5+12.3+12.1+12.2}{7}=12.271](https://tex.z-dn.net/?f=%5Cbar%20X%3D%20%5Cfrac%7B12.1%2B12.3%2B12.4%2B12.5%2B12.3%2B12.1%2B12.2%7D%7B7%7D%3D12.271)
And for the standard deviation we can use the following formula:
![s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}](https://tex.z-dn.net/?f=%20s%3D%20%5Csqrt%7B%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20%28X_i%20-%5Cbar%20X%29%5E2%7D%7Bn-1%7D%7D)
And after replace we got:
![s = 0.1496](https://tex.z-dn.net/?f=%20s%20%3D%200.1496)
And as we can ee we got a small value for the deviation <1 on this case.
Step-by-step explanation:
For example if we have the following data:
12.1, 12.3,12.4,12.5,12.3,12.1,12.2
We see that the data are similar for all the observations so we would expect a small standard deviation
If we calculate the sample mean we can use the following formula:
![\bar X=\frac{\sum_{i=1}^n X_i}{n}](https://tex.z-dn.net/?f=%5Cbar%20X%3D%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20X_i%7D%7Bn%7D)
And replacing we got:
![\bar X= \frac{12.1+12.3+12.4+12.5+12.3+12.1+12.2}{7}=12.271](https://tex.z-dn.net/?f=%5Cbar%20X%3D%20%5Cfrac%7B12.1%2B12.3%2B12.4%2B12.5%2B12.3%2B12.1%2B12.2%7D%7B7%7D%3D12.271)
And for the standard deviation we can use the following formula:
![s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}](https://tex.z-dn.net/?f=%20s%3D%20%5Csqrt%7B%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20%28X_i%20-%5Cbar%20X%29%5E2%7D%7Bn-1%7D%7D)
And after replace we got:
![s = 0.1496](https://tex.z-dn.net/?f=%20s%20%3D%200.1496)
And as we can ee we got a small value for the deviation <1 on this case.