Answer:
The standard deviation of the age distribution is 6.2899 years.
Step-by-step explanation:
The formula to compute the standard deviation is:
![SD=\sqrt{\frac{1}{n}\sum\limits^{n}_{i=1}{(x_{i}-\bar x)^{2}}}](https://tex.z-dn.net/?f=SD%3D%5Csqrt%7B%5Cfrac%7B1%7D%7Bn%7D%5Csum%5Climits%5E%7Bn%7D_%7Bi%3D1%7D%7B%28x_%7Bi%7D-%5Cbar%20x%29%5E%7B2%7D%7D%7D)
The data provided is:
X = {19, 19, 21, 25, 25, 28, 29, 30, 31, 32, 40}
Compute the mean of the data as follows:
![\bar x=\frac{1}{n}\sum\limits^{n}_{i=1}{x_{i}}](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7B1%7D%7Bn%7D%5Csum%5Climits%5E%7Bn%7D_%7Bi%3D1%7D%7Bx_%7Bi%7D%7D)
![=\frac{1}{11}\times [19+19+21+...+40]\\\\=\frac{299}{11}\\\\=27.182](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B11%7D%5Ctimes%20%5B19%2B19%2B21%2B...%2B40%5D%5C%5C%5C%5C%3D%5Cfrac%7B299%7D%7B11%7D%5C%5C%5C%5C%3D27.182)
Compute the standard deviation as follows:
![SD=\sqrt{\frac{1}{n}\sum\limits^{n}_{i=1}{(x_{i}-\bar x)^{2}}}](https://tex.z-dn.net/?f=SD%3D%5Csqrt%7B%5Cfrac%7B1%7D%7Bn%7D%5Csum%5Climits%5E%7Bn%7D_%7Bi%3D1%7D%7B%28x_%7Bi%7D-%5Cbar%20x%29%5E%7B2%7D%7D%7D)
![=\sqrt{\frac{1}{11-1}\times [(19-27.182)^{2}+(19-27.182)^{2}+...+(40-27.182)^{2}]}}\\\\=\sqrt{\frac{395.6364}{10}}\\\\=6.28996\\\\\approx 6.2899](https://tex.z-dn.net/?f=%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B11-1%7D%5Ctimes%20%5B%2819-27.182%29%5E%7B2%7D%2B%2819-27.182%29%5E%7B2%7D%2B...%2B%2840-27.182%29%5E%7B2%7D%5D%7D%7D%5C%5C%5C%5C%3D%5Csqrt%7B%5Cfrac%7B395.6364%7D%7B10%7D%7D%5C%5C%5C%5C%3D6.28996%5C%5C%5C%5C%5Capprox%206.2899)
Thus, the standard deviation of the age distribution is 6.2899 years.