The standard deviation of the grade points average of the college seniors is 0.57.
The given parameters:
3.2, 1.9 ,2.7, 2.4, 2.8, 2.9, 3.8, 3.0, 2.5, 3.3, 1.8, 2.5, 3.7, 2.8, 2.0, 3.2, 2.3, 2.1, 2.5, 1.9
The sum of the given data is calculated as follows;
∑x = 3.2 + 1.9 + 2.7 + 2.4 + 2.8 + 2.9 + 3.8 + 3.0 + 2.5 + 3.3 + 1.8 + 2.5 + 3.7 + 2.8 + 2.0 + 3.2 + 2.3 + 2.1 + 2.5 + 1.9
∑x = 53.3
The mean of the distribution is calculated as follows;
![\bar x = \frac{\Sigma x}{N} \\\\\bar x = \frac{53.3}{20} \\\\\bar x = 2.665](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B%5CSigma%20x%7D%7BN%7D%20%5C%5C%5C%5C%5Cbar%20x%20%3D%20%5Cfrac%7B53.3%7D%7B20%7D%20%5C%5C%5C%5C%5Cbar%20x%20%3D%202.665)
The square of the difference between the grade point and the mean;
![\Sigma (x - \bar x)^2 = (3.2- 2.665)^2 + (1.9-2.665)^2 + (2.7- 2.665)^2 + (2.4-2.665)^2 \\\\+(2.8-2.665)^2 +(2.9 - 2.665)^2 + (3.8 - 2.665)^2 + (3-2.665)^2+ \\\\(2.5 -2.665)^2 + (3.3 - 2.665)^2 + (1.8-2.665)^2 + (2.5 - 2.665)^2 + \\\\(3.7-2.665)^2 + (2.8 - 2.665)^2 + (2-2.665)^2 + (3.2-2.665)^2 + \\\\(2.3 - 2.665)^2 + (2.1-2.665)^2 + (2.5-2.665)^2 + (1.9 - 2.665)^2\\\\\Sigma (x - \bar x)^2 = 6.5055](https://tex.z-dn.net/?f=%5CSigma%20%28x%20-%20%5Cbar%20x%29%5E2%20%3D%20%283.2-%202.665%29%5E2%20%2B%20%281.9-2.665%29%5E2%20%2B%20%282.7-%202.665%29%5E2%20%2B%20%282.4-2.665%29%5E2%20%20%5C%5C%5C%5C%2B%282.8-2.665%29%5E2%20%2B%282.9%20-%202.665%29%5E2%20%2B%20%283.8%20-%202.665%29%5E2%20%2B%20%283-2.665%29%5E2%2B%20%5C%5C%5C%5C%282.5%20-2.665%29%5E2%20%2B%20%283.3%20-%202.665%29%5E2%20%2B%20%281.8-2.665%29%5E2%20%2B%20%282.5%20-%202.665%29%5E2%20%2B%20%5C%5C%5C%5C%283.7-2.665%29%5E2%20%2B%20%282.8%20-%202.665%29%5E2%20%2B%20%282-2.665%29%5E2%20%2B%20%283.2-2.665%29%5E2%20%2B%20%5C%5C%5C%5C%282.3%20-%202.665%29%5E2%20%2B%20%282.1-2.665%29%5E2%20%2B%20%282.5-2.665%29%5E2%20%2B%20%281.9%20-%202.665%29%5E2%5C%5C%5C%5C%5CSigma%20%28x%20-%20%5Cbar%20x%29%5E2%20%3D%206.5055)
The standard deviation of the grade points is calculated as follows;
![\sigma = \sqrt{\frac{\Sigma (x-\bar x)^2}{N} } \\\\\sigma = \sqrt{\frac{6.5055}{20} } \\\\\sigma = 0.57](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7B%5Cfrac%7B%5CSigma%20%28x-%5Cbar%20x%29%5E2%7D%7BN%7D%20%7D%20%5C%5C%5C%5C%5Csigma%20%3D%20%5Csqrt%7B%5Cfrac%7B6.5055%7D%7B20%7D%20%7D%20%5C%5C%5C%5C%5Csigma%20%3D%200.57)
Thus, the standard deviation of the grade points average of the college seniors is 0.57.
Learn more about standard deviation here: brainly.com/question/12402189