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;
[tex]\bar x = \frac{\Sigma x}{N} \\\\\bar x = \frac{53.3}{20} \\\\\bar x = 2.665[/tex]
The square of the difference between the grade point and the mean;
[tex]\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[/tex]
The standard deviation of the grade points is calculated as follows;
[tex]\sigma = \sqrt{\frac{\Sigma (x-\bar x)^2}{N} } \\\\\sigma = \sqrt{\frac{6.5055}{20} } \\\\\sigma = 0.57[/tex]
Thus, the standard deviation of the grade points average of the college seniors is 0.57.
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