Question

the grade point averages (gpa) for 12 randomly selected college students are shown on the right. complete parts (a) through (c) below. assume the population is normally distributed. 2.4 3.1 2.9 1.6 0.9 4.0 2.3 1.2 3.5 0.3 2.1 3.4 (a) find the sample mean. \\( \overline { x } = 2.31 \\) (round to two decimal places as needed.) (b) find the sample standard deviation. \\( s = \square \\) (round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for sample standard deviation

The formula for sample standard deviation \(s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}\), where \(n\) is the sample size, \(x_{i}\) are the data points, and \(\bar{x}\) is the sample mean. Here \(n = 12\) and \(\bar{x}=2.31\).

Step2: Calculate \((x_{i}-\bar{x})^{2}\) for each data point

For \(x_1 = 2.4\): \((2.4 - 2.31)^{2}=(0.09)^{2}=0.0081\)
For \(x_2 = 3.1\): \((3.1 - 2.31)^{2}=(0.79)^{2}=0.6241\)
For \(x_3 = 2.9\): \((2.9 - 2.31)^{2}=(0.59)^{2}=0.3481\)
For \(x_4 = 1.6\): \((1.6 - 2.31)^{2}=(- 0.71)^{2}=0.5041\)
For \(x_5 = 0.9\): \((0.9 - 2.31)^{2}=(-1.41)^{2}=1.9881\)
For \(x_6 = 4.0\): \((4.0 - 2.31)^{2}=(1.69)^{2}=2.8561\)
For \(x_7 = 2.3\): \((2.3 - 2.31)^{2}=(-0.01)^{2}=0.0001\)
For \(x_8 = 1.2\): \((1.2 - 2.31)^{2}=(-1.11)^{2}=1.2321\)
For \(x_9 = 3.5\): \((3.5 - 2.31)^{2}=(1.19)^{2}=1.4161\)
For \(x_{10}=0.3\): \((0.3 - 2.31)^{2}=(-2.01)^{2}=4.0401\)
For \(x_{11}=2.1\): \((2.1 - 2.31)^{2}=(-0.21)^{2}=0.0441\)
For \(x_{12}=3.4\): \((3.4 - 2.31)^{2}=(1.09)^{2}=1.1881\)

Step3: Sum up \((x_{i}-\bar{x})^{2}\)

\(\sum_{i = 1}^{12}(x_{i}-\bar{x})^{2}=0.0081 + 0.6241+0.3481 + 0.5041+1.9881+2.8561+0.0001+1.2321+1.4161+4.0401+0.0441+1.1881=14.259\)

Step4: Calculate the sample standard deviation

\(s=\sqrt{\frac{14.259}{12 - 1}}=\sqrt{\frac{14.259}{11}}\approx\sqrt{1.2963}\approx1.14\)

Answer:

\(s = 1.14\)