Question

the numbers of regular - season wins for 10 football teams in a given season are given below. determine the range, mean, variance, and standard deviation of the population data set. 2, 6, 15, 5, 13, 6, 13, 10, 2, 10
the range is 13. (simplify your answer.)
the population mean is 8.2 (simplify your answer. round to the nearest tenth as needed.)
the population variance is (simplify your answer. round to the nearest tenth as needed.)

Explanation:

Step1: Recall variance formula

The formula for population variance $\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\mu)^{2}}{n}$, where $x_{i}$ are the data - points, $\mu$ is the population mean, and $n$ is the number of data - points. Here $n = 10$ and $\mu=8.2$.

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

For $x_1 = 2$: $(2 - 8.2)^{2}=(-6.2)^{2}=38.44$
For $x_2 = 6$: $(6 - 8.2)^{2}=(-2.2)^{2}=4.84$
For $x_3 = 15$: $(15 - 8.2)^{2}=(6.8)^{2}=46.24$
For $x_4 = 5$: $(5 - 8.2)^{2}=(-3.2)^{2}=10.24$
For $x_5 = 13$: $(13 - 8.2)^{2}=(4.8)^{2}=23.04$
For $x_6 = 6$: $(6 - 8.2)^{2}=(-2.2)^{2}=4.84$
For $x_7 = 13$: $(13 - 8.2)^{2}=(4.8)^{2}=23.04$
For $x_8 = 10$: $(10 - 8.2)^{2}=(1.8)^{2}=3.24$
For $x_9 = 2$: $(2 - 8.2)^{2}=(-6.2)^{2}=38.44$
For $x_{10}=10$: $(10 - 8.2)^{2}=(1.8)^{2}=3.24$

Step3: Sum up $(x_{i}-\mu)^{2}$

$\sum_{i = 1}^{10}(x_{i}-\mu)^{2}=38.44 + 4.84+46.24+10.24+23.04+4.84+23.04+3.24+38.44+3.24 = 195.6$

Step4: Calculate the population variance

$\sigma^{2}=\frac{\sum_{i = 1}^{10}(x_{i}-\mu)^{2}}{n}=\frac{195.6}{10}=19.6$

Answer:

$19.6$