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, 8, 15, 2, 12, 9, 12, 8, 2, 8. the range is 13 (simplify your answer.) the population mean is 7.8 (simplify your answer. round to the nearest tenth as needed.) the population variance is (simplify your answer. round to the nearest tenth as needed.) the population standard deviation is (simplify your answer. round to the nearest tenth as needed.)

Explanation:

Step1: Recall range formula

Range = Max - Min. The data set is 2, 8, 15, 2, 12, 9, 12, 8, 2, 8. Max = 15, Min = 2. So Range = 15 - 2 = 13.

Step2: Calculate population mean

The formula for population mean $\mu=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. Here $n = 10$, and $\sum_{i=1}^{10}x_{i}=2 + 8+15 + 2+12+9+12+8+2+8=78$. So $\mu=\frac{78}{10}=7.8$.

Step3: Calculate population variance

The formula for population variance $\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\mu)^{2}}{n}$.
$(2 - 7.8)^{2}=(- 5.8)^{2}=33.64$, $(8 - 7.8)^{2}=0.04$, $(15 - 7.8)^{2}=51.84$, $(2 - 7.8)^{2}=33.64$, $(12 - 7.8)^{2}=17.64$, $(9 - 7.8)^{2}=1.44$, $(12 - 7.8)^{2}=17.64$, $(8 - 7.8)^{2}=0.04$, $(2 - 7.8)^{2}=33.64$, $(8 - 7.8)^{2}=0.04$.
$\sum_{i = 1}^{10}(x_{i}-7.8)^{2}=33.64+0.04 + 51.84+33.64+17.64+1.44+17.64+0.04+33.64+0.04 = 199.6$.
So $\sigma^{2}=\frac{199.6}{10}=19.96\approx20.0$.

Step4: Calculate population standard deviation

The formula for population standard deviation $\sigma=\sqrt{\sigma^{2}}$. Since $\sigma^{2}=19.96$, $\sigma=\sqrt{19.96}\approx4.5$.

Answer:

The population variance is 20.0.
The population standard deviation is 4.5.