fill in the blank 1 point\nthe number of wins for each team in a league last year are listed. approximate…

Explanation:

Step1: Calculate the sum of data

The data set is (4, 10, 6, 10, 4, 3, 3, 4, 3, 4, 5, 7, 3, 6, 4, 9, 4, 3, 5, 3). The sum (S=\sum_{i = 1}^{n}x_{i}=4 + 10+6 + 10+4+3+3+4+3+4+5+7+3+6+4+9+4+3+5+3=105).

Step2: Calculate the population mean (\mu)

There are (n = 20) data - points. The population mean (\mu=\frac{S}{n}=\frac{105}{20}=5.25).

Step3: Calculate the squared - differences ((x_{i}-\mu)^{2})

For (x_1 = 4), ((4 - 5.25)^{2}=(-1.25)^{2}=1.5625); for (x_2 = 10), ((10 - 5.25)^{2}=(4.75)^{2}=22.5625); and so on for all 20 data - points.

Step4: Calculate the sum of squared - differences (\sum_{i = 1}^{n}(x_{i}-\mu)^{2})

(\sum_{i = 1}^{20}(x_{i}-5.25)^{2}=1.5625+22.5625+0.5625+22.5625+1.5625+4.515625+4.515625+1.5625+4.515625+1.5625+0.0625+3.0625+4.515625+0.5625+1.5625+13.265625+1.5625+4.515625+0.0625+4.515625 = 105.75)

Step5: Calculate the population standard deviation (\sigma)

The formula for the population standard deviation is (\sigma=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\mu)^{2}}{n}}). Substituting (n = 20) and (\sum_{i = 1}^{n}(x_{i}-\mu)^{2}=105.75), we get (\sigma=\sqrt{\frac{105.75}{20}}\approx\sqrt{5.2875}\approx2.3)

Answer:

Population mean: (5.25), Population standard deviation: approximately (2.3)