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, 10, 15, 2, 13, 6, 13, 8, 4, 6 the range is (simplify your answer.)

Explanation:

Step1: Find the maximum and minimum values

The data set is \(2, 10, 15, 2, 13, 6, 13, 8, 4, 6\). The maximum value \(x_{max}=15\) and the minimum value \(x_{min} = 2\).

Step2: Calculate the range

The formula for the range \(R\) of a data - set is \(R=x_{max}-x_{min}\). Substitute \(x_{max}=15\) and \(x_{min}=2\) into the formula: \(R = 15 - 2=13\).

Step3: Calculate the mean \(\mu\)

The formula for the mean of a population \(\mu=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), where \(n = 10\) and \(x_{i}\) are the data - points. \(\sum_{i=1}^{10}x_{i}=2 + 10+15+2+13+6+13+8+4+6=79\). So, \(\mu=\frac{79}{10}=7.9\).

Step4: Calculate the variance \(\sigma^{2}\)

The formula for the population variance \(\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\mu)^{2}}{n}\).
\((2 - 7.9)^{2}=(-5.9)^{2}=34.81\), \((10 - 7.9)^{2}=(2.1)^{2}=4.41\), \((15 - 7.9)^{2}=(7.1)^{2}=50.41\), \((2 - 7.9)^{2}=(-5.9)^{2}=34.81\), \((13 - 7.9)^{2}=(5.1)^{2}=26.01\), \((6 - 7.9)^{2}=(-1.9)^{2}=3.61\), \((13 - 7.9)^{2}=(5.1)^{2}=26.01\), \((8 - 7.9)^{2}=(0.1)^{2}=0.01\), \((4 - 7.9)^{2}=(-3.9)^{2}=15.21\), \((6 - 7.9)^{2}=(-1.9)^{2}=3.61\).
\(\sum_{i = 1}^{10}(x_{i}-7.9)^{2}=34.81+4.41+50.41+34.81+26.01+3.61+26.01+0.01+15.21+3.61 = 198.9\).
So, \(\sigma^{2}=\frac{198.9}{10}=19.89\).

Step5: Calculate the standard deviation \(\sigma\)

The formula for the population standard deviation \(\sigma=\sqrt{\sigma^{2}}\). Since \(\sigma^{2}=19.89\), \(\sigma=\sqrt{19.89}\approx4.46\).

Answer:

Range: \(13\)
Mean: \(7.9\)
Variance: \(19.89\)
Standard Deviation: \(\approx4.46\)