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, 7, 15, 3, 15, 9, 11, 10, 2, 7 the range is (simplify your answer.)

Explanation:

Step1: Find the range

The range is calculated as \( \text{Range}=\text{Max}-\text{Min} \).
From the data set \( \{2,7,15,3,15,9,11,10,2,7\} \), the maximum value \( \text{Max} = 15 \) and the minimum value \( \text{Min}=2 \).
So, \( \text{Range}=15 - 2 \).

Step2: Calculate the mean

The formula for the population mean \( \mu=\frac{\sum_{i = 1}^{N}x_{i}}{N} \), where \( N = 10 \) (the number of data points) and \( x_{i} \) are the data values.
\(\sum_{i=1}^{10}x_{i}=2 + 7+15+3+15+9+11+10+2+7=81\).
Then \( \mu=\frac{81}{10}=8.1 \).

Step3: Calculate the variance

The formula for the population variance \( \sigma^{2}=\frac{\sum_{i = 1}^{N}(x_{i}-\mu)^{2}}{N} \).
\((2 - 8.1)^{2}=(-6.1)^{2}=37.21\), \((7 - 8.1)^{2}=(-1.1)^{2}=1.21\), \((15 - 8.1)^{2}=(6.9)^{2}=47.61\), \((3 - 8.1)^{2}=(-5.1)^{2}=26.01\), \((15 - 8.1)^{2}=47.61\), \((9 - 8.1)^{2}=(0.9)^{2}=0.81\), \((11 - 8.1)^{2}=(2.9)^{2}=8.41\), \((10 - 8.1)^{2}=(1.9)^{2}=3.61\), \((2 - 8.1)^{2}=37.21\), \((7 - 8.1)^{2}=1.21\).
\(\sum_{i = 1}^{10}(x_{i}-\mu)^{2}=37.21+1.21 + 47.61+26.01+47.61+0.81+8.41+3.61+37.21+1.21=210.9\).
\(\sigma^{2}=\frac{210.9}{10}=21.09\).

Step4: Calculate the standard deviation

The formula for the population standard deviation \( \sigma=\sqrt{\sigma^{2}} \).
Since \( \sigma^{2}=21.09 \), then \( \sigma=\sqrt{21.09}\approx4.6 \).

Answer:

The range is \(13\), the mean is \(8.1\), the variance is \(21.09\), and the standard deviation is approximately \(4.6\).