Respuesta :
The numbers of regular-season wins for 10 football teams in a given season are given below
[tex]2,10,15,4,14,7,14,8,2,10[/tex]We are asked to find the range, mean, variance, and standard deviation of the population data set.
Range:
The range is the difference between the maximum value and the minimum value in a data set.
From the given data set,
Maximum value = 15
Minimum value = 2
[tex]\begin{gathered} \text{Range}=\text{maximum}-\text{minimum} \\ \text{Range}=15-2 \\ \text{Range}=13 \end{gathered}[/tex]Therefore, the range is 13
Mean:
The population mean is given by
[tex]\mu=\frac{\sum^{}_{}X}{N}[/tex]Where X is the terms in the data set and N is the number of terms in the data set.
[tex]\begin{gathered} \mu=\frac{2+10+15+4+14+7+14+8+2+10}{10} \\ \mu=\frac{86}{10} \\ \mu=8.6 \end{gathered}[/tex]Therefore, the population mean is 8.6
Variance:
The population variance is given by
[tex]\sigma^2=\frac{\sum^{}_{}(X-\mu)^2}{N}[/tex]Where X is the terms in the data set, μ is the mean, and N is the number of terms in the data set.
[tex]\begin{gathered} \sigma^2=\frac{\sum^{}_{}(X-\mu)^2}{N} \\ \sigma^2=\frac{(2-8.6)^2+(10-8.6)^2+(15-8.6)^2+(4-8.6)^2+(14-8.6)^2+(7-8.6)^2+(14-8.6)^2+(8-8.6)^2++(2-8.6)^2++(10-8.6)^2}{10} \\ \sigma^2=\frac{214.4}{10} \\ \sigma^2=21.4 \end{gathered}[/tex]Therefore, the population variance is 21.4
Standard deviation:
The population standard deviation is given by
[tex]\begin{gathered} \sigma^{}=\sqrt[]{\frac{\sum^{}_{}(X-\mu)^2}{N}} \\ \sigma=\sqrt[]{\sigma^2} \end{gathered}[/tex]Since we have already find the population variance, we can simply find take the square root of variance.
[tex]\begin{gathered} \sigma=\sqrt[]{\sigma^2} \\ \sigma=\sqrt[]{21.4} \\ \sigma=4.6 \end{gathered}[/tex]Therefore, the population standard deviation is 4.6
