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Jazzie · Answer 1 · 2023-10-25T04:27:50+0000

The numbers of regular-season wins for 10 football teams in a given season are given below

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

$\begin{gathered} \text{Range}=\text{maximum}-\text{minimum} \\ \text{Range}=15-2 \\ \text{Range}=13 \end{gathered}$

Therefore, the range is 13

Mean:

The population mean is given by

$\mu=\frac{\sum^{}_{}X}{N}$

Where X is the terms in the data set and N is the number of terms in the data set.

$\begin{gathered} \mu=(2+10+15+4+14+7+14+8+2+10)/(10) \\ \mu=(86)/(10) \\ \mu=8.6 \end{gathered}$

Therefore, the population mean is 8.6

Variance:

The population variance is given by

$\sigma^2=\frac{\sum^{}_{}(X-\mu)^2}{N}$

Where X is the terms in the data set, μ is the mean, and N is the number of terms in the data set.

$\begin{gathered} \sigma^2=\frac{\sum^{}_{}(X-\mu)^2}{N} \\ \sigma^2=((2-8.6)^2+(10-8.6)^2+(15-8.6)^2+(4-8.6)^2+(14-8.6)^2+(7-8.6)^2+(14-8.6)^2+(8-8.6)^2++(2-8.6)^2++(10-8.6)^2)/(10) \\ \sigma^2=(214.4)/(10) \\ \sigma^2=21.4 \end{gathered}$

Therefore, the population variance is 21.4

Standard deviation:

The population standard deviation is given by

$\begin{gathered} \sigma^{}=\sqrt[]{\frac{\sum^{}_{}(X-\mu)^2}{N}} \\ \sigma=\sqrt[]{\sigma^2} \end{gathered}$

Since we have already find the population variance, we can simply find take the square root of variance.

$\begin{gathered} \sigma=\sqrt[]{\sigma^2} \\ \sigma=\sqrt[]{21.4} \\ \sigma=4.6 \end{gathered}$

Therefore, the population standard deviation is 4.6

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