In Exercises 5–20, find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as “minutes”) in your results. (The same data were used in Section 3-1, where we found measures of center. Here we find measures of variation.) Then answer the given questions.

Football Player Weights Listed below are the weights in pounds of 11 players randomly selected from the roster of the Seattle Seahawks when they won Super Bowl XLVIII (the same players from the preceding exercise). Are the measures of variation likely to be typical of all NFL players?

189 254 235 225 190 305 195 202 190 252 305

Weights (in pounds) of 11 players (represented as n) are provided.

The following are the measures of dispersion that are generally computed for a sample of data:

The difference between the greatest and the smallest values for a given set of data is called therange.

The ratio of the squared difference of a set of data from its mean to the value of the sample size minus one is called thesample variance.It can be mathematically written as

$s^2=\frac{{\sum }_{1=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1}$

The under root value of the sample variance is same as the value of the sample standard deviation with the original units.

The range of weights is obtained as

$$\begin{gathered} \text{Range}=\text{Greatest}\text{Value}-\text{Smallest}\text{Value} \\ =305-189 \\ =116.0\text{pounds} \end{gathered}$$

Therefore, the range of weights is equal to 116.0 pounds.

The mean weight of the players is calculated as

$$\begin{gathered} \overline{x}=\frac{{\sum }_{1=1}^n{x}_i}{n} \\ =\frac{189+254+...+305}{11} \\ =\frac{2542}{11} \\ \approx 231.1 \end{gathered}$$

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Thus, the mean weight of the players is 231.1 pounds.

The variance $\left(s^2\right)$of the sample of weights is

$$\begin{gathered} s^2=\frac{{\sum }_{1=1}^n{\left(x_i-\overline{x}\right)}^2}{n-1} \\ =\frac{{\left(189-231.1\right)}^2+{\left(254-231.1\right)}^2+...+{\left(305-231.1\right)}^2}{11-1} \\ =\frac{19236.909}{10} \\ \approx 1923.7 \end{gathered}$$

Therefore, the variance of weights is equal to 1923.7 pounds squared.

The standard deviation of the sample of weights is equal to

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{1923.7} \\ \approx 43.9 \end{gathered}$$

Therefore, the standard deviation of weights is equal to 43.9 pounds.

Here, the players are chosen randomly from the same team (Seattle Seahawks). So, there is a possibility that the players of this team may possess the same features in common.

So, the calculated values cannot be used to represent the entire population of players.

Therefore, the measures of variation are not likely to be considered as the typical of all NFL players.