91影视

The given data shows the annual costs of tuition and fees of the top 10 colleges of the United States.

Therange measures the overall variation in a dataset. It is computed as follows:

$$\begin{gathered} \text{Range}=\text{Maximum}\text{Value}-\text{Minimum}\text{Value} \\ =49138-46785 \\ =2353.0\text{dollars} \end{gathered}$$

Hence, the value of the range is 2353.0 dollars.

Thevariance measures the square of the deviation between the values from the mean value. It is computed as follows:

$s^2=\frac{{鈭�}_{i=1}^n{\left(x_i-\stackrel{炉}{x}\right)}^2}{n-1}$

Thestandard deviation measures the deviation between the values from the mean value. It is computed as follows:

$s=\sqrt{s^2}$

The sample mean is computed as follows:

$$\begin{gathered} \stackrel{炉}{x}=\frac{{鈭�}_{i=1}^n{x}_i}{n} \\ =\frac{49+138+....+785}{10} \\ =47327.4\text{dollars} \end{gathered}$$

Thus, the sample mean is equal to 47327.4 dollars.

The variance is calculated as follows, substituting the values:

$$\begin{gathered} s^2=\frac{{鈭�}_{i=1}^n{\left(x_i-\stackrel{炉}{x}\right)}^2}{n-1} \\ =\frac{{\left(49-47327.4\right)}^2+{\left(138-47327.4\right)}^2+...+{\left(785-47327.4\right)}^2}{10-1} \\ =527910.5\text{dollars} \end{gathered}$$

Hence, the value of variance is 527910.5 dollars square.

The standard deviation is calculated as follows:

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{527910.5} \\ =726.6\text{dollars} \end{gathered}$$

Hence, the value of standard deviation is 726.6 dollars.

The sample fees are taken from the top 10 colleges in the US. They are randomly selected samples taken from the population of all fees concerning all US colleges.

Thus, the values of the variation from the sample are not appropriate representative of the costs of the entire population of US colleges.