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A sample consisting of the highest amounts of the net worths of 10 celebrities (n) is given.

The range of a sample is computed by deducting the minimum value from the maximum value.

$$\begin{gathered} \text{Range}=\text{Largest}\text{Value}-\text{Lowest}\text{Value} \\ =250-150 \\ =100.0\text{million}\text{dollars} \end{gathered}$$

Therefore, the range of net worths is equal to $100.0 million.

Thevariance$\left(s^2\right)$ and thestandard deviation (s)assess the amount of dispersion present in a set of values.

The sample variance is calculated as shown below:

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

Here,

x represents the observations, and

$\stackrel{炉}{x}$is the sample mean.

The sample mean is calculated as

.$$\begin{gathered} \stackrel{炉}{x}=\frac{{鈭�}_{1=1}^n{x}_i}{n} \\ =\frac{250+200+...+150}{10} \\ =\frac{1720}{10} \\ 鈮�172 \end{gathered}$$

Thus, the sample mean is $172.0 million.

Now, the variance is calculated as

$$\begin{gathered} s^2=\frac{{鈭�}_{i=1}^n{\left(x_i-\stackrel{炉}{x}\right)}^2}{n-1} \\ =\frac{{\left(250-172\right)}^2+{\left(200-172\right)}^2+...+{\left(150-172\right)}^2}{10-1} \\ =\frac{9310}{9} \\ =1034.4 \end{gathered}$$

.

Therefore, the variance of the net worths is equal to $1034.4${\left(\text{million}\text{dollars}\right)}^2$.

The sample standard deviation is calculated as shown below

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{1034.4} \\ 鈮�32.2 \end{gathered}$$

Therefore, the standard deviation of net worths is equal to $32.2 million.

As the sample comprises only those celebrities who have the highest amounts of net worth, the measures of variation are not typical for all the celebrities as they do not represent the entire group of celebrities.