measures of dispersion\nten friends kept track of how many times they sent a text over a one - hour period…

Explanation:

Step1: Recall range formula

The range of a data - set is given by $Range = \text{Max value}-\text{Min value}$.

Step2: Identify max and min values

In the data set ${4,4,4,6,7,8,10,11,18,19}$, the minimum value is $4$ and the maximum value is $19$.

Step3: Calculate the range

$Range=19 - 4=15$.

Step4: Recall sample standard - deviation formula

The sample standard deviation formula is $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$, where $n$ is the number of data points, $x_{i}$ are the individual data points, and $\bar{x}$ is the mean. Here, $n = 10$ and $\bar{x}=9.1$.

Step5: Calculate $(x_{i}-\bar{x})^{2}$ for each data point

For $x_1 = 4$: $(4 - 9.1)^{2}=(-5.1)^{2}=26.01$. Since there are 3 values of 4, the total contribution is $3\times26.01 = 78.03$. For $x_4=6$: $(6 - 9.1)^{2}=(-3.1)^{2}=9.61$. For $x_5 = 7$: $(7 - 9.1)^{2}=(-2.1)^{2}=4.41$. For $x_6 = 8$: $(8 - 9.1)^{2}=(-1.1)^{2}=1.21$. For $x_7 = 10$: $(10 - 9.1)^{2}=(0.9)^{2}=0.81$. For $x_8 = 11$: $(11 - 9.1)^{2}=(1.9)^{2}=3.61$. For $x_9 = 18$: $(18 - 9.1)^{2}=(8.9)^{2}=79.21$. For $x_{10}=19$: $(19 - 9.1)^{2}=(9.9)^{2}=98.01$.

Step6: Calculate the sum $\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}$

$\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=78.03+9.61 + 4.41+1.21+0.81+3.61+79.21+98.01=274.9$.

Step7: Calculate the sample standard deviation

$s=\sqrt{\frac{274.9}{10 - 1}}=\sqrt{\frac{274.9}{9}}\approx\sqrt{30.5444}\approx5.53$.

Answer:

Range: 15 Sample standard deviation: 5.53