Question

measures of dispersion
ten friends kept track of how many times they sent a text over a one - hour period. the results are in the table below.
number of texts 4 4 4 6 7 8 10 11 18 19
they determine that the mean of the data set is 9.1, but they also want to know the range and standard deviation.
calculate the range of the data set.
texts
calculate the sample standard deviation of the data set.
texts
round your result to the two decimal places as needed.
question help: message instructor

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