find the standard deviation for the given set of data. round your answer to the nearest tenth.\n{3, 5, 6, 7…

Explanation:

Step1: Calculate the mean

The mean $\bar{x}$ of a data set ${x_1,x_2,\cdots,x_n}$ is given by $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$. Here, $n = 7$, and $\sum_{i=1}^{7}x_i=3 + 5+6+7+9+11+22=63$. So, $\bar{x}=\frac{63}{7}=9$.

Step2: Calculate the squared differences

For each data point $x_i$, calculate $(x_i-\bar{x})^2$:

• When $x_1 = 3$, $(3 - 9)^2=(-6)^2 = 36$.
• When $x_2 = 5$, $(5 - 9)^2=(-4)^2 = 16$.
• When $x_3 = 6$, $(6 - 9)^2=(-3)^2 = 9$.
• When $x_4 = 7$, $(7 - 9)^2=(-2)^2 = 4$.
• When $x_5 = 9$, $(9 - 9)^2=0^2 = 0$.
• When $x_6 = 11$, $(11 - 9)^2=2^2 = 4$.
• When $x_7 = 22$, $(22 - 9)^2=13^2 = 169$.

Step3: Calculate the variance

The variance $s^2$ (for a sample) is given by $s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}$. $\sum_{i = 1}^{7}(x_i-\bar{x})^2=36+16+9+4+0+4+169 = 248$. So, $s^2=\frac{248}{7-1}=\frac{248}{6}\approx41.33$.

Step4: Calculate the standard deviation

The standard deviation $s=\sqrt{s^2}$. $s=\sqrt{\frac{248}{6}}\approx\sqrt{41.33}\approx6.4$.

Answer:

$6.4$