Question

find (a) the range and (b) the standard deviation of the set of data. 14, 32, 8, 17, 14, 29, 26 (a) the range is 24. (simplify your answer.) (b) the standard deviation is (round to the nearest hundredth as needed.)

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 $\sum_{i=1}^{7}x_i=14 + 32+8+17+14+29+26=140$.
So, $\bar{x}=\frac{140}{7}=20$.

Step2: Calculate the variance

The variance $s^{2}$ of a sample data set is $s^{2}=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^{2}}{n - 1}$.
$(14 - 20)^{2}=(-6)^{2}=36$, $(32 - 20)^{2}=12^{2}=144$, $(8 - 20)^{2}=(-12)^{2}=144$, $(17 - 20)^{2}=(-3)^{2}=9$, $(14 - 20)^{2}=36$, $(29 - 20)^{2}=9^{2}=81$, $(26 - 20)^{2}=6^{2}=36$.
$\sum_{i = 1}^{7}(x_i-\bar{x})^{2}=36+144+144+9+36+81+36=486$.
Then $s^{2}=\frac{486}{7 - 1}=\frac{486}{6}=81$.

Step3: Calculate the standard deviation

The standard deviation $s=\sqrt{s^{2}}$.
Since $s^{2}=81$, then $s=\sqrt{81}=9.00$

Answer:

(a) The range is $24$.
(b) The standard deviation is $9.00$.