Question

find the mean and standard deviation for the following data set. round to the nearest tenth.
11, 14, 15, 15, 16, 16, 17, 18, 19, 23
use the mean and standard deviation to select the correct statement.
(1 point)

16.6 is a typical value for the data set. 3.7 is the typical distance that a value in the data set is from 16.6.
16.4 is a typical value for the data set. 3.0 is the typical distance between any pair of values in the data set.
16.6 is a typical value for the data set. 3.7 is the typical distance between any pair of values in the data set.
16.4 is a typical value for the data set. 3.0 is the typical distance that a value in the data set is from 16.4

Explanation:

Step1: Calculate the mean

The mean formula is $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. Here $n = 10$ and $\sum_{i=1}^{10}x_{i}=11 + 14+15+15+16+16+17+18+19+23=166$. So $\bar{x}=\frac{166}{10}=16.6$.

Step2: Calculate the variance

The variance formula is $s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$.
$(11 - 16.6)^{2}=(-5.6)^{2}=31.36$; $(14 - 16.6)^{2}=(-2.6)^{2}=6.76$; $(15 - 16.6)^{2}=(-1.6)^{2}=2.56$ (twice, so $2\times2.56 = 5.12$); $(16 - 16.6)^{2}=(-0.6)^{2}=0.36$ (twice, so $2\times0.36=0.72$); $(17 - 16.6)^{2}=(0.4)^{2}=0.16$; $(18 - 16.6)^{2}=(1.4)^{2}=1.96$; $(19 - 16.6)^{2}=(2.4)^{2}=5.76$; $(23 - 16.6)^{2}=(6.4)^{2}=40.96$.
$\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=31.36+6.76 + 5.12+0.72+0.16+1.96+5.76+40.96=92.8$.
$s^{2}=\frac{92.8}{9}\approx10.31$.

Step3: Calculate the standard deviation

The standard deviation $s=\sqrt{s^{2}}=\sqrt{10.31}\approx3.2$ (using a calculator, more accurately, if we use the formula for sample standard - deviation in a more precise calculation:

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The mean $\bar{x} = 16.6$ and the standard deviation $s\approx3.7$. The mean is a typical value of the data - set, and the standard deviation is the typical distance that a value in the data - set is from the mean.

Answer:

16.6 is a typical value for the data set. 3.7 is the typical distance that a value in the data set is from 16.6.