for the following set of data, find the sample standard deviation, to the nearest thousandth.\n| data |…

Explanation:

Step1: Calculate the sample mean $\bar{x}$

First, find the sum of the products of data - value and frequency $\sum_{i = 1}^{n}x_if_i$ and the total frequency $n=\sum_{i = 1}^{n}f_i$. $\sum_{i = 1}^{n}x_if_i=14\times5 + 15\times9+16\times7 + 24\times4+27\times5+29\times1+31\times4$ $=70 + 135+112 + 96+135+29+124$ $=601$. $n=5 + 9+7 + 4+5+1+4=35$. $\bar{x}=\frac{\sum_{i = 1}^{n}x_if_i}{n}=\frac{601}{35}\approx17.1714$.

Step2: Calculate the sum of squared - deviations $\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2$

For $x_1 = 14,f_1 = 5$: $(14 - 17.1714)^2\times5=( - 3.1714)^2\times5 = 9.9978\times5 = 49.989$. For $x_2 = 15,f_2 = 9$: $(15 - 17.1714)^2\times9=( - 2.1714)^2\times9 = 4.714\times9 = 42.426$. For $x_3 = 16,f_3 = 7$: $(16 - 17.1714)^2\times7=( - 1.1714)^2\times7 = 1.372\times7 = 9.604$. For $x_4 = 24,f_4 = 4$: $(24 - 17.1714)^2\times4=(6.8286)^2\times4 = 46.630\times4 = 186.52$. For $x_5 = 27,f_5 = 5$: $(27 - 17.1714)^2\times5=(9.8286)^2\times5 = 96.601\times5 = 483.005$. For $x_6 = 29,f_6 = 1$: $(29 - 17.1714)^2\times1=(11.8286)^2\times1 = 139.91$. For $x_7 = 31,f_7 = 4$: $(31 - 17.1714)^2\times4=(13.8286)^2\times4 = 191.23\times4 = 764.92$. $\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2=49.989+42.426 + 9.604+186.52+483.005+139.91+764.92$ $=1676.374$.

Step3: Calculate the sample standard deviation $s$

The formula for the sample standard deviation of grouped data is $s=\sqrt{\frac{\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2}{n - 1}}$. Here, $n = 35$, so $s=\sqrt{\frac{1676.374}{35 - 1}}=\sqrt{\frac{1676.374}{34}}\approx\sqrt{49.3051}\approx7.022$.

Answer:

$7.022$