for the following set of data, find the population standard deviation, to the nearest thousandth.\ndata…

Explanation:

Step1: Calculate the mean $\mu$

First, find the sum of the products of data - value and frequency $\sum_{i = 1}^{n}x_if_i$ and the total frequency $\sum_{i = 1}^{n}f_i$. $\sum_{i = 1}^{n}x_if_i=9\times7 + 14\times8+21\times7 + 27\times8+35\times6+38\times2+39\times5+45\times8$ $=63 + 112+147+216+210+76+195+360$ $=1389$. $\sum_{i = 1}^{n}f_i=7 + 8+7 + 8+6+2+5+8=51$. The mean $\mu=\frac{\sum_{i = 1}^{n}x_if_i}{\sum_{i = 1}^{n}f_i}=\frac{1389}{51}\approx27.235$.

Step2: Calculate the sum of squared - differences times frequency $\sum_{i = 1}^{n}f_i(x_i-\mu)^2$

For $x_1 = 9,f_1 = 7$: $f_1(x_1-\mu)^2=7\times(9 - 27.235)^2=7\times(- 18.235)^2=7\times332.515225 = 2327.606575$. For $x_2 = 14,f_2 = 8$: $f_2(x_2-\mu)^2=8\times(14 - 27.235)^2=8\times(-13.235)^2=8\times175.105225 = 1400.8418$. For $x_3 = 21,f_3 = 7$: $f_3(x_3-\mu)^2=7\times(21 - 27.235)^2=7\times(-6.235)^2=7\times38.875225 = 272.126575$. For $x_4 = 27,f_4 = 8$: $f_4(x_4-\mu)^2=8\times(27 - 27.235)^2=8\times(-0.235)^2=8\times0.055225 = 0.4418$. For $x_5 = 35,f_5 = 6$: $f_5(x_5-\mu)^2=6\times(35 - 27.235)^2=6\times7.765^2=6\times60.305225 = 361.83135$. For $x_6 = 38,f_6 = 2$: $f_6(x_6-\mu)^2=2\times(38 - 27.235)^2=2\times10.765^2=2\times115.881225 = 231.76245$. For $x_7 = 39,f_7 = 5$: $f_7(x_7-\mu)^2=5\times(39 - 27.235)^2=5\times11.765^2=5\times138.401225 = 692.006125$. For $x_8 = 45,f_8 = 8$: $f_8(x_8-\mu)^2=8\times(45 - 27.235)^2=8\times17.765^2=8\times315.595225 = 2524.7618$. $\sum_{i = 1}^{n}f_i(x_i-\mu)^2=2327.606575+1400.8418+272.126575 + 0.4418+361.83135+231.76245+692.006125+2524.7618=7811.3785$.

Step3: Calculate the population standard deviation $\sigma$

The formula for the population standard deviation of a frequency - distribution is $\sigma=\sqrt{\frac{\sum_{i = 1}^{n}f_i(x_i-\mu)^2}{\sum_{i = 1}^{n}f_i}}$. $\sigma=\sqrt{\frac{7811.3785}{51}}\approx\sqrt{153.164284}\approx12.376$.

Answer:

$12.376$