Question

for the following set of data, find the population standard deviation, to the nearest thousandth.
101, 79, 117, 99, 96, 112, 87, 75

Explanation:

Step1: Calculate the mean

Let the data set be $x_1 = 101,x_2=79,x_3 = 117,x_4=99,x_5=96,x_6=112,x_7=87,x_8=75$. The mean $\mu=\frac{\sum_{i = 1}^{n}x_i}{n}$, where $n = 8$.
$\mu=\frac{101 + 79+117+99+96+112+87+75}{8}=\frac{766}{8}=95.75$

Step2: Calculate the squared - differences

$(x_1-\mu)^2=(101 - 95.75)^2=(5.25)^2 = 27.5625$
$(x_2-\mu)^2=(79 - 95.75)^2=(-16.75)^2=280.5625$
$(x_3-\mu)^2=(117 - 95.75)^2=(21.25)^2 = 451.5625$
$(x_4-\mu)^2=(99 - 95.75)^2=(3.25)^2 = 10.5625$
$(x_5-\mu)^2=(96 - 95.75)^2=(0.25)^2 = 0.0625$
$(x_6-\mu)^2=(112 - 95.75)^2=(16.25)^2 = 264.0625$
$(x_7-\mu)^2=(87 - 95.75)^2=(-8.75)^2 = 76.5625$
$(x_8-\mu)^2=(75 - 95.75)^2=(-20.75)^2 = 430.5625$

Step3: Calculate the variance

The population variance $\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_i-\mu)^2}{n}$.
$\sum_{i = 1}^{8}(x_i-\mu)^2=27.5625+280.5625+451.5625+10.5625+0.0625+264.0625+76.5625+430.5625 = 1541.5$
$\sigma^{2}=\frac{1541.5}{8}=192.6875$

Step4: Calculate the standard deviation

The population standard deviation $\sigma=\sqrt{\sigma^{2}}$.
$\sigma=\sqrt{192.6875}\approx13.881$

Answer:

$13.881$