Question

question: for the following set of data, find the population standard deviation, to the nearest hundredth. 55, 94, 57, 110, 134, 147, 151, 128

Explanation:

Step1: Calculate the mean

Let the data - set be \(x_1 = 55,x_2 = 94,x_3 = 57,x_4 = 110,x_5 = 134,x_6 = 147,x_7 = 151,x_8 = 128\).
The mean \(\mu=\frac{\sum_{i = 1}^{n}x_i}{n}\), where \(n = 8\).
\(\sum_{i=1}^{8}x_i=55 + 94+57+110+134+147+151+128=876\).
\(\mu=\frac{876}{8}=109.5\).

Step2: Calculate the squared - differences

\((x_1-\mu)^2=(55 - 109.5)^2=(- 54.5)^2 = 2970.25\)
\((x_2-\mu)^2=(94 - 109.5)^2=(-15.5)^2 = 240.25\)
\((x_3-\mu)^2=(57 - 109.5)^2=(-52.5)^2 = 2756.25\)
\((x_4-\mu)^2=(110 - 109.5)^2=(0.5)^2 = 0.25\)
\((x_5-\mu)^2=(134 - 109.5)^2=(24.5)^2 = 600.25\)
\((x_6-\mu)^2=(147 - 109.5)^2=(37.5)^2 = 1406.25\)
\((x_7-\mu)^2=(151 - 109.5)^2=(41.5)^2 = 1722.25\)
\((x_8-\mu)^2=(128 - 109.5)^2=(18.5)^2 = 342.25\)

Step3: Calculate the variance

The population variance \(\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_i-\mu)^2}{n}\).
\(\sum_{i = 1}^{8}(x_i-\mu)^2=2970.25+240.25+2756.25 + 0.25+600.25+1406.25+1722.25+342.25=9038\).
\(\sigma^{2}=\frac{9038}{8}=1129.75\).

Step4: Calculate the standard deviation

The population standard deviation \(\sigma=\sqrt{\sigma^{2}}\).
\(\sigma=\sqrt{1129.75}\approx33.61\).

Answer:

\(33.61\)