Question

4What is the standard deviation of the following data set rounded to the nearest tenth?52.1, 45.5, 51, 48.8, 43.610.36.910.43.2

Answer 1

Given:

The data set is 52.1, 45.5, 51, 48.8, 43.6.

Here, number of points are n=5.

The mean of this data is,

`\begin{gathered} \mu=\frac{\sum^{}_{}x_i}{n} \\ =(52.1+45.5+51+48.8+43.6.)/(5) \\ =48.2 \end{gathered}`

The variance is calculated as,

`\begin{gathered} \sigma^2=\frac{\sum^{}_{}(x_i-\mu)^2}{n} \\ =((52.1-48.2)^2+(45.5-48.2)^2+(51-48.2)^2+(48.8-48.2)^2+(43.6-48.2)^2)/(5) \\ =(51.86)/(5) \\ =10.372 \end{gathered}`

The standard deviation is,

`S\mathrm{}D\mathrm{}=\sqrt[]{10.372}=3.22`