Question

In Exercises 3.201-3.206, we have provided simple data sets for you to practice the basics of finding a

a. population mean

b. population standard deviation

3.205$4,2,0,2,2$

Short answer

a. The population mean of given data is $2$.

b.The population standard deviation of given data is $1.264$.

Step-by-step solution

Part(a) Step 1: Given information 

We have been given that the data set is $4,2,0,2,2$.

We need to find out population mean.

Part(a) Step 2: Simplify

The formula to find the population mean is:

$\mu =\frac{{\displaystyle \sum _{i=1}^N{x}_i}}{N}$

Where,

$\mu =$Population mean

$\sum _{i=1}^N{x}_i=$The sum of all the individual items in the group

$N=$The number of items in the group

Putting the given data set $4,2,0,2,2$in population mean formula,

$$\begin{gathered} \mu =\frac{4+2+0+2+2}{5} \\ \mu =\frac{10}{5}=2 \end{gathered}$$

Part(b) step 1: Given information 

We have been given that the data set is $4,2,0,2,2$.

We need to find out population standard deviation.

Part(b) Step 2: Simplify

The formula to find the population standard deviation is:

$\sigma =\sqrt{\frac{{\displaystyle \sum _{i=1}^N{\left(x_i-\mu \right)}^2}}{N}}$

Where,

Population standard deviation

Each value from the population

The number of items in the population group

Population mean

Putting the given data set $4,2,0,2,2$and $\mu =2$in population standard deviation formula,

$$\begin{gathered} \sigma =\sqrt{\frac{{\left(4-2\right)}^2+{\left(2-2\right)}^2+{\left(0-2\right)}^2+{\left(2-2\right)}^2+{\left(2-2\right)}^2}{5}} \\ \sigma =\sqrt{\frac{4+0+4+0+0}{5}} \\ \sigma =\sqrt{\frac{8}{5}}=\sqrt{1.6} \\ \sigma =1.264 \end{gathered}$$