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

$3.2051,9,8,4,3$

Short answer

a. The population mean of given data is $5$.

b. The population standard deviation of given data is$3.033$.

Step-by-step solution

Part(a) Step 1: Given information 

We have been given that the data set is $1,9,8,4,3$.

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 $1,9,8,4,3$in population mean formula,

role="math" localid="1650726647545" $$\begin{gathered} \mu =\frac{1+9+8+4+3}{5} \\ \mu =\frac{25}{5}=5 \end{gathered}$$

Part(b) step 1: Given information 

We have been given that the data set is $1,9,8,4,3$.

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,

$\sigma =$Population standard deviation

$x_i=$Each value from the population

$N=$The number of items in the population group

$\mu =$Population mean

Putting the given data set $1,9,8,3,4$and $\mu =5$in population standard deviation formula,

$$\begin{gathered} \sigma =\sqrt{\frac{{\left(1-5\right)}^2+{\left(9-5\right)}^2+{\left(8-5\right)}^2+{\left(4-5\right)}^2+{\left(3-5\right)}^2}{5}} \\ \sigma =\sqrt{\frac{16+16+9+1+4}{5}} \\ \sigma =\sqrt{\frac{46}{5}}=\sqrt{9.2} \\ \sigma =3.033 \end{gathered}$$