Question

The data set shows the ages of children who are chosen for roles in a play 12,14,1,10,3. (A) what is the x of the data set. (B) The table below will help you find the sum of the squares of differences between each data value in the mean use the table to organize your work and answersWhat is the standard deviation of the data set? Use the sum from part B and show your work round final answer to the tenths place.

Answer 1

Answer:

Step-by-step explanation:

The mean of the data set is:

`\begin{gathered} \bar{x}=(12+14+1+10+3)/(5) \\ \\ =(40)/(5)=8 \end{gathered}`

For x = 12:

`\begin{gathered} x-\bar{x}=12-8=4 \\ (x-\bar{x})^2=4^2=16 \end{gathered}`

For x = 14:

`\begin{gathered} x-\bar{x}^{}=14-8=6 \\ (x-\bar{x})^2=6^2=36 \end{gathered}`

For x = 1:

`\begin{gathered} x-\bar{x}^{}=1-8=-7 \\ (x-\bar{x})^2=(-7)^2=49 \end{gathered}`

For x = 10:

`\begin{gathered} x-\bar{x}^{}=10-8=2 \\ (x-\bar{x})^2=2^2=4 \end{gathered}`

For x = 3:

`\begin{gathered} x-\bar{x}^{}=3-8^{}=-5 \\ (x-\bar{x})^2=(-5)^2=25 \end{gathered}`

Sum is: S = 16 + 36 + 49 + 4 + 25 = 130

Standard Deviation:

`\begin{gathered} SD=\sqrt[]{(S)/(5)}=\sqrt[]{(130)/(5)} \\ \\ =\sqrt[]{26}=5.1 \end{gathered}`