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Find the sum of squares of the deviations from the mean of a series of ACT scores listed below.31, 22, 24, 15, 23The sum of squares is(Type an integer or a decimal.)

User KAmol
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1 Answer

10 votes
10 votes

Solution

Given the following scores below


31,22,24,15,23

To find the sum of squares of the deviations,

Firstly, we find the mean, the formula to find the mean of ungrouped data is


\begin{gathered} \bar{x}=(\sum_X)/(n) \\ Where\text{ n is the number of terms} \end{gathered}

Where n = 5

Substitute the given data into the formula to find the mean of ungrouped data


\begin{gathered} \bar{x}=(31+22+24+15+23)/(5) \\ \bar{x}=(115)/(5)=23 \\ \bar{x}=23 \end{gathered}

Secondly, we subtract the mean from each of the values given as shown below


\begin{gathered} 31-23=8 \\ 22-23=-1 \\ 24-23=1 \\ 15-23=-8 \\ 23-23=0 \end{gathered}

Then, to find the sum of squares of the deviations

We add the square of each of the deviations deduced above


\begin{gathered} =(8)^2+(-1)^2+(1)^2+(-8)^2+(0)^2 \\ =64+1+1+64+0 \\ =130 \end{gathered}

Hence, the sum of squares is 130

User Davidcelis
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