Answer :
Certainly! Let's go through the given problem step-by-step.
We have the data items: [tex]\(24, 40, 56, 96, 144, 216\)[/tex] and the mean ([tex]\(\overline{x}\)[/tex]) is [tex]\(96\)[/tex].
### Part (a): Finding the deviation from the mean for each of the data items.
The deviation from the mean for a data item is calculated as:
[tex]\[ \text{Deviation} = \text{Data item} - \text{Mean} \][/tex]
Let's calculate the deviation for each data item:
1. For 24:
[tex]\[ \text{Deviation} = 24 - 96 = -72 \][/tex]
2. For 40:
[tex]\[ \text{Deviation} = 40 - 96 = -56 \][/tex]
3. For 56:
[tex]\[ \text{Deviation} = 56 - 96 = -40 \][/tex]
4. For 96:
[tex]\[ \text{Deviation} = 96 - 96 = 0 \][/tex]
5. For 144:
[tex]\[ \text{Deviation} = 144 - 96 = 48 \][/tex]
6. For 216:
[tex]\[ \text{Deviation} = 216 - 96 = 120 \][/tex]
Now, we can fill these values into the table:
\begin{tabular}{|c|c|c|c|c|c|}
\hline 24 & 40 & 56 & 96 & 144 & 216 \\
\hline -72 & -56 & -40 & 0 & 48 & 120 \\
\hline
\end{tabular}
### Part (b): Finding the sum of the deviations in part (a).
The sum of the deviations is simply the sum of the values we just calculated:
[tex]\[ \text{Sum of deviations} = (-72) + (-56) + (-40) + 0 + 48 + 120 \][/tex]
Let's add these up step by step:
[tex]\[ (-72) + (-56) = -128 \][/tex]
[tex]\[ -128 + (-40) = -168 \][/tex]
[tex]\[ -168 + 0 = -168 \][/tex]
[tex]\[ -168 + 48 = -120 \][/tex]
[tex]\[ -120 + 120 = 0 \][/tex]
Thus, the sum of the deviations is [tex]\(0\)[/tex].
So, the answers are:
a.
\begin{tabular}{|c|c|c|c|c|c|}
\hline 24 & 40 & 56 & 96 & 144 & 216 \\
\hline -72 & -56 & -40 & 0 & 48 & 120 \\
\hline
\end{tabular}
b. The sum of the deviations in part (a) is [tex]\(0\)[/tex].
We have the data items: [tex]\(24, 40, 56, 96, 144, 216\)[/tex] and the mean ([tex]\(\overline{x}\)[/tex]) is [tex]\(96\)[/tex].
### Part (a): Finding the deviation from the mean for each of the data items.
The deviation from the mean for a data item is calculated as:
[tex]\[ \text{Deviation} = \text{Data item} - \text{Mean} \][/tex]
Let's calculate the deviation for each data item:
1. For 24:
[tex]\[ \text{Deviation} = 24 - 96 = -72 \][/tex]
2. For 40:
[tex]\[ \text{Deviation} = 40 - 96 = -56 \][/tex]
3. For 56:
[tex]\[ \text{Deviation} = 56 - 96 = -40 \][/tex]
4. For 96:
[tex]\[ \text{Deviation} = 96 - 96 = 0 \][/tex]
5. For 144:
[tex]\[ \text{Deviation} = 144 - 96 = 48 \][/tex]
6. For 216:
[tex]\[ \text{Deviation} = 216 - 96 = 120 \][/tex]
Now, we can fill these values into the table:
\begin{tabular}{|c|c|c|c|c|c|}
\hline 24 & 40 & 56 & 96 & 144 & 216 \\
\hline -72 & -56 & -40 & 0 & 48 & 120 \\
\hline
\end{tabular}
### Part (b): Finding the sum of the deviations in part (a).
The sum of the deviations is simply the sum of the values we just calculated:
[tex]\[ \text{Sum of deviations} = (-72) + (-56) + (-40) + 0 + 48 + 120 \][/tex]
Let's add these up step by step:
[tex]\[ (-72) + (-56) = -128 \][/tex]
[tex]\[ -128 + (-40) = -168 \][/tex]
[tex]\[ -168 + 0 = -168 \][/tex]
[tex]\[ -168 + 48 = -120 \][/tex]
[tex]\[ -120 + 120 = 0 \][/tex]
Thus, the sum of the deviations is [tex]\(0\)[/tex].
So, the answers are:
a.
\begin{tabular}{|c|c|c|c|c|c|}
\hline 24 & 40 & 56 & 96 & 144 & 216 \\
\hline -72 & -56 & -40 & 0 & 48 & 120 \\
\hline
\end{tabular}
b. The sum of the deviations in part (a) is [tex]\(0\)[/tex].