A group of data items and their mean are given.

[tex]\[ 24, 40, 56, 96, 144, 216; \text{ Mean } = 96 \][/tex]

a. Find the deviation from the mean for each of the data items.
b. Find the sum of the deviations in part (a).

a. Type the deviation from the mean for each of the data items.

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|}
\hline 24 & 40 & 56 & 96 & 144 & 216 \\
\hline \square & \square & \square & \square & \square & \square \\
\hline
\end{tabular}
\][/tex]

(Simplify your answers.)

b. The sum of the deviations in part (a) is [tex]\(\square\)[/tex]. (Simplify your answer.)



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].