Question

The mean of 5 observations is 5. If three of the observations are 1, 2, 6 and the other two observations are such that each is greater than 5, then the mean deviation from the mean of the observations is

• A. 2.8
• B. 2.6
• C. 2.5
• D. 2.4

Hint:

Mean deviation is independent of the specific values of $x$ and $y$ as long as $x+y=16$ and they are on opposite sides of the mean or maintain the same sum of absolute differences.

Answer 1

The Correct Option is A

Step 1: Recall the mean deviation definition.
Mean deviation about the mean is the average of the absolute gaps from the mean, $\dfrac1n\sum|x_i-\bar x|$. Here $n=5$ and $\bar x=5$.
Step 2: Use the mean to find the missing pair sum.
The five values are $1,2,6,x,y$ with mean $5$, so their total is $25$. Thus $1+2+6+x+y=25$, giving $x+y=16$.
Step 3: Note the deviations of the unknowns.
Both $x,y$ exceed $5$, so $|x-5|=x-5$ and $|y-5|=y-5$. Their combined deviation is $(x-5)+(y-5)=x+y-10=16-10=6$, independent of the exact split.
Step 4: Deviations of the known three.
$|1-5|=4$, $|2-5|=3$, $|6-5|=1$. These add to $4+3+1=8$.
Step 5: Total absolute deviation.
Adding the known and unknown parts, the sum of all five deviations is $8+6=14$.
Step 6: Divide by $n$.
\[ \text{Mean deviation}=\frac{14}{5}=2.8. \] This matches option 1.
\[ \boxed{2.8} \]