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

Concept: The mean deviation from the mean of a set of observations $x_1, x_2, \dots, x_n$ is defined as: \[ \text{Mean Deviation} = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}| \] where $\bar{x}$ is the mean of the observations.

Step 1: Find the values of the two unknown observations.
Let the five observations be $1, 2, 6, x, y$.
The mean is $\bar{x} = 5$. Thus, $\frac{1+2+6+x+y}{5} = 5$.
$9 + x + y = 25 \implies x + y = 16$.
We are given that each unknown observation is greater than 5. Let $x = 6 + a$ and $y = 6 + b$ (since they must be greater than 5). However, simply testing integers satisfying $x+y=16$ where $x, y > 5$, we find possible pairs like $(6, 10), (7, 9), (8, 8)$.
Note: For any pair $(x, y)$ summing to 16 where $x, y > 5$, the mean deviation will be constant. Let's use $x=7$ and $y=9$.

Step 2: Calculate the absolute deviations from the mean \(\bar{x} = 5\).
Observations: $1, 2, 6, 7, 9$.
Deviations $|x_i - 5|$:
$|1 - 5| = 4$
$|2 - 5| = 3$
$|6 - 5| = 1$
$|7 - 5| = 2$
$|9 - 5| = 4$

Step 3: Find the mean of these absolute deviations.
Sum of deviations $= 4 + 3 + 1 + 2 + 4 = 14$.
Mean Deviation $= \frac{14}{5} = 2.8$.
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Mean Deviation = 2.8 minipage center