Let the median and the mean deviation about the median of 7 observation 170, 125, 230, 190, 210, a, b be 170 and \(\frac{205}{7}\)respectively. Then the mean deviation about the mean of these 7 observations is :
- 31
- 28
- 30
- 32
The Correct Option is C
Approach Solution - 1
Given that the median is 170, the observations are arranged as:
125, a, b, 170, 190, 210, 230
The mean deviation about the median is given by:
\[ \frac{0 + |45| + |60| + |20| + |40| + |170 - a| + |170 - b|}{7} = \frac{205}{7} \]
From this, we find:
\(|170 - a| + |170 - b| = 300 \implies a + b = 300\)
Now, the mean of the observations is:
\[ \text{Mean} = \frac{125 + a + b + 170 + 190 + 210 + 230}{7} = \frac{125 + 300 + 170 + 190 + 210 + 230}{7} = 175 \]
The mean deviation about the mean is:
\[ \frac{|125 - 175| + |a - 175| + |b - 175| + |170 - 175| + |190 - 175| + |210 - 175| + |230 - 175|}{7} \]
Simplifying:
\[ \frac{50 + |a - 175| + |b - 175| + 5 + 15 + 35 + 55}{7} = 30 \]
Approach Solution -2
Find the mean deviation about the mean for the 7 observations: 170, 125, 230, 190, 210, a, b, given that their median is 170 and mean deviation about the median is \( \frac{205}{7} \).
Concept Used:
For n observations \( x_1, x_2, \dots, x_n \):
- Median is the middle value when arranged in order.
- Mean deviation about median = \( \frac{1}{n} \sum |x_i - M| \), where M is median.
- Mean deviation about mean = \( \frac{1}{n} \sum |x_i - \bar{x}| \), where \( \bar{x} \) is mean.
Step-by-Step Solution:
Step 1: Arrange known observations and use median information.
Observations: 125, 170, 190, 210, 230, a, b (n=7).
Median is 4th observation when arranged in order = 170.
So in sorted order: \( x_{(1)}, x_{(2)}, x_{(3)}, 170, x_{(5)}, x_{(6)}, x_{(7)} \).
Given numbers: 125, 170, 190, 210, 230 must fit into these positions.
Since median=170, 170 must be the 4th observation. So first 3 observations ≤ 170, last 3 ≥ 170.
125 is smallest ⇒ \( x_{(1)} = 125 \).
We have another 170? Possibly a or b = 170? But then median still 170 if fourth smallest is 170.
Let's assign: \( x_{(1)} = 125 \), \( x_{(4)} = 170 \).
Remaining known: 170, 190, 210, 230 and a, b.
We need \( x_{(2)}, x_{(3)} \le 170 \), \( x_{(5)}, x_{(6)}, x_{(7)} \ge 170 \).
So possible: \( x_{(2)} = 170 \) (second 170), \( x_{(3)} = ? \) must be ≤ 170 ⇒ maybe a or b ≤ 170.
\( x_{(5)} = 190 \), \( x_{(6)} = 210 \), \( x_{(7)} = 230 \).
So sorted order: 125, 170, (a or b ≤ 170), 170, 190, 210, 230.
Thus a, b are: one is ≤ 170 (goes in position 3), other is 170 (goes in position 2 or 4? Actually position 4 is fixed as 170, so other 170 must be in position 2).
So a = 170, b ≤ 170.
Actually: positions: 1:125, 2:170, 3: b (≤170), 4:170, 5:190, 6:210, 7:230.
So a = 170, b ≤ 170.
Step 2: Use mean deviation about median information.
Mean deviation about median (170) = \( \frac{205}{7} \).
\[ \frac{1}{7} \left[ |125-170| + |170-170| + |b-170| + |170-170| + |190-170| + |210-170| + |230-170| \right] = \frac{205}{7} \] \[ \frac{45 + 0 + |b-170| + 0 + 20 + 40 + 60}{7} = \frac{205}{7} \] \[ 45 + |b-170| + 20 + 40 + 60 = 205 \] \[ |b-170| + 165 = 205 \] \[ |b-170| = 40 \] Since b ≤ 170, 170 - b = 40 ⇒ b = 130.
Step 3: Observations are: 125, 130, 170, 170, 190, 210, 230.
Compute mean:
\[ \bar{x} = \frac{125 + 130 + 170 + 170 + 190 + 210 + 230}{7} \] \[ = \frac{1225}{7} = 175 \]
Step 4: Mean deviation about mean:
\[ \frac{1}{7} \left[ |125-175| + |130-175| + |170-175| + |170-175| + |190-175| + |210-175| + |230-175| \right] \] \[ = \frac{1}{7} \left[ 50 + 45 + 5 + 5 + 15 + 35 + 55 \right] \] \[ = \frac{1}{7} \left[ 210 \right] = 30 \]
Therefore, the mean deviation about the mean is 30.
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