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Question

The number of ‘likes’ on the last four photos posted on Facebook is 3, 15, 21, and 13. The mean absolute deviation value about the mean of ‘likes’ is:

This question was previously asked in
SSC CGL 2019 (Tier 2) GS Finance & Economics Previous Year Paper (17-Nov-2020)
The correct answer is

5

Understanding Mean Absolute Deviation (MAD)

The Mean Absolute Deviation (MAD) is a measure of variability or dispersion in a dataset. It tells us, on average, how far each data point is from the mean of the dataset. A higher MAD means the data points are more spread out, while a lower MAD means they are clustered closer to the mean.

In this question, we are given the number of 'likes' on four photos: 3, 15, 21, and 13. We need to calculate the mean absolute deviation about the mean for these 'likes'.

Calculating the Mean of 'Likes'

The first step in calculating the Mean Absolute Deviation is to find the mean (average) of the data points. The mean ($\bar{x}$) is calculated by summing all the data points and dividing by the total number of data points.

The number of 'likes' are $x_1 = 3$, $x_2 = 15$, $x_3 = 21$, and $x_4 = 13$. The number of data points ($n$) is 4.

The formula for the mean is: $$\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$$

Let's calculate the sum of the 'likes': $$\sum x_i = 3 + 15 + 21 + 13 = 52$$

Now, calculate the mean: $$\bar{x} = \frac{52}{4} = 13$$

The mean number of 'likes' is 13.

Calculating Absolute Deviations from the Mean

Next, we find the absolute deviation for each data point. The absolute deviation for a data point $x_i$ is the absolute difference between the data point and the mean ($\bar{x}$), which is $|x_i - \bar{x}|$. We use the absolute value because we are interested in the distance from the mean, regardless of whether the data point is greater or less than the mean.

Let's calculate the absolute deviation for each number of 'likes':

  • For 3 likes: $|3 - 13| = |-10| = 10$
  • For 15 likes: $|15 - 13| = |2| = 2$
  • For 21 likes: $|21 - 13| = |8| = 8$
  • For 13 likes: $|13 - 13| = |0| = 0$

We can summarize these in a table:

Number of 'Likes' ($x_i$) Mean ($\bar{x}$) Difference ($x_i - \bar{x}$) Absolute Deviation ($|x_i - \bar{x}|$)
3 13 3 - 13 = -10 |-10| = 10
15 13 15 - 13 = 2 |2| = 2
21 13 21 - 13 = 8 |8| = 8
13 13 13 - 13 = 0 |0| = 0

Calculating the Mean Absolute Deviation (MAD)

The final step is to calculate the Mean Absolute Deviation (MAD). This is the mean of the absolute deviations. We sum the absolute deviations and divide by the number of data points ($n$).

The sum of the absolute deviations is: $$10 + 2 + 8 + 0 = 20$$

The formula for the Mean Absolute Deviation is: $$\text{MAD} = \frac{\sum_{i=1}^{n} |x_i - \bar{x}|}{n}$$

Substitute the sum of absolute deviations and the number of data points: $$\text{MAD} = \frac{20}{4} = 5$$

Final Mean Absolute Deviation Value

The mean absolute deviation value about the mean of the 'likes' is 5. This means that, on average, the number of likes for each photo deviates by 5 from the mean number of likes (which is 13).

Revision Table: Key Concepts in MAD Calculation

Here’s a quick summary of the steps involved in calculating the Mean Absolute Deviation:

Step Description Calculation for 'Likes' Data
1 Calculate the Mean ($\bar{x}$) Sum of data points / Number of data points = (3+15+21+13) / 4 = 13
2 Calculate Absolute Deviations Find $|x_i - \bar{x}|$ for each data point: $|3-13|=10, |15-13|=2, |21-13|=8, |13-13|=0$
3 Sum Absolute Deviations Add all absolute deviations: 10 + 2 + 8 + 0 = 20
4 Calculate MAD Sum of absolute deviations / Number of data points = 20 / 4 = 5

Additional Information on Measures of Dispersion

Mean Absolute Deviation is one way to measure how spread out data is. Other common measures of dispersion include:

  • Range: The difference between the highest and lowest values in the dataset. For our 'likes' data, Range = 21 - 3 = 18.
  • Variance: The average of the squared differences from the mean. It gives more weight to points that are further from the mean.
  • Standard Deviation: The square root of the variance. It is widely used because it is in the same units as the original data.

Understanding these measures helps in analyzing the variability within a dataset, providing insights beyond just the average value.

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Important Questions from Mean Deviation

  1. What is the mean deviation about the mean ?

  2. The mean deviation about median of 10 observations is 15. If each observation is multiplied by $-3$, then find the new mean deviation about median of resulting observations.
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  4. If the mean deviation 1, 1 + d, 1 + 2d, ..., 1 + 100d from their mean is 255, then d is equal to

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