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:
5
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'.
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.
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':
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 |
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: $$\text{MAD} = \frac{20}{4} = 5$$
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).
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 |
Mean Absolute Deviation is one way to measure how spread out data is. Other common measures of dispersion include:
Understanding these measures helps in analyzing the variability within a dataset, providing insights beyond just the average value.
If the mean deviation of a set of observations is 15, then the value of quartile deviation is:
If the number of observations in a series is 15, then the third quartile is equal to:
The mean deviation from the average A is minimum if A represents
What is the mean deviation about the mean ?
Let xi, i = 1, 2, ..., n be n observations and wi = pxi + k, i = 1, 2, ..., n where p and k are constants. If the mean of xi's is 48 and standard deviation is 12, whereas the mean of wi's is 55 and standard deviation is 15, then the value of p and k should be
If the mean deviation 1, 1 + d, 1 + 2d, ..., 1 + 100d from their mean is 255, then d is equal to
The mean of 5 observation is 5 and their variance is 124. If three of the observations are 1, 2, 6, then the mean deviation from the mean of the data is