The key idea is to calculate the Mean Deviation About the Median for a discrete frequency distribution.

First, we find the median ($M$) of the distribution. The total number of students is $N = \sum f_i = 2+3+8+3+4 = 20$. Since $N$ is even, the median is the average of the $\left(\frac{N}{2}\right)$-th and $\left(\frac{N}{2}+1\right)$-th observations, i.e., the 10th and 11th observations.

The cumulative frequencies are:

• For $x=10$, $cf=2$
• For $x=11$, $cf=2+3=5$
• For $x=12$, $cf=5+8=13$ Both the 10th and 11th observations fall within the group where $x=12$. Thus, the median $M = 12$.

Next, we calculate the sum of the products of frequencies and the absolute deviations from the median, $\sum f_i |x_i - M|$:

• For $x=10$: $2 \times |10 - 12| = 2 \times 2 = 4$
• For $x=11$: $3 \times |11 - 12| = 3 \times 1 = 3$
• For $x=12$: $8 \times |12 - 12| = 8 \times 0 = 0$
• For $x=14$: $3 \times |14 - 12| = 3 \times 2 = 6$
• For $x=15$: $4 \times |15 - 12| = 4 \times 3 = 12$ So, $\sum f_i |x_i - M| = 4 + 3 + 0 + 6 + 12 = 25$.

Finally, the mean deviation about the median is given by the formula:

When we talk about "mean deviation," we are looking for a measure of dispersion, or how spread out the data points are from a central value. The "about the median" part specifies that this central value should be the median. The median is particularly useful here because the sum of absolute deviations is minimized when taken about the median. This makes the mean deviation about the median a robust measure of dispersion, less affected by extreme values than the mean deviation about the mean.

Here's how to calculate it step-by-step for the given discrete frequency distribution:

1. Understand the Data and Calculate Total Frequency ($N$) We are given a discrete frequency distribution, where $x_i$ represents the marks obtained and $f_i$ represents the number of students (frequency) for each mark. First, we need to find the total number of observations, $N$, which is the sum of all frequencies.

2. Calculate Cumulative Frequencies ($cf$)

   To find the median for a frequency distribution, we need to determine the cumulative frequencies. The cumulative frequency for a given $x_i$ is the sum of its frequency and the frequencies of all preceding $x_i$ values.

   Marks ($x_i$)	No. of students ($f_i$)	Cumulative Frequency ($cf$)
   10	2	2
   11	3	$2+3=5$
   12	8	$5+8=13$
   14	3	$13+3=16$
   15	4	$16+4=20$

3. Find the Median ($M$)

   Since $N = 20$ (an even number), the median is the average of the $(N/2)$-th observation and the $(N/2 + 1)$-th observation.

   • The $(N/2)$-th observation is the $(20/2) = 10$-th observation.
   • The $(N/2 + 1)$-th observation is the $(20/2 + 1) = 11$-th observation.

   Looking at the cumulative frequency column:

   • The 10th observation falls in the group where the cumulative frequency is 13 (i.e., $x_i = 12$).
   • The 11th observation also falls in the group where the cumulative frequency is 13 (i.e., $x_i = 12$).

   Therefore, the 10th observation is 12, and the 11th observation is 12.

   The median $M = \frac{10\text{th observation} + 11\text{th observation}}{2} = \frac{12 + 12}{2} = 12$.

   Watch out

   For discrete data, if $N$ is even, the median is the average of the two middle values. If $N$ is odd, the median is simply the value of the $((N+1)/2)$-th observation. Do not use the formula for continuous data (like $L + \frac{N/2 - cf}{f} \times h$) here.

4. Calculate Absolute Deviations from the Median ($|x_i - M|$)

   Now that we have the median $M=12$, we find the absolute difference between each mark ($x_i$) and the median.

   | Marks ($x_i$) | No. of students ($f_i$) | Cumulative Frequency ($cf$) | $|x_i - M|$ |

   | :------------ | :---------------------- | :-------------------------- | :---------- |

   | 10 | 2 | 2 | $|10-12|=2$ |

   | 11 | 3 | 5 | $|11-12|=1$ |

   | 12 | 8 | 13 | $|12-12|=0$ |

   | 14 | 3 | 16 | $|14-12|=2$ |

   | 15 | 4 | 20 | $|15-12|=3$ |

5. Calculate $f_i |x_i - M|$

   Multiply each absolute deviation by its corresponding frequency. This accounts for how many times each deviation occurs.

   | Marks ($x_i$) | No. of students ($f_i$) | $|x_i - M|$ | $f_i |x_i - M|$ |

   | :------------ | :---------------------- | :---------- | :-------------- |

   | 10 | 2 | 2 | $2 \times 2 = 4$ |

   | 11 | 3 | 1 | $3 \times 1 = 3$ |

   | 12 | 8 | 0 | $8 \times 0 = 0$ |

   | 14 | 3 | 2 | $3 \times 2 = 6$ |

   | 15 | 4 | 3 | $4 \times 3 = 12$ |

6. Sum $f_i |x_i - M|$

   Add up all the values from the $f_i |x_i - M|$ column.

7. Calculate the Mean Deviation about the Median The formula for mean deviation about the median for a discrete frequency distribution is:

   $MD(M) = \frac{\sum f_i |x_i - M|}{\sum f_i}$

   Substitute the values we calculated: