6. Construct a discrete frequency table and find the mean deviation from the median for the following data set:

19, 23, 30, 29, 11, 21, 26, 36, 41, 42, 49, 52, 56, 58, 53, 27, 20, 34, 62, 22, 23, 25, 27, 36, 42, 42, 52, 50, 58, 53, 30, 11, 29, 26, 41, 36, 58, 53, 19, 23, 30



Answer :

Sure! Let's break this problem into two primary tasks:

1. Construct a discrete frequency table from the given data.
2. Compute the mean deviation from the median of the data.

### Step 1: Constructing the Discrete Frequency Table

The given data is:
[tex]\[ 19, 23, 30, 29, 11, 21, 26, 36, 41, 42, 49, 52, 56, 58, 53, 27, 20, 34, 62, 22, 23, 25, 27, 36, 42, 42, 52, 50, 58, 53, 30, 11, 29, 26, 41, 36, 58, 53, 19, 23, 30 \][/tex]

First, let's sort the data in ascending order:
[tex]\[ 11, 11, 19, 19, 20, 21, 22, 23, 23, 23, 25, 26, 26, 27, 27, 29, 29, 30, 30, 30, 34, 36, 36, 36, 41, 41, 42, 42, 42, 49, 50, 52, 52, 53, 53, 53, 56, 58, 58, 58, 62 \][/tex]

Next, count the frequency of each unique value:

[tex]\[ \begin{array}{|c|c|} \hline \text{Value} & \text{Frequency} \\ \hline 11 & 2 \\ 19 & 2 \\ 20 & 1 \\ 21 & 1 \\ 22 & 1 \\ 23 & 3 \\ 25 & 1 \\ 26 & 2 \\ 27 & 2 \\ 29 & 2 \\ 30 & 3 \\ 34 & 1 \\ 36 & 3 \\ 41 & 2 \\ 42 & 3 \\ 49 & 1 \\ 50 & 1 \\ 52 & 2 \\ 53 & 3 \\ 56 & 1 \\ 58 & 3 \\ 62 & 1 \\ \hline \end{array} \][/tex]

This is our discrete frequency table.

### Step 2: Finding the Mean Deviation from the Median

#### Finding the Median

1. Arrange the data in ascending order (already done above).
2. Locate the middle value. Since there are 41 values (an odd number), the median is the 21st value:

From the sorted list, the 21st value is 34. Thus, the median of the data set is 34.

#### Calculating the Mean Deviation from the Median

1. Find the absolute deviations from the median:

[tex]\[ \begin{aligned} &|11 - 34|, |11 - 34|, |19 - 34|, |19 - 34|, |20 - 34|, |21 - 34|, |22 - 34|, |23 - 34|, |23 - 34|, |23 - 34|, \\ &|25 - 34|, |26 - 34|, |26 - 34|, |27 - 34|, |27 - 34|, |29 - 34|, |29 - 34|, |30 - 34|, |30 - 34|, |30 - 34|, \\ &|34 - 34|, |36 - 34|, |36 - 34|, |36 - 34|, |41 - 34|, |41 - 34|, |42 - 34|, |42 - 34|, |42 - 34|, |49 - 34|, \\ &|50 - 34|, |52 - 34|, |52 - 34|, |53 - 34|, |53 - 34|, |53 - 34|, |56 - 34|, |58 - 34|, |58 - 34|, |58 - 34|, |62 - 34| \end{aligned} \][/tex]

[tex]\[ \begin{aligned} &|23|, |23|, |15|, |15|, |14|, |13|, |12|, |11|, |11|, |11|, \\ &|9|, |8|, |8|, |7|, |7|, |5|, |5|, |4|, |4|, |4|, \\ &|0|, |2|, |2|, |2|, |7|, |7|, |8|, |8|, |8|, |15|, \\ &|16|, |18|, |18|, |19|, |19|, |19|, |22|, |24|, |24|, |24|, |28| \end{aligned} \][/tex]

2. Sum these absolute deviations:

[tex]\[ 23 + 23 + 15 + 15 + 14 + 13 + 12 + 11 + 11 + 11 + 9 + 8 + 8 + 7 + 7 + 5 + 5 + 4 + 4 + 4 + 0 + 2 + 2 + 2 + 7 + 7 + 8 + 8 + 8 + 15 + 16 + 18 + 18 + 19 + 19 + 19 + 22 + 24 + 24 + 24 + 28 = 493 \][/tex]

3. Divide the sum by the number of data points:

[tex]\[ \frac{493}{41} \approx 12.02 \][/tex]

Therefore, the mean deviation from the median is approximately 12.02.