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November 20, 2024The median is the value that divides a given number of observations into precisely two halves. The median of ungrouped data is estimated depending on whether several terms are odd or even and whether the data is sorted in ascending or descending order. A measure of central tendency is the central position of a set of values in data. The mean, median, and mode are the most widely used measures of central tendency.
The sum of the values of all observations divided by the total number of observations is the mean or average of a set of observations. In this article, we will study the definition, properties, and uses of the median and the mean and median differences.
The median is the variable’s value that divides the distribution into two equal parts when the data is arranged in ascending or descending order. The number of observations above it is equal to the number of observations below it. When the data is arranged in ascending or descending order, the median of ungrouped data is calculated as below:
When the number of observations \((n)\) is odd, then the median is the value of the \({\left( {\frac{{n + 1}}{2}} \right)^{{\text{th}}}}\) observation.
When the number of observations \((n)\) is even, the median is the mean of the \({\left( {\frac{n}{2}} \right)^{{\text{th}}}}\) and the \({\left( {\frac{n}{2} + 1} \right)^{{\text{th}}}}\) observation.
If the values \(x_i\) in the raw data are arranged to increase or decrease in magnitude, then the middle-most value in the arrangement is called the median.
Thus, for the ungrouped data \(x_1,\;x_2,\;x_3,\;……,\;x_n\) the median is computed by using the following steps.
In the case of a discrete frequency distribution \(\frac {x_i}{f_i} ; i = 1,\;2,\;3,…..,\;n,\) we calculate the median using the below steps:
To calculate the median of grouped or continuous frequency data, we use the below steps.
Mean | Median |
Mean is the arithmetic average of a set of numbers. | The median distinguishes the higher sample from the lower value, usually from a probability distribution. |
The mean’s application is for normal distributions. | Deviated distributions find the most common use of the median. |
Several extrinsic factors limit the usage of the mean. | For measuring data with uneven distributions, it is far more solid and dependable. |
The mean can be found by adding all of the values together and dividing the sum by the number of values. | The median can be found by ordering all of the numbers in ascending or descending way in the set and then looking for the number in the middle of the distribution. |
It’s extremely sensitive to data that’s out of the ordinary. | It isn’t particularly sensitive to outlier data. |
It specifies the data set’s central value. | It specifies the midpoint of the data set’s centre of gravity. |
Q.1. Find the median of the following data.
\(19,\;25,\;59,\;48,\;35,\;31,\;30,\;32,\;51\)
Ans: Arranging the given data in ascending order, we get
\(19,\;25,\;30,\;31,\;32,\;35,\;48,\;51,\;59\)
Here, the number of observations \(= 9\) which is odd.
Since the number of observations is odd. Therefore,
Median \(=\) value of \({\left( {\frac{{9 + 1}}{2}} \right)^{{\text{th}}}}\)
Median \(=\) value of \(5^{\rm{th}}\) observation \(= 32\)
Hence, median \(= 32.\)
Q.2. Obtain the median for the following frequency distribution:
\(x\) | \(F\) |
\(1\) | \(8\) |
\(2\) | \(10\) |
\(3\) | \(11\) |
\(4\) | \(16\) |
\(5\) | \(20\) |
\(6\) | \(25\) |
\(7\) | \(15\) |
\(8\) | \(9\) |
\(9\) | \(6\) |
Ans:
\(x\) | \(F\) | \(cf\) |
\(1\) | \(8\) | \(8\) |
\(2\) | \(10\) | \(18\) |
\(3\) | \(11\) | \(29\) |
\(4\) | \(16\) | \(45\) |
\(5\) | \(20\) | \(65\) |
\(6\) | \(25\) | \(90\) |
\(7\) | \(15\) | \(105\) |
\(8\) | \(9\) | \(114\) |
\(9\) | \(6\) | \(120\) |
\(N = 120\) |
Here, \(N = 120 \Rightarrow \frac{N}{2} = 60\)
We find that the cumulative frequency just greater than \(\frac {N}{2}\) i.e., \(60\) is \(65\), and the value of \(x\) corresponding to \(65\) is \(5\). Therefore, median \(= 5.\)
Q.3. The median of the following data is \(525\). Find the values of \(x\) and \(y\), if the total frequency is \(100\).
Class Interval | Frequency |
\(0 – 100\) | \(2\) |
\(100 – 200\) | \(5\) |
\(200 – 300\) | \(x\) |
\(300 – 400\) | \(12\) |
\(400 – 500\) | \(17\) |
\(500 – 600\) | \(20\) |
\(600 – 700\) | \(y\) |
\(700 – 800\) | \(9\) |
\(800 – 900\) | \(7\) |
\(900 – 1000\) | \(4\) |
Ans: It is given that, total frequency \(N = 100\)
Class Interval | Frequency | Cumulative frequency\((cf)\) |
\(0 – 100\) | \(2\) | \(2\) |
\(100 – 200\) | \(5\) | \(7\) |
\(200 – 300\) | \(x\) | \(7 + x\) |
\(300 – 400\) | \(12\) | \(19 + x\) |
\(400 – 500\) | \(17\) | \(36 + x\) |
\(500 – 600\) | \(20\) | \(56 + x\) |
\(600 – 700\) | \(y\) | \(56 + x + y\) |
\(700 – 800\) | \(9\) | \(65 + x + y\) |
\(800 – 900\) | \(7\) | \(72 + x + y\) |
\(900 – 1000\) | \(4\) | \(76 + x + y\) |
Total \(= 100\) |
We have \(N = ∑ f_i = 100\)
\(⇒ 76 + x + y = 100 ⇒ x + y = 24\)
It is given that the median is \(525\). It lies in class \(500 – 600\)
Therefore, \(l = 500,\;h = 100,\;f = 20,\;F = 36 + x\) and \(N = 100\)
median \(l + \left[ {\frac{{\frac{N}{2} – F}}{f}} \right] \times h\)
\( \Rightarrow 525 = 500 + \frac{{50 – \left( {36 + x} \right)}}{{20}} \times 100\)
\( \Rightarrow 525 – 500 = \left( {14 – x} \right) \times 5\)
\( \Rightarrow 25 = 70 – 5x \Rightarrow 5x = 45 \Rightarrow x = 9\)
Substututing \(x = 9\) in \(x + y = 24\), we get \(y = 15\)
Hence, \(x = 9\) and \(y = 15.\)
Q.4. Compute the median for the following cumulative frequency distribution:
Less than \(20\) | Less than \(30\) | Less than \(40\) | Less than \(50\) | Less than \(60\) | Less than \(70\) | Less than \(80\) | Less than \(90\) | Less than \(100\) |
\(0\) | \(4\) | \(16\) | \(30\) | \(46\) | \(66\) | \(82\) | \(92\) | \(100\) |
Ans: We are given the cumulative frequency distribution. So, we first construct a frequency table from the given cumulative frequency distribution and then we will make necessary computations to compute the median.
Class Intervals | Frequency \((f)\) | Cumulative frequency \((c.f)\) |
\(20 – 30\) | \(4\) | \(4\) |
\(30 – 40\) | \(12\) | \(16\) |
\(40 – 50\) | \(14\) | \(30\) |
\(50 – 60\) | \(16\) | \(46\) |
\(60 – 70\) | \(20\) | \(66\) |
\(70 – 80\) | \(16\) | \(82\) |
\(80 – 90\) | \(10\) | \(92\) |
\(90 – 100\) | \(8\) | \(100\) |
\(N = \sum {{f_i} = 100}\) |
Here, \(N = \sum {f_i} = 100\). Therefore, \(\frac{N}{2} = 50\)
We observe that the cumulative frequency just greater than \(\frac{N}{2} = 50\) is \(66\), and the corresponding class is \(60 – 70.\)
So, \(l = 60,\;f = 20,\;F = 46\) and \(h = 10\)
Now, median \( = l + \left[ {\frac{{\frac{N}{2} – F}}{f}} \right] \times h\)
\(⇒\) median \(= 60 + \frac {(50 – 46)}{20} × 10 = 62\)
Therefore, median \(= 62.\)
Q.5. Find the median of the daily wages of ten workers from the following data:
\(20,\;25,\;17,\;18,\;8,\;15,\;22,\;11,\;9,\;14\)
Ans: Arranging the wages in ascending order of magnitude, we have
\(8,\;9,\;11,\;14,\;15,\;17,\;18,\;20,\;22,\;25\)
Since there are \(10\) observations, therefore, the median is the average of \({\left( {\frac{{10}}{2}} \right)^{{\text{th}}}}\) and \({\left( {\frac{{10}}{2} + 1} \right)^{{\text{th}}}}\) observations.
Hence, median \(= \frac{15+17}{2} =16.\)
In this article, we will learn the mean and median. Also, we have learnt the method to find the median for grouped frequency distribution, ungrouped data and discrete frequency distribution. Also, we have studied the difference between mean and median, uses of median and properties of median and solved some example problems on the median.
Q.1. What is a median?
Ans: The variable’s value that divides the distribution into two equal parts, i.e., the value of the variable such that the number of observations above it is equal to the number of observations below it is known as the median.
Q.2. How to find the median?
Ans: When the data is arranged in ascending or descending order, the median of ungrouped data is calculated as below:
When the number of observations \((n)\) is odd, then the median is the value of the \({\left( {\frac{{n + 1}}{2}} \right)^{{\text{th}}}}\) observation.
When the number of observations \((n)\) is even, the median is the mean of the \({\left( {\frac{{n}}{2}} \right)^{{\text{th}}}}\) and the \({\left( {\frac{{n}}{2}} + 1 \right)^{{\text{th}}}}\) observation.
Q.3. How to calculate the median of grouped data?
Ans: To calculate the median of grouped data, we have
Median \( = l + \left[ {\frac{{\frac{N}{2} – F}}{f}} \right] \times h\)
Where \(l =\) lower frequency of the median class
\(f =\) frequency of the median class
\(h =\) size of the median class
\(F =\) Cumulative frequency of the class preceding the median class
\(N = \sum\limits_{i = 1}^n {{f_i}} \)
Q.4. How to calculate the median of ungrouped data?
Ans: If \(n\) is odd, then median \(=\) value of \({\left( {\frac{{n + 1}}{2}} \right)^{{\text{th}}}}\) observation. If \(n\) is even, then median \(=\) mean of the \({\left( {\frac{{n}}{2}} \right)^{{\text{th}}}}\) and the \({\left( {\frac{{n}}{2}} + 1\right)^{{\text{th}}}}\) observation. Where \(n\) is the number of observations.
Q.5. What is the mean vs median?
Ans: The mean is the average of given data, whereas the median is the middle value of the given data.
Learn the Concepts of Mean and Median
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