Median Example Problems

How do you Calculate the Median

Median of a distribution is the value of the variable which divides the distribution into two equal parts i.e. it is the value of the variable such that the number of observations above it is equal to the number of observations below it.
If the values xi in the raw data. are arranged in order of increasing or decreasing magnitude, then the middle, most value in the arrangement is called the median.
Algorithm :
Step I : Arrange the observations (values of the variate) in ascending or descending order of magnitude.
Step II  : Determine the total number of observations, say, n.
Step III  : If n is odd, then
Median = value of \({\left( {\frac{{n + 1}}{2}} \right)^{th}}\) observation
If n is even, then
Median = \(\frac{{Value\;of\;{{\left( {\frac{n}{2}} \right)}^{th}}observation + \;Value\;of\,{{\left( {\frac{n}{2} + 1} \right)}^{th}}observation}}{2}\)

The median can be calculated graphically while mean cannot be.
The sum of the absolute deviations taken from the median is less than the sum of the absolute deviations taken from any other observation in the data.
Median is not affected by extreme values.

Median Example Problems with Solutions

Example 1:    Find the median of the following data :
25, 34, 31, 23, 22, 26, 35, 28, 20, 32
Solution:    Arranging the data in ascending order, we get20, 22, 23, 25, 26, 28, 31, 32, 34, 35
Here, the number of observations n = 10 (even).
∴ Median = \(\frac{{Value\;of\;{{\left( {\frac{10}{2}} \right)}^{th}}observation + \;Value\;of\,{{\left( {\frac{10}{2} + 1} \right)}^{th}}observation}}{2}\)
⇒ Median = \(\frac{{Value\;of\;{5^{th}}observation\; + Value\;of\;{6^{th}}observation}}{2}\)
∴  Median = \(\frac{{26 + 28}}{2}\) = 27
Hence, median of the given data is 27.

Example 2:    Find the median of the following values :
37, 31, 42, 43, 46, 25, 39, 45, 32
Solution:    Arranging the data in ascending order, we have 25, 31, 32, 37, 39, 42, 43, 45, 46
Here, the number of observations n = 9 (odd)
∴ Median = Value of \({\left( {\frac{{9 + 1}}{2}} \right)^{th}}\) observation
= Value of 5th observation = 39.

Example 3:    The median of the observations 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 arranged in ascending order is 24. Find the value of x.
Solution:    Here, the number of observations n = 10. Since n is even, therefore
Median =  \(\frac{{{{\left( {\frac{n}{2}} \right)}^{th}}observation\; + {{\left( {\frac{n}{2} + 1} \right)}^{th}}observation}}{2}\)
⇒   24 =  \(\frac{{{5^{th}}observation + {6^{th}}observation}}{2}\)
⇒   24 =  \(\frac{{(x + 2) + (x + 4)}}{2}\)
⇒   24 =  \(\frac{{2x + 6}}{2}\)
⇒   24 = x + 3   ⇒ x = 21.
Hence,  x = 21.

Example 4:    Find the median of the following data : 19, 25, 59, 48, 35, 31, 30, 32, 51. If 25 is replaced by 52, what will be the new median.
Solution:    Arranging the given data in ascending order, we have 19, 25, 30, 31, 32, 35, 48, 51, 59
Here, the number of observations n = 9 (odd)
Since the number of observations is odd. Therefore.
Median = Value of \({\left( {\frac{{9 + 1}}{2}} \right)^{th}}\) observations
⇒   Median = value of 5^th observation = 32.
Hence, Median = 32
If 25 is replaced by 52, then the new observations arranged in ascending order are :
19, 30, 31, 32, 35, 48, 51, 52, 59
∴ New median = Value of 5^th observation = 35.

Example 5:    Calculate the median for the following  distribution.

Solution:    The cumulative frequency table is constructed as shown below :

Here, n = 23, which is odd
Median  = \({t_{\frac{{23 + 1}}{2}}}\) = t₁₂ = 49
(i.e. weight of the 12th student when the weights have been arranged in order)

Example 6:    The following data have been arranged in desending orders of magnitude 75, 70, 68, x + 2, x – 2, 50, 45, 40
If the median of the data is 60, find the value of x.
Solution:    The number of observations are 8, the median will be the average of 4th and 5th number
⇒   Median = \(\frac{{(x + 2) + (x-2)}}{2}\)
⇒   60 = \(\frac{{2x}}{2}\)  ⇒ x = 60

Example 7:    Find the median of the following data
(i)   17, 27, 37, 13, 18, 25, 32, 34, 23
(ii)  24, 37, 19, 41, 28, 32, 29, 31, 33, 21
Solution:    (i)   The scores when arranged in ascending order are
13, 17, 18, 23, 25, 27, 32, 34, 37
Here, the number of scores n = 9 (odd)
∴    Median = \({t_{\frac{{9 + 1}}{2}}}\) = t₅ = 25
(ii)  The scores when arraged in ascending order are
19, 21, 24, 28, 29, 31, 33, 34, 37, 41.
Total number of scores = 10, which is even. So there will be two middle-terms which are t₅ = 29 and t₆ = 31.
∴  Median = \(\frac{{{t_5} + {t_6}}}{2} = \frac{{29 + 31}}{2}\) = 30

Example 8:    Find the median of the following data :
(i)   8, 10, 5, 7, 12, 15, 11
(ii)  12, 14, 10, 7, 15, 16
Solution:    (i)   8, 10, 5, 7, 12, 15, 11
These numbers are arranged in an order
5, 7, 8, 10, 11, 12, 15
The number of observations = 7 (odd)
⇒   Median = \(\frac{7+1}{2}\) = 4th term
⇒   Median = 10
(ii)  12, 14, 10, 7, 15, 16
These numbers are arranged in an order
7, 10, 12, 14, 15, 16
The number of observations = 6 (even)
The medians will be mean of  = 3rd and 4th terms i.e., 12 and 14
⇒  The median = \(\frac{12+14}{2}\) = 13

Example 9:    The following data have been arranged in desending orders of magnitude 75, 70, 68, x + 2, x – 2, 50, 45, 40
If the median of the data is 60, find the value of x.
Solution:    The number of observations are 8, the median will be the average of 4th and 5th number
⇒   Median = \(\frac{{(x + 2) + (x–2)}}{2}\)
⇒   60 = \(\frac{2x}{2}\)
⇒  x = 60

Example 10:    Find the median of 6, 8, 9, 10, 11, 12 and 13.
Solution:    Total number of terms = 7
The middle terms = \(\frac{1}{2}\) (7 + 1) = 4th
Median = Value of the 4th term = 10.
Hence, the median of the given series is 10.

Example 11:    Find the median of 21, 22, 23, 24, 25, 26, 27 and 28.
Solution:    Total number of terms = 8
Median
= Value of \(\frac{1}{2}\left[ {\frac{8}{2}th\,\,term + \,\left( {\frac{8}{2} + 1} \right)th\,term} \right]\)
= Value of \(\frac{1}{2}\) [4th term + 5th term]
= \(\frac{1}{2}\) [24 + 25] = \(\frac{49}{2}\) = 24.5

Example 12:    The number of runs scored by 11 players of a cricket team of school are 5, 19, 42, 11, 50, 30, 21, 0, 52, 36, 27. Find the median.
Solution:    Let us arrange the value in ascending order
0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52
∴  Median M = \({\left( {\frac{{n + 1}}{2}} \right)^{th}}\) value
= \({\left( {\frac{{11 + 1}}{2}} \right)^{th}}\) value = 6th value
Now 6th value in data is 27
∴        Median = 27 runs.