
The table shows the salaries of 280 persons.
Salary (in thousand) No. of persons 5-10 49 10-15 133 15-20 63 20-25 15 25-30 6 30-35 7 35-40 4 40-45 2 45-50 1
Calculate the median salary of the data.
| Salary (in thousand) | No. of persons |
| 5-10 | 49 |
| 10-15 | 133 |
| 15-20 | 63 |
| 20-25 | 15 |
| 25-30 | 6 |
| 30-35 | 7 |
| 35-40 | 4 |
| 40-45 | 2 |
| 45-50 | 1 |
Answer
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Hint: Given a distribution of the data of salaries of persons, we are asked to find the median salary of the data, which is the same as finding the median of the distribution. Now to solve this problem we have to know about cumulative frequency which is the sum of all the given frequencies which in this case is the sum of no. of persons. The median is the middle value of the data when arranged in the order of the highest to lowest, here the salaries are already arranged in the increasing order.
Complete step by step solution:
First calculating the cumulative frequency $(cf)$ of the given data:
Here the median of the data is given by:
$ \Rightarrow $Median $ = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$
Where $l$= lower limit of the median class.
$n$= number of observations.
$cf$= cumulative frequency of class preceding the median class.
$f$= frequency of median class.
$h$= class size.
Here $n = 280$; hence $\dfrac{n}{2} = 140$
Thus the median class here is 10-15, hence the lower limit of the median class is 10.
$\therefore l = 10$
The cumulative frequency of the preceding class of the median class is 49.
$\therefore cf = 49$
Frequency of the median class is 133.
$\therefore f = 133$
The class size is 5.
$\therefore h = 5$
Now substituting all the obtained values in the median expression, as given below :
$ \Rightarrow $Median $ = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$
$ \Rightarrow $Median $ = 10 + \left( {\dfrac{{140 - 49}}{{133}}} \right) \times 5$
$ \Rightarrow $Median $ = 10 + \left( {\dfrac{{91}}{{133}}} \right) \times 5$
$ \Rightarrow $Median $ = 10 + 3.42$
$\therefore $ Median = 13.42
The median salary (in thousand) is 13.42.
Hence the median salary is 13,420
Median salary of the data is 13,420.
Note: Here while solving the median expression, in the numerator we have to substitute the value of the cumulative frequency of the preceding median class but not of the median class, and in the denominator we have to substitute the value of the frequency of the median class, but not the cumulative frequency. These two points are very important to remember.
Complete step by step solution:
First calculating the cumulative frequency $(cf)$ of the given data:
| Salary (in thousand) | No. of persons $(f)$ | $(cf)$ |
| 5-10 | 49 | 49 |
| 10-15 | 133 | 49+133 = 182 |
| 15-20 | 63 | 182+63 = 245 |
| 20-25 | 15 | 245+15 = 260 |
| 25-30 | 6 | 260+6 = 266 |
| 30-35 | 7 | 266+7 = 273 |
| 35-40 | 4 | 273+4 = 277 |
| 40-45 | 2 | 277+2 = 279 |
| 45-50 | 1 | 279+1 =280 |
Here the median of the data is given by:
$ \Rightarrow $Median $ = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$
Where $l$= lower limit of the median class.
$n$= number of observations.
$cf$= cumulative frequency of class preceding the median class.
$f$= frequency of median class.
$h$= class size.
Here $n = 280$; hence $\dfrac{n}{2} = 140$
Thus the median class here is 10-15, hence the lower limit of the median class is 10.
$\therefore l = 10$
The cumulative frequency of the preceding class of the median class is 49.
$\therefore cf = 49$
Frequency of the median class is 133.
$\therefore f = 133$
The class size is 5.
$\therefore h = 5$
Now substituting all the obtained values in the median expression, as given below :
$ \Rightarrow $Median $ = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$
$ \Rightarrow $Median $ = 10 + \left( {\dfrac{{140 - 49}}{{133}}} \right) \times 5$
$ \Rightarrow $Median $ = 10 + \left( {\dfrac{{91}}{{133}}} \right) \times 5$
$ \Rightarrow $Median $ = 10 + 3.42$
$\therefore $ Median = 13.42
The median salary (in thousand) is 13.42.
Hence the median salary is 13,420
Median salary of the data is 13,420.
Note: Here while solving the median expression, in the numerator we have to substitute the value of the cumulative frequency of the preceding median class but not of the median class, and in the denominator we have to substitute the value of the frequency of the median class, but not the cumulative frequency. These two points are very important to remember.
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