Calculate the median of income used by employee for the following data:
Income 4-14 14-24 24-34 34-44 Employee 12 10 8 4
A) 18
B) 19
C) 20
D) 21
| Income | 4-14 | 14-24 | 24-34 | 34-44 |
| Employee | 12 | 10 | 8 | 4 |
Answer
614.7k+ views
Hint: We need to find the value of N or $\sum {f(x)} $, to begin with, to evaluate the value of $\dfrac{N}{2}$, since the sum of total number of employees is even number, to locate the median class and then find the cumulative frequency, simply apply the formula for median for a given distribution.
Cumulative Frequency is the successive total of all frequencies class-wise.
Complete step by step answer:
Now, to solve and find the median we will first need the cumulative frequency table from the given data,
So, for that let us construct a new table:
Now, coming to the formula for Median of a given frequency distribution:
Median = $l + \left( {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right) \times h$,
Where $N = \sum {f = } $Total number of frequency,
cf=Cumulative frequency of the class preceding the median class
l= Lower limit of the median class
f=Frequency of the median class
h= class width.
Now, according to the question here , $N = \sum {f = } $34, which means $\dfrac{N}{2} = \dfrac{{34}}{2} = 17$,
Hence, by rule of median, the median class will be the one which contains $\dfrac{N}{2} = \dfrac{{34}}{2} = 17$ and from the table of data 17 belongs to the class interval 14-24.
Thus, the median class is 14-24.
Next, coming to the formula for calculating the value of median,
The values of the variables in the formula as per the question will be:
l = 14
cf = cumulative frequency of the class preceding the median class = 12
f= Frequency of the median class = 10
h= class width = 10.
So, putting all these values the Median will now become:
$
= l + \left( {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right) \times h \\
= 14 + \left( {\dfrac{{\dfrac{{34}}{2} - 12}}{{10}}} \right) \times 10 \\
= 14 + \left( {\dfrac{{17 - 12}}{{10}}} \right) \times 10 \\
= 14 + \left( {\dfrac{5}{{10}}} \right) \times 10 \\
= 14 + 5 \\
= 19 \\
$
Therefore, the median income earned by the employees is 19.
Hence the correct answer is option B.
Note: Median usually means the middle value. The middle value depends on the number of items. If the number of observations is even, then the median is given as:
$Median = \left( {\dfrac{{n + 1}}{2}} \right)th$ observation while for an even number of observations as: $Median = \left( {\dfrac{{{{\dfrac{n}{2}}^{th}}obs. + {{\dfrac{{n + 1}}{2}}^{th}}obs.}}{2}} \right)$.
For a frequency distribution data, Correct median class needs to be determined and cumulative frequency needs to be calculated to solve the sum. If the median class is incorrect then the whole sum will be incorrect. Cumulative frequency is used to determine the number of frequencies above or below a particular value.
Cumulative Frequency is the successive total of all frequencies class-wise.
Complete step by step answer:
Now, to solve and find the median we will first need the cumulative frequency table from the given data,
So, for that let us construct a new table:
| Income | Employee(Frequency $f$) | Cumulative Frequency |
| 4-14 | 12 | 12 |
| 14-24 | 10 | 12+10=22 |
| 24-34 | 8 | 22+8=30 |
| 34-44 | 4 | 30+4=34 |
| $\sum {f = } $N=34 |
Now, coming to the formula for Median of a given frequency distribution:
Median = $l + \left( {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right) \times h$,
Where $N = \sum {f = } $Total number of frequency,
cf=Cumulative frequency of the class preceding the median class
l= Lower limit of the median class
f=Frequency of the median class
h= class width.
Now, according to the question here , $N = \sum {f = } $34, which means $\dfrac{N}{2} = \dfrac{{34}}{2} = 17$,
Hence, by rule of median, the median class will be the one which contains $\dfrac{N}{2} = \dfrac{{34}}{2} = 17$ and from the table of data 17 belongs to the class interval 14-24.
Thus, the median class is 14-24.
Next, coming to the formula for calculating the value of median,
The values of the variables in the formula as per the question will be:
l = 14
cf = cumulative frequency of the class preceding the median class = 12
f= Frequency of the median class = 10
h= class width = 10.
So, putting all these values the Median will now become:
$
= l + \left( {\dfrac{{\dfrac{N}{2} - cf}}{f}} \right) \times h \\
= 14 + \left( {\dfrac{{\dfrac{{34}}{2} - 12}}{{10}}} \right) \times 10 \\
= 14 + \left( {\dfrac{{17 - 12}}{{10}}} \right) \times 10 \\
= 14 + \left( {\dfrac{5}{{10}}} \right) \times 10 \\
= 14 + 5 \\
= 19 \\
$
Therefore, the median income earned by the employees is 19.
Hence the correct answer is option B.
Note: Median usually means the middle value. The middle value depends on the number of items. If the number of observations is even, then the median is given as:
$Median = \left( {\dfrac{{n + 1}}{2}} \right)th$ observation while for an even number of observations as: $Median = \left( {\dfrac{{{{\dfrac{n}{2}}^{th}}obs. + {{\dfrac{{n + 1}}{2}}^{th}}obs.}}{2}} \right)$.
For a frequency distribution data, Correct median class needs to be determined and cumulative frequency needs to be calculated to solve the sum. If the median class is incorrect then the whole sum will be incorrect. Cumulative frequency is used to determine the number of frequencies above or below a particular value.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Discuss the various forms of bacteria class 11 biology CBSE

