
The median of 230 observations of the following frequency distribution is 46. Find a and b:
Class 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Frequency 12 30 a 65 b 25 18
| Class | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Frequency | 12 | 30 | a | 65 | b | 25 | 18 |
Answer
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Hint: From given information, we will get cumulative frequency. To derive the values of a and b, firstly we have to use median’s equation for given observations. In a given problem, the value of median is given. By using the given median’s value and median formula, we get two equations. On solving these equations, we will get values of a and b.
Complete step-by-step answer:
From given expression, we can add one more row of cumulative frequency as follows:
From given information, we can write
Total observation are = 230
Median = 46
So, median class will be 40-50
Thus, 46 lies between 40-50
Now, the total number of observations will be equal to the cumulative frequency.
\[\begin{array}{l}
150 + a + b = 230\\
a + b = 230 - 150\\
a + b = 80{\rm{ }}...{\rm{(i)}}
\end{array}\]
Now, for the second equation we will use the median formula.
\[median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right)h\]
But, we have
\[\begin{array}{l}
l = 40\\
cf = 42 + a\\
f = 65\\
\dfrac{n}{2} = 230/2 = 115\\
h = 10
\end{array}\]
Now, we can put all these values in equation
\[46 = 40 + \left( {\dfrac{{115 - \left( {42 + a} \right)}}{{65}}} \right)10\]
On simplification, we get
\[\begin{array}{l}
\Rightarrow 46 - 40 + \left( {\dfrac{{73 - a}}{{65}}} \right)10\\
\Rightarrow 6 \times 65 = 730 - 10a\\
\Rightarrow - 10a = - 340\\
\Rightarrow a = 34
\end{array}\]
Now, put the value of a in equation (i), we get
\[\begin{array}{l}
\Rightarrow b = 80 - a\\
\Rightarrow b = 80 - 34 = 46
\end{array}\]
So, the value of a = 34 and b = 46
Thus the value of a is 34 and b is 46.
Note: Students should know how to calculate cumulative frequency. Students have to take care while calculating cumulative frequency for each observation.
Students make mistakes while using the median formula.
After getting the value of a, one can forget to put the value of in Eq.(i) to get the value of b. Here, we require the value of a as well as b.
Complete step-by-step answer:
From given expression, we can add one more row of cumulative frequency as follows:
| Class | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Frequency | 12 | 30 | a | 65 | B | 25 | 18 |
| C.F. | 12 | 42 | 42+a | 107+a | 107+a+b | 132+a+b | 150+a+b |
From given information, we can write
Total observation are = 230
Median = 46
So, median class will be 40-50
Thus, 46 lies between 40-50
Now, the total number of observations will be equal to the cumulative frequency.
\[\begin{array}{l}
150 + a + b = 230\\
a + b = 230 - 150\\
a + b = 80{\rm{ }}...{\rm{(i)}}
\end{array}\]
Now, for the second equation we will use the median formula.
\[median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right)h\]
But, we have
\[\begin{array}{l}
l = 40\\
cf = 42 + a\\
f = 65\\
\dfrac{n}{2} = 230/2 = 115\\
h = 10
\end{array}\]
Now, we can put all these values in equation
\[46 = 40 + \left( {\dfrac{{115 - \left( {42 + a} \right)}}{{65}}} \right)10\]
On simplification, we get
\[\begin{array}{l}
\Rightarrow 46 - 40 + \left( {\dfrac{{73 - a}}{{65}}} \right)10\\
\Rightarrow 6 \times 65 = 730 - 10a\\
\Rightarrow - 10a = - 340\\
\Rightarrow a = 34
\end{array}\]
Now, put the value of a in equation (i), we get
\[\begin{array}{l}
\Rightarrow b = 80 - a\\
\Rightarrow b = 80 - 34 = 46
\end{array}\]
So, the value of a = 34 and b = 46
Thus the value of a is 34 and b is 46.
Note: Students should know how to calculate cumulative frequency. Students have to take care while calculating cumulative frequency for each observation.
Students make mistakes while using the median formula.
After getting the value of a, one can forget to put the value of in Eq.(i) to get the value of b. Here, we require the value of a as well as b.
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