Question

Find the values of x and y if the median for the following data is 31.
classes- |0-10| 10-20 | 20-30 |
frequency-| 5 | x | 6 |
classes |30-40 | 40-50|50-60
frequency| y | 6 | 5 ​

Answer 1

${\large{\pmb{\sf{\bigstar \: {\underline{Required \: Solution...}}}}}}$

★ Firstly let us see that how to solve this question, some steps are given below that tell us how to find median of grouped data, let us do this question!

Step 1. Firstly we have to make a table with 3 columns.

• First column is made for the class interval
• Second column is made for the frequency
• Third column is made for the cumulative frequency.

Step 2. Now we have to write the class intervals and frequency in the respective columns.

Step 3. Now let us write the cumulative frequency in the column.

Step 4. Now we have to find the sum of frequencies (∑f)

Step 5. Now we have to find n/2, find the class whose cumulative frequency is greater than and nearest to n/2 that's median.

Step 6. Now we use the formula that is given below:

${\small{\underline{\boxed{\sf{\rightarrow \quad l \: = \Bigg(\dfrac{\dfrac{n}{2}-cf}{f} \Bigg) \times h \quad}}}}}$

~ Now let us make class interval table, it is given below(By BrainlyButterfliee):

$\begin{gathered}\boxed{\begin{array}{c|c|c|c|c|c|c}\bf Class\: interval&\sf 0-10&\sf 10-20&\sf 20-30 &\sf 30-40&\sf 40-50&\sf 50-60&\\\frac{\qquad \qquad \qquad}{}&\frac{\qquad \quad \quad}{}&\frac{\qquad \quad\quad}{}&\frac{\qquad \quad \quad}{}&\frac{\qquad \quad \quad}{}&\frac{\qquad \quad\quad}{}&\frac{\qquad \quad\quad}{}\\\bf Frequency&\sf 5&\sf x&\sf 6&\sf y&\sf 6&\sf 5\end{array}}\end{gathered}$

~ Now let us make a frequency table, by following the above mentioned steps(By BrainlyButterfliee):

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{|c|c|c|} \bf \: \: Classes&\: \bf \: Frequency & \bf \: C.frequency\\\frac{\qquad \qquad \qquad}{}&\frac{\qquad \quad \quad}{}&\frac{\qquad \quad\quad}{}{} \\ \\ \sf 0-10 & \sf 5 & \sf 5 \\ \sf 10-20 & \sf x & \sf 5+x\\ \sf 20-30 & \sf 11+x & \sf 10\\ \sf 30-40 & \sf 11+x+y & \sf 18\\ \sf 40-50 & \sf 6 &\sf 17+x+y\\ \sf 50-60 & \sf 5 & \sf 22+x+y \\ \end{array}}\end{gathered}\end{gathered}$

~ Now according to the question,

${\sf{:\implies Total \: = 40}}$

${\sf{:\implies 22 + x + y = 40}}$

${\sf{\therefore \: x + y = 40 - 22}}$

${\sf{Hence, \: x + y = 18 \dots Eq. \: 1}}$

${\sf{Median \: = 30 - 40}}$

~ Now we have to use the below mentioned formula:

${\small{\underline{\boxed{\sf{\rightarrow \quad Median \: = l \: = \Bigg(\dfrac{\dfrac{n}{2}-cf}{f} \Bigg) \times h \quad}}}}}$

• n/2 is 40/2 = 20 here
• l is 30 here
• Median is 31 here
• cf is (11+x) here
• f is y here
• h is 10 here

${\small{\underline{\boxed{\sf{Median \: = l \: = \Bigg(\dfrac{\dfrac{n}{2}-cf}{f} \Bigg) \times h}}}}} \\ \\ :\implies \sf Median \: = l \: = \Bigg(\dfrac{\dfrac{n}{2}-cf}{f} \Bigg) \times h \\ \\ :\implies \sf 31 \: = 30 \: + \Bigg(\dfrac{20-(11+x)}{y} \Bigg) \times 10 \\ \\ :\implies \sf 31 \: = 30 \: + \Bigg(\dfrac{9-x}{y} \Bigg) \times 10 \\ \\ :\implies \sf 1 \: = \Bigg(\dfrac{9-x}{y} \Bigg) \times 10$

${\sf{\therefore \: 10x + y = 90 \dots Eq. \: 2}}$

~ Now let's subtract Eq. 1 from Eq. 2

• x = 8

• y = 10