Question

The median of the following data is 525. Find the values of x and y, if the total frequency is 100.

Class interval	Class interval frequency
0-100	2
100-200	5
200-300	X
300-400	12
400-500	17
500-600	20
600-700	Y
700-800	9
800-900	7
900-1000	4

Answer 1

According to the question we have to determine the values of x and y. So, first of all we have to find the cumulative frequency for the given class interval but before that we have to understand about the cumulative frequency as explained below:
Cumulative frequency: The cumulative frequency can be calculated or obtained by adding each given frequency from a frequency distribution table to the sum of its predecessors and its last value will be always equal to the total for all observations, since all the frequencies will already have been added to the previous total.
Now, we have to find the linear expression in terms of x and y with the help of sum of all the frequencies as mentioned in the question which is 100 and after that we have to find the class interval which lies for the given median which is 525.
Now, the obtained class interval we can obtain the value of upper limit and lower limit for the obtained class interval.
Same as, we have to find the height of the obtained class interval and the frequency and cumulative for that obtained class interval.
And now, we have to find the value of sum of all the cumulative frequencies obtained (N) which is equal to $\sum {{f_i}}$
Now, to find the median we have to use the formula as given below:

Formula used: Median$= l + \dfrac{{\dfrac{N}{2} - F}}{f} \times h...................(a)$
Where, l is the lower limit of the class interval, N is the sum of all the frequencies, f is the frequency, F is the cumulative for that class interval and h is the height of the obtained class interval.

Complete step-by-step solution:
Given,
Median of the given data = 525
Step 1: First of all we have to find the cumulative frequency for the given class interval and frequencies as mentioned in the solution hint.

Class interval	Class interval frequency$({f_i})$	Cumulative frequency
0-100	2	$2 + 0 = 0$
100-200	5	$2 + 5 = 7$
200-300	X	$7 + x$
300-400	12	$7 + x + 12 = 19 + x$
400-500	17	$19 + x + 17 = 36 + x$
500-600	20	$36 + x + 20 = 56 + x$
600-700	Y	$56 + x + y$
700-800	9	$56 + x + y + 9 = 65 + x + y$
800-900	7	$65 + x + y + 7 = 72 + x + y$
900-1000	4	$72 + x + y + 4 = 76 + x + y$

Step 2: Now, we have to obtain the expression in terms of x and y with the help of sum of frequencies.
$\Rightarrow N = \sum {{f_i}}$
On substituting all the values,
$\Rightarrow 76 + x + y = 100 \\\ \Rightarrow x + y = 100 - 76 \\\ \Rightarrow x + y = 24$
Step 3: Now, we to find the lower limit with the help of the median which is 525 hence, the required interval is 500-600
So, the lower limit is l = 500
And the required frequency for that interval is f = 20
Step 4: Now, we have to find the height of the interval obtained which is 500-600 hence,
$\Rightarrow h = 600 - 500 \\\ \Rightarrow h = 100$
And cumulative frequency for the required class interval is $F = 36 + x$
Step 5: Now, we have to find the median for the given data with the help of the formula (a) as mentioned in the solution hint. Hence, on substituting all the values in the formula (a),
Median $\Rightarrow 525 = 500 + \dfrac{{\dfrac{{100}}{2} - (36 + x)}}{{20}}$
On solving the expression obtained just above,
$\Rightarrow 525 - 500 = \dfrac{{50 - (36 + x)}}{{20}} \times 100 \\\ \Rightarrow 25 = (14 - x) \times 5 \\\ \Rightarrow 25 = 70 - 5x \\\ \Rightarrow 5x = 70 - 25 \\\ \Rightarrow x = \dfrac{{45}}{5} \\\ \Rightarrow x = 9$
Step 6: Now, to find the value of y we have to substitute the value of x as obtained in step 5 in the expression (1) as obtained in step 2. Hence,
$\Rightarrow 9 + y = 24 \\\ \Rightarrow y = 24 - 9 \\\ \Rightarrow y = 15$

Hence, with the help of the formula (a) we have obtained the values $x = 9$ and $y = 15$

Note: Median is the set of data with an odd and even number of values and median is a value separating the higher half from the lower half of a given data sample.
Cumulative frequency can be calculated or obtained by adding each given frequency from a frequency distribution table to the sum of its predecessors and its last value will be always equal to the total for all observations, since all the frequencies will already have been added to the previous total.