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Q2:

If the median of the distribution given below is 28.5, find the values of $x$ and $y$. 

If the median of the distribution given ...

Solution :

Given: A frequency distribution table with a total frequency $N = 60$ and a median value of $28.5$.

Class IntervalFrequency ($f$)
0-105
10-20$x$
20-3020
30-4015
40-50$y$
50-605
Total60

To find: The values of $x$ and $y$.

Step 1: Constructing the Cumulative Frequency Table

Class IntervalFrequency ($f$)Cumulative Frequency ($cf$)
0-1055
10-20$x$$5 + x$
20-3020$25 + x$
30-4015$40 + x$
40-50$y$$40 + x + y$
50-605$45 + x + y$

Step 2: Establishing the first linear equation

Since the total frequency is given as $60$, we have:

$45 + x + y = 60$
$x + y = 60 - 45$
$x + y = 15$ --- (Equation 1)

Step 3: Identifying the Median Class

The median is given as $28.5$. Since $28.5$ lies between $20$ and $30$, the median class is $20-30$.

From the median class $20-30$:

  • Lower limit ($l$) = $20$
  • Frequency of median class ($f$) = $20$
  • Cumulative frequency of the class preceding the median class ($cf$) = $5 + x$
  • Class size ($h$) = $10$
  • $N/2 = 60/2 = 30$

Step 4: Applying the Median Formula

The formula for the median is: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$

Substituting the known values:

$28.5 = 20 + \left( \frac{30 - (5 + x)}{20} \right) \times 10$
$28.5 - 20 = \left( \frac{30 - 5 - x}{20} \right) \times 10$
$8.5 = \frac{25 - x}{2}$
$8.5 \times 2 = 25 - x$
$17 = 25 - x$
$x = 25 - 17$
$x = 8$

Step 5: Solving for $y$

Substitute $x = 8$ into Equation 1:

$8 + y = 15$
$y = 15 - 8$
$y = 7$

Final Answer: The values are $x = 8$ and $y = 7$.


More Questions from Class 10 Mathematics Statistics EXERCISE 13.3


CBSE Solutions for Class 10 Mathematics Statistics


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