Q) If the median of the following frequency distribution is 32.5. Find the values of X and Y :

(Q 32 – 30/2/2 – CBSE 2026 Question Paper)

Ans:

Step 1: Let’s re-organize the data in the frequency table to find out each part:

Step 2: Given that the total of frequencies = 40

∴  31 + X + Y = 40

∴  X + Y = 40 – 31

∴  X + Y = 9 ……… (i)

Step 3: To find the median, we need to identify middle value of the data:

• First, we add a column for cumulative frequency in the frequency table to find the median. Its shown in last column.
• Next, Total number of frequencies = 31 + X + Y . It shown in the last row of middle column.
• Next, we need to identify Median Class. The Median class is the class where the cumulative frequency crosses 50% of the total of frequencies. Here in the above data table, Cumulative frequency of 14 + X  is not crossing 50% of (31 + X + Y) at class “20-30” but 26 + X  is crossing 50% of (31 + X + Y) at class “30-40”.
• Hence, our Median class = (30 – 40)

Step 4: Next, To find the median, we use the formula:

Median = L + h

Here:

L = Lower boundary of the median class = 30

n = Total number of Classes = 40

= Cumulative frequency of the class before the median class = 14 + X

f = Frequency of the median class = 12

h = Class width = 40 – 30 = 10

hence, the Median = 30 + h

∴ 32.5 = 30 + (10)

∴ 2.5 =  (10)

∴ 2.5 x =  20 – (14 + X)

∴ 3 =  6 – X

∴ X = 6 – 3 

∴ X = 3

Step 4: By substituting the value of X in equation (i), we get:

X + Y = 9

∴ Y = 9 – X = 9 – 3

∴ Y = 6

Therefore, the values of frequencies, X and Y are  3 and 6 respectively.

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