Answer :

Given total frequency of the given data is 100.

i.e., f = 100………..(i)

i.e., sum of frequency = 100

From table adding all the frequencies, we get

2 + 5 + x + 12 + 17 + 20 + Y + 9 + 7 + 4 = 100

⇒ x + Y + 76 = 100

⇒ x + Y = 100 – 76

⇒ x + Y = 24………..(ii)

Now we will rewrite the table with cumulative frequencies to find the median

So, from above table it is clear that the median class is 500 – 600

And we know, the median is calculated using the formula

Where l = lower limit of median class = 500

h = class interval

= 600 – 500

= 100

cf = cumulative frequency of the class before the median class

= 36 + x

f = frequency of the median class

= 20

n = ∑f_{i}

= 100

And given median = 525

Substituting these values in the above equation of median we get

⇒ 25 = (50 × 5) – (36 × 5) – 5x

⇒ 5x = 250 – 180 – 25

⇒ 5x = 45

⇒ x = 9

Substituting the value of x in equation (i), we get

⇒ x + Y = 24

⇒ 9 + Y = 24

⇒ Y = 24 – 9

⇒ Y = 15

**Hence the values of x and Y are 9 and 15 respectively.**

**OR**

First, we will rewrite the table with cumulative frequencies.

Now we will draw less than table by writing cumulative frequencies under the number of student column.

Now we will draw less than type ogive for the above data,

We first draw the coordinate axes, with lower limits of the profit along the horizontal axis, and the cumulative frequency along the vertical axes.

Then, using less than table data, we plot the points (10,5), (20, 8), (30, 12), (40,15), (50, 18), (60, 22), (70, 29), (80, 38), (90, 45), (100, 53). We join these points with a smooth curve to get the ‘less than’ ogive, as shown below.

Now from the graph, we see that total frequency = 53

⇒ N = ∑f_{i} = 53

So, we will draw a line y = 26.5 parallel to x-axis, the intersection of this line with the less than ogive curve gives the x coordinate.

And this x – coordinate is the median of the given data.

**Hence the required median mark of the given data is 66.4.**

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