Q. 24.7( 12 Votes )

# The following dat

Answer :

We have been given data of weight in less than type and data of number of students.

We have

Taking weight as x-axis and number of students as y-axis.

Thus plotting the points on a graph, we get

Now we need to plot median in the graph and then verify it using the formula.

So, note that total number of students are 35.

⇒ N = 35

Find N/2.

⇒

Draw a line parallel to x axis, passing through 17.2 on y-axis intersecting the less than ogive.

We have graphically,

We can notice from the graph, the line intersecting the ogive touches the y-axis at 46.7.

Now, let us solve theoretically,

Let us make a table showing frequencies and class intervals.

Since,

Observe, cf = 28 is just greater than 17.5.

Thus, median class = 46 – 48

Median is given by

Where,

L = Lower class limit of median class = 46

N/2 = 17.5

cf = cumulative frequency of the class preceding median class = 12

f = frequency of the median class = 16

h = class interval of the median class = 2

Substituting these values in the formula of median, we get

⇒

⇒ Median = 46 + 0.6875

⇒ Median = 46.6875

**Thus, the median of data is 46.6875 and it is verified.**

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