Q. 24.7( 12 Votes )
The following dat
We have been given data of weight in less than type and data of number of students.
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
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.
Observe, cf = 28 is just greater than 17.5.
Thus, median class = 46 – 48
Median is given by
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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