Weight (in kg)Number of studentsLess than 380Less than 403Less than 425Less than 449Less than 4614Less than 4828Less than 5032Less than 5235
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula."">

# During the medical check-up of 35 students of a class, their weights were recorded as follows:Weight (in kg)Number of studentsLess than 380Less than 403Less than 425Less than 449Less than 4614Less than 4828Less than 5032Less than 5235Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula."

Given:

During the medical check-up of 35 students of a class, their weights were recorded.

To do:

We have to draw a less than type ogive for the given data and obtain the median weight from the graph and verify the result by using the formula.

Solution:

We first prepare the cumulative  frequency distribution table by less than method as given below:

Represent weights along X-axis and cumulative frequency along Y-axis.

Plot the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35) on the graph and join them in a free hand to get an ogive as shown.

Here,

$N = 35$

$\frac{N}{2}=\frac{35}{2}=17.5$

From the point, $P$ (17.5) on Y-axis draw a line parallel to X-axis meeting the curve and then draw a line parallel to the Y-axis meeting the X-axis at $Q$.

From the figure, the median is 46.5 kg (approx.)

Therefore,
46-48 is the median class
Here,

$l= 46, h = 2,f= 14, F= 14$

This implies,

Median $=l+\frac{\frac{N}{2}-F}{f} \times h$

$=46+\frac{17.5-14}{14} \times 2$

$=46+\frac{3.5 \times 2}{14}$

$=46+0.5$

$=46.5$

Hence verified.

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Updated on: 10-Oct-2022

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